An Introduction to the Stieltjes Constants

Some series and integrals involving the Riemann zeta function, binomial coefficients and the harmonic numbers

Volume II(b)

Donal F. Connon

18 February 2008

Abstract

In this series of seven papers, predominantly by means of elementary analysis, we establish a number of identities related to the Riemann zeta function, including the following:

∞ n 11⎛⎞n kp+1 γ p ()uuk=−⎜⎟(1)log( − + ) ∑∑k pn++11nk==00⎝⎠

x (1)− p+1 γςς()udu=− [(1)pp++ (0,)x (1) (0)] ∫ p p +1 1

⎛⎞11 2 1⎡ ⎛⎞1⎤ γγ11⎜⎟=−[2 15log 2 − 6 γπγπ log 2] −++−Γ⎢ 4log 2 3log 4log ⎜⎟⎥ ⎝⎠42 2⎣ ⎝⎠4⎦

∞ ⎡⎤1 1 log log(1/t ) −=dtγγ() u+ 2 γ () u+ γ log u+ log2 u ∫ ⎢⎥u 01 1 ⎣⎦1− tt log(1/ t ) t

∞ n 11⎛⎞n k 221212 ∑∑⎜⎟(−+=−−++ 1)kk log ( 1)γ1 log(2πγ ) πlog (2 π ) nk==00n +12⎝⎠k 242

∞ n ks∞ −−+1(1)yu n ⎛⎞n xxtuextxt⎛ ⎞ tdusy⎜⎟ s ==2 Φ⎜,,+ 1 ⎟ ∑∑k (ky+−Γ ) (1 t ) ( s ) ∫ ⎡⎤−u (1− t ) (1 − t ) nk==11⎝⎠ 0 11−+()xe t ⎝ ⎠ ⎣⎦ where ()u is the Stieltjes constant and is the Hurwitz-Lerch zeta function. γ p Φ(,,zsu )

Whilst this paper is mainly expository, some of the formulae reported in it are believed to be new, and the paper may also be of interest specifically due to the fact that most of the various identities have been derived by elementary methods.

CONTENTS OF VOLUMES I TO VI: Volume/page

SECTION:

1. Introduction I/11

2. An integral involving cot x I/17

The Riemann-Lebesgue lemma I/21

3. Results obtained from the basic identities I/24

Some stuff on Stirling numbers of the first kind I/69

Euler, Landen and Spence polylogarithm identities I/86

An application of the binomial theorem I/117

Summary of harmonic number series identities I/154

4. Elementary proofs of the Flajolet and Sedgewick identities II(a)/5

Some identities derived from the Hasse/Sondow equations II(a)/13

A connection with the gamma, beta and psi functions II(a)/26

An application of the digamma function to the derivation of Euler sums II(a)/31

Gauss hypergeometric summation II(a)/48

Stirling numbers revisited II(a)/51

Logarithmic series for the digamma and gamma functions II(a)/64

An alternative proof of Alexeiewsky’s theorem II(a)/74

A different view of the crime scene II(a)/91

An easy proof of Lerch’s identity II(a)/109

Stirling’s approximation for logΓ (u ) II(a)/114

The Gosper/Vardi functional equation II(a)/125

A logarithmic series for log A II(a)/137

Asymptotic formula for logGu (+ 1) II(a)/139

Gosper’s integral II(a)/144

The vanishing integral II(a)/147

Another trip to the land of G II(a)/165 x Evaluation of ∫ππuudun cot II(a)/169 0 An observation by Glasser II(a)/188

An introduction to Stieltjes constants II(b)/5

A possible connection with the Fresnel integral II(b)/16

Evaluation of some Stieltjes constants II(b)/21

A connection with logarithmic integrals II(b)/67

A hitchhiker’s guide to the Riemann hypothesis II(b)/84

A multitude of identities and theorems II(b)/101

Various identities involving polylogarithms III/5

Sondow’s formula for γ III/42

Evaluation of various logarithmic integrals III/61

Some integrals involving polylogarithms III/66

A little bit of log(π / 2) III/88

Alternative derivations of the Glaisher-Kinkelin constants III/108

Some identities involving harmonic numbers IV/5

Some integrals involving logΓ (x ) IV/77

Another determination of logG (1/ 2) IV/88

An application of Kummer’s Fourier series for logΓ (1+ x ) IV/92

Some Fourier series connections IV/109

Further appearances of the Riemann functional equation IV/124

More identities involving harmonic numbers and polylogarithms IV/133

5. An application of the Bernoulli polynomials V/5

6. Trigonometric integral identities involving:

- Riemann zeta function V/7

- Barnes double gamma function V/31

- Sine and cosine integrals V/45

- Yet another derivation of Gosper’s integral V/108

7. Some applications of the Riemann-Lebesgue lemma V/141

8. Some miscellaneous results V/145

APPENDICES (Volume VI):

A. Some properties of the Bernoulli numbers and the Bernoulli polynomials

B. A well-known integral

C. Euler’s reflection formula for the gamma function and related matters π 2 ∞ 1 D. A very elementary proof of = ∑ 2 8(21)n=0 n +

E. Some aspects of Euler’s constant γ and the gamma function

F. Elementary aspects of Riemann’s functional equation for the zeta function

ACKNOWLEDGEMENTS

REFERENCES

4 AN INTRODUCTION TO THE STIELTJES CONSTANTS

We recall Hasse’s formula for the Hurwitz zeta function from (4.3.106a) that for Re (s )≠ 1 ∞ 1(1)n ⎛⎞n − k (1)(,)ssu−= ς ∑∑⎜⎟ s−1 nk==00nuk++1()⎝⎠k and note that for s > 1

∞∞11n ⎛⎞n (1)− k (1)s −= ∑∑∑s ⎜⎟ s−1 nnk===000()nu+++ n 1⎝⎠k () uk

∞ 1(1)n ⎛⎞n − k Hence, in passing, we see that is positive for s > 1 and is ∑∑⎜⎟ s−1 nk==00nuk++1()⎝⎠k monotonically decreasing with u (which is not intuitively obvious).

We have the limit

∞∞n 11⎛⎞n k (4.3.200) lim(ssu−= 1)ςδ ( , ) ( −== 1)n,0 1 s→1 ∑∑⎜⎟ ∑ nk==00nn++11⎝⎠k n = 0

This limit is independent of u and we therefore have

∂∂ lim[(ssu−= 1)ςς ( , )] lim [( ssu −= 1) ( , )] 0 ∂∂uuss→→11

We also note that

∂∂∞ 1 (4.3.200a) (,su ) s( s 1,) u ςς==∑ s −+ ∂∂+uunun=0 () and hence as before we have

∂ lim [(1)(,)]ssusssu−=−−+=ςς lim[( 1)( 1,)]0 ss→→11∂u

As a result of (4.3.200) we may apply L’Hôpital’s rule to the limit

⎡⎤⎡⎤(s−− 1)(,)1ςς su (s − 1)(,)′ su +ς (,) su lim ⎢⎥⎢⎥= lim ss→→11⎣⎦⎣⎦s −11

5 From (4.3.107a) we see that

∞ 1 n ⎛⎞n log(uk+ ) (4.3.201) (1)(,)(,)ssusu−+=−−ςς′ (1)k ∑∑⎜⎟ s−1 nk==00nu++1()⎝⎠k k and hence we have ∞ 1 n ⎛⎞n lim([]ssusu−+=−−+ 1)(,ςς′ ) (, ) (1)log(k uk ) s→1 ∑∑⎜⎟ nk==00n +1 ⎝⎠k

Referring back to (4.3.74) in Volume II(a) we may then deduce that

(4.3.202) lim[ (ssusuu−+=− 1)ςς′ ( , ) ( , )] ψ ( ) s→1 and therefore conclude that

⎡⎤⎡⎤(1)(,)1ssu−−ς 1 (4.3.203) lim ⎢⎥⎢⎥=−=−limςψ (su , ) () u ss→→11⎣⎦⎣⎦ss−−11

The above limit is derived in a more complex manner in [126, p.91] and [135, p.271] using Hermite’s integral formula for the Hurwitz zeta function. A further derivation is contained in Volume VI.

With u =1 we immediately see that

⎡⎤⎡⎤11 (4.3.204) lim⎢⎥⎢⎥ς (ss ,1)−= limςψγ ( ) −=−= (1) ss→→11⎣⎦⎣⎦ss−−11 which is derived in an alternative way in (4.4.99m) in Volume III. The above limits then imply that

(4.3.204a) lim[ςς (su , )−=−− ( s )] γψ ( u ) s→1

In [135, p.266] we see that

⎡⎤ς (,su ) (4.3.204b) lim⎢⎥=− 1 s→1 ⎣⎦Γ−(1s )

This is easily demonstrated as follows. We have

ς (,su ) (1− s )(,ς su ) = Γ−(1sss ) (1 −Γ− ) (1 )

6 and hence we see that

⎡⎤⎡⎤(1−−ssu )ςς ( , ) (1 ssu ) ( , ) lim⎢⎥⎢⎥= lim=− 1 ss→→11⎣⎦⎣⎦(1−Γ−ss ) (1 ) Γ−(2 s )

L’Hôpital’s rule then gives us

⎡⎤ς ′(,su ) (4.3.205) lim⎢⎥= 1 s→1 ⎣⎦Γ−′(1s )

Differentiating (4.3.202) with respect to u gives us

d ⎡⎤1 lim⎢⎥ςψ (su , )−=−′ ( u ) du s→1 ⎣⎦s −1 and hence d limςψ (su , )=− ′ ( u ) du s→1

We have by interchanging the derivative and the limit operations

d ∂ limςς (su , )= lim ( su , ) duss→→11∂ u

=−+lim[ssς ( 1, u )] =−ς (2, u ) s→1 giving the well-known result ς (2,uu )=ψ ′ ( )

This may also be derived by differentiating (4.3.200a).

From (4.3.203) we see that

⎡⎤1 lim⎢⎥ςψ (su , )−+ ( u ) = 0 s→1 ⎣⎦s −1 and we have the Laurent expansion of the Hurwitz zeta function ς (su , ) about s =1

∞ p 1(1)− p (4.3.206) ςγ(,su )=+∑ p ()(u s − 1) sp−1!p=0

where γ p (u ) are known as the generalised Stieltjes constants. We have

(4.3.206a) γ 0 ()uu=−ψ () and γ 0 (1)= −=ψγ (1) .

We have for p ≥ 0

p pp⎡ () (1)!− p ⎤ (4.3.207) γςp ()usu=− (1)lim⎢ (,)− p+1 ⎥ s→1 ⎣ (1)s − ⎦ and, in particular, we have

⎡⎤1 (4.3.208) γς0 ()usuu=−=− lim(,)⎢⎥ ψ () s→1 ⎣⎦s −1

⎡ 1 ⎤ ′ γς1()usu=− lim⎢ (,) + 2 ⎥ s→1 ⎣ (1)s − ⎦

Applying L’Hôpital’s rule again gives us

⎡⎤1 ςψ(,su )−+ () u ⎢⎥⎡⎤1 s −1 ′ (4.3.208a) lim⎢⎥=+=− lim⎢⎥ςγ (su , )2 1 ( u ) ss→→11⎢⎥ss−−1(1)⎣⎦ ⎣⎦

Using (4.3.202) we see that

lim[ (ssusuu−++= 1)ςςψ′ ( , ) ( , ) ( )] 0 s→1 and applying L’Hôpital’s rule gives us

⎡⎤(ssusuu−++ 1)(,ςςψ′′ ) (, ) ()⎡⎤ ( ssusu −+ 1)(, ς′ ) 2 ς′ (, ) lim⎢⎥= lim ⎢⎥ ss→→11⎣⎦s −11⎣⎦

From (4.3.201) we see that

∂+∞ 1ln ⎛⎞n og2 ()ku [(ssusu 1)′ ( , ) ( , )] ( 1) k −+=ςς ∑∑⎜⎟ − s−1 ∂++snuknk==001()⎝⎠k and thus we have

∞ 1 n ⎛⎞n (4.3.209) lim([]ssusu−+= 1)(,ςς′′ ) 2 ′ (, ) ( − 1)log(k 2 ku + ) s→1 ∑∑⎜⎟ nk==00n +1 ⎝⎠k

Accordingly we see that

8 ⎡⎤(1)(,)(,)()ssusuu−++ςςψ′ ∞ 1n ⎛⎞n lim= (−+ 1)k log2 (ku ) s→1 ⎢⎥∑∑⎜⎟ ⎣⎦sn−+11nk==00⎝⎠k and the limit may be written as

⎡⎤1(,)()1ςψsu+ u ′ lim⎢⎥ς (su , ) ++22 − s→1 ⎣⎦(1)sss−−− 1 (1)

⎡⎤⎡⎤1(,)()1ςψsu+ u ′ =++lim⎢⎥⎢⎥ς (su , )22 lim − ss→→11⎣⎦⎣⎦(1)sss−−− 1 (1)

=−2()γ1 u where we have used (4.3.207) and (4.3.208a).

Hence we obtain

∞ n 11⎛⎞n k 2 (4.3.210) γ1()uk=− ∑∑⎜⎟ (1)log(− +u ) 21nk==00n + ⎝⎠k

The corresponding Stieltjes constants γ pp= γ (1) for the Riemann zeta function ς (s ) are given by p pp⎡⎤() (1)!− p γςp =−(1)lim⎢⎥ ()s − p+1 s→1 ⎣⎦(1)s − and hence we have

∞ n 1 ⎛⎞n k (4.3.211) γγγ00(1)===−∑∑⎜⎟ ( − 1) log(k + 1) nk==00n +1 ⎝⎠k

(where we have employed the Ser/Sondow formula (4.4.92a) for γ noting that no contribution comes from the term in (4.4.92a) with n = 0 ). Furthermore we see that

∞ n 11⎛⎞n k 2 (4.3.212) γγ11(1)==−∑∑⎜⎟ ( − 1) log (k + 1) 21nk==00n + ⎝⎠k

We note from The On-Line Encyclopedia of Integer Sequences that

γ1 =−0.728158458...

9 On suspects from (4.3.212) that a pattern is developing and this is investigated below in a more generalised setting.

Suppose that we have the following functional equation for suitably analytic functions

(safsugsu−= ) ( , ) ( , ) and suppose that limgsu ( , )= gau ( , ) exists. We then have sa→

lim[ (safsugau−−= ) ( , ) ( , )] 0 sa→

Applying L’Hôpital’s rule results in

⎡⎤()(,)(,)safsugau−− ∂ lim=−− lim[] (safsugau ) ( , ) ( , ) sa→→⎣⎦⎢⎥sa−∂sa s

∂ ==limgsu ( , ) g′ ( au , ) sa→ ∂s

Therefore we have

⎡⎤gau(,) ∂ limf (su , )−= limgsu ( , ) = g′ ( au , ) sa→→⎣⎦⎢⎥sa−∂sa s and this may be written as

⎡⎤gau(,) ∂ lim(,)fsu−− gau′′ (,) = lim gsu (,) −= gau (,) 0 sa→→⎣⎦⎢⎥sa−∂sa s

A further application of L’Hôpital’s rule produces

⎡⎤gau(,) fsu(, )−− gau′ (, ) ⎢⎥⎡ gau(,)⎤ sa− ′ lim ⎢⎥=+lim⎢ fsu ( , ) 2 ⎥ sa→→⎢⎥sa−−sa⎣ () sa⎦ ⎣⎦

∂∂⎡⎤∂2 =−=limgsu ( , ) g′ ( au , ) lim gsu ( , ) ∂∂ss⎣⎦⎢⎥sa→→sa∂s2

Therefore we may deduce that

10 ⎡⎤gau(,) ∂2 ′ lim⎢⎥f (su , )+=22 limgsu ( , ) sa→→⎣⎦()sa−∂sa s and the process may be continued indefinitely to result in

pp+1 ⎡⎤()p (1)!(,)−∂pgau lim⎢⎥f (su , ) −=pp++11limgsu ( , ) sa→→⎣⎦()sa−∂ sa s as may be readily proved by mathematical induction.

We also have where f (,su ) is simply of the form f ()s

pp+1 ⎡⎤()p (1)!()− pga d (4.3.213) lim⎢⎥f (sg ) −=pp++11lim (s ) sa→→⎣⎦()sa− sa ds

In the case of the Hurwitz zeta function we let

f (,su )= ς (, su ) and, with reference to (4.3.106a), we see that

∞ 1(1)n ⎛⎞n − k gsu(, )= ∑∑⎜⎟ s−1 nk==00nuk++1()⎝⎠k

∞ n 1 ⎛⎞n k gu(1, )=−=∑∑⎜⎟ ( 1) 1 nk==00n +1 ⎝⎠k

∂−+pk++11∞ 1(1)log()n ⎛⎞n puk gsu(, )=− (1)p+1 ps+−11∑∑⎜⎟ ∂++snuknk==001()⎝⎠k

⎡⎤(1)!− p p ∞ 1 n ⎛⎞n limς ()ppkp (su , ) −=−−+( 1) ++1 ( 1) log1 (1k ) s→1 ⎢⎥p+1 ∑∑⎜⎟ ⎣⎦(1)sn−+nk==001⎝⎠k

From (4.3.207) we had

p pp⎡ () (1)!− p ⎤ γςp ()us=− (1)lim⎢ (,)u− p+1 ⎥ s→1 ⎣ (1)s − ⎦ and therefore we may determine the following expression for the Stieltjes constants

11 ∞ n 11⎛⎞n kp+1 γ p ()uu=−∑∑⎜⎟ (1)log( − +k ) pn++11nk==00⎝⎠k

∞ n 11⎛⎞n kp+1 γγpp==−(1)∑∑⎜⎟ ( − 1) log (1 +k ) pn++11nk==00⎝⎠k

∞ n 1 ⎛⎞n k γψ0 (uu )=−∑∑⎜⎟ ( − 1) log( +k ) =− (u ) nk==00n +1 ⎝⎠k

An alternative derivation of the series for the generalised Stieltjes constants is shown below. We have ∞ p 1(1)− p ςγ(,su )=+∑ p ()(u s − 1) sp−1!p=0 and therefore

∞ p (1)− p+1 (1)(,)1ssu−=+ςγ∑ p ()(1)us − p=0 p!

Differentiation with respect to s gives us

∞∞pp ∂−(1) pp(1)− [(s− 1)ςγ ( su , )] =∑∑pp( ups )(+−= 1)( 1) γγ0 ( u ) + ( ups )(+− 1)( 1) ∂sppp==01!! p and as s →1 we have using (4.3.202)

−=ψ ()uuγ 0 ()

A further differentiation gives us

∂−2 ∞ (1)p [(ssu 1) ( , )] ( upps )( 1) ( 1) p−1 2 −=ςγ∑ p +− ∂spp=0 !

∞ p (1)− p−1 =−2()γγ1 uu +∑ p ()(1)(pp+s − 1) p=2 p!

As s →1 we have

∞ 1 n ⎛⎞n lim[] (ssusu−+= 1)ςς′′ ( , ) 2 ′ ( , ) ( − 1)k log2 (ku + ) s→1 ∑∑⎜⎟ nk==00n +1 ⎝⎠k

12 and therefore

∞ n 1 ⎛⎞n k 2 ∑∑⎜⎟(1)log(−+=−ku ) 2γ1 () u nk==00n +1 ⎝⎠k

More generally we have

∂−mp∞ (1) [(ssu 1) ( , )] ( uppppms )( 1) ( 1)...( 2 )( 1) pm+−1 m −=ςγ∑ p +−+−− ∂spp=0 !

∞ p (1)− pm+−1 =+∑ γ p (up )( 1) pp (− 1)...( p+ 2− ms )(− 1) pm=−1 p!

∞ p m−1 (1)− pm+−1 =−mu( 1)γγmp−1 ( ) +∑ ( uppppms )(+ 1) ( − 1)...( +− 2 )( − 1) pm= p!

Therefore we see that

m ∂ m−1 lim [(ssum−=− 1)ςγ ( , )] ( 1)m−1 ( u ) s→1 ∂sm

The Hasse identity is

∞ 1(1)n ⎛⎞n − k (1)(,)ssu−= ς ∑∑⎜⎟ s−1 nk==00nuk++1()⎝⎠k and therefore we have an alternative derivative expression

∂m ∞ 1 n ⎛⎞n lim [(ssu−=− 1)ς ( , )] ( 1)mkm ( − 1) log (uk + ) s→1 m ∑∑⎜⎟ ∂+snnk==001 ⎝⎠k

This then gives us for m ≥ 0

∞ n 11⎛⎞n km+1 (4.3.214) γ m ()uu= −−+∑∑⎜⎟ (1)log (k ) mn++11nk==00⎝⎠k

In particular we have

∞ n 1 ⎛⎞n k (4.3.215) γψ0 (uu )=−∑∑⎜⎟ ( − 1) log( +k ) =− (u ) nk==00n +1 ⎝⎠k

13 The identity (4.3.215) was noted by Coffey in [45e] (and this concurs with (4.4.99aix) of Volume III where the summation started at n =1)

∞ n 1 ⎛⎞n k ∑∑⎜⎟(−+=− 1) log(uk )ψ ( u ) log u nk==10n +1 ⎝⎠k 

We have the following summation

∞∞pp ∞n (1)−− (1)⎧ 1 1 ⎛⎞n kp+1 ⎫ ∑∑∑∑γ p ()uu=−⎨ ⎜⎟ (1)log( − +k )⎬ pp==00pppn!!11⎩⎭++ nk == 00⎝⎠k

∞∞n pp++11 1(⎛⎞n k −1)log()uk+ =−∑∑⎜⎟(1) ∑ nk==00np++1(⎝⎠k p = 0 1)!

∞ n 1 ⎛⎞n k =−−+−∑∑⎜⎟( 1)() exp[ log(uk )] 1 nk==00n +1 ⎝⎠k

∞ n 11⎛⎞n k ⎛⎞ =−−∑∑⎜⎟(1)⎜⎟ 1 nk==00nuk++1 ⎝⎠k ⎝⎠

∞ 1(1)n ⎛⎞n − k =−1 +∑∑⎜⎟ nk==00nuk+1 ⎝⎠k +

Therefore we obtain

∞ (1)− p ′ (4.3.216) ∑ γψp ()uu= ()− 1 p=0 p!

This equation was also derived by Coffey [45c] in 2005 using a different method. The result is in fact most easily seen by letting s = 2 in the Laurent expansion for ς (su , )

∞ p 1(1)− p ςγ(,su )=+∑ p ()(u s − 1) sp−1!p=0

This gives us ∞ (1)− p ςγ(2,uu )=+ 1∑ p ( ) p=0 p!

For s > 1 we may use the following form of the Hurwitz zeta function

14 ∞ 1 ′ ςψ(2,uu )==∑ 2 ( ) n=0 ()nu+ and the result follows immediately.

Using the same method as above, it is easily seen that

∞ γ ()u 3 ∑ p = − u p=0 p!2 and this may also be confirmed by letting s = 0 in the Laurent expansion for ς (su , ) . We then see that

∞ γ 1 (4.3.217) ∑ p = p=0 p!2

We may slightly generalise (4.3.216) as follows by including a factor of t p .

∞∞pp ∞n (1)−−pp(1)⎧ 1 1 ⎛⎞n kp+1 ⎫ ∑∑∑∑tuγ p ()=− t⎨ ⎜⎟(1)log( −+uk )⎬ pp==00pppn!!11⎩⎭++ n == 00k⎝⎠k

∞∞n pp++11 p + 1 11⎛⎞n k (1)log()− tuk+ =−∑∑⎜⎟(1) ∑ tnnk==00++1(1)!⎝⎠k p = 0 p

∞ n 11⎛⎞n k =−−+−∑∑⎜⎟( 1)() exp[tuk log( )] 1 tnnk==00+1 ⎝⎠k

11∞ n ⎛⎞n ⎛⎞1 (1)k 1 =−−∑∑⎜⎟ ⎜⎟t tnnk==00++1[]⎝⎠k ⎝⎠ uk

11∞ 1n ⎛⎞n (1)− k

=− + ∑∑⎜⎟ t ttnk==00 n++1()⎝⎠k uk

Therefore we obtain (which is simply a defining expression for the Stieltjes constants)

∞ p (1)− p 1 (4.3.217a) ∑ tuγςp ()= ( tu+− 1,) p=0 p! t and with t =1 this reverts to (4.3.216)

15 ∞ (1)− p ′ ∑ γςp ()uu= (2,)−= 1 ψ () u − 1 p=0 p!

A POSSIBLE CONNECTION WITH THE FRESNEL INTEGRAL

In 1955 Briggs [35b] showed that

∞ ∞ log p x (4.3.217b) γπ= 2cos2kx dx p ∑∫ k =1 1 x and we have the summation

∞∞∞(1)−−pp (1) ∞ log px ttkppγπ= 2cos2xdx ∑∑∑p ∫ ppk===001pp!!1 x

∞∞∞ cos 2π kx (− 1)pp t log p x = 2∑∑∫ dx kp==101 xp!

∞ ∞ cos 2π kx exp[− t log x ] = 2∑∫ dx k =1 1 x

∞ ∞ cos 2π kx = 2 dx ∑∫ t+1 k=1 1 x

Then using (4.3.217a) we see that

1cos2∞ ∞ π kx ς (1)2tdx+−= ∑∫ t+1 txk=1 1

∞∞∞ cos 2ππkx 1 cos 2 kx =−22dx dx ∑∑∫∫tt++11 kk==1100xx

∞∞∞ cos ku 1 cos 2π kx =−2(2π )t du 2 dx ∑∑∫∫tt++11 kk==1100ux

In a manner similar to (F.15) in Volume VI it may be shown that for 0 < q < 1

16 ∞ cosu π du = ∫ q 0 uqqΓ()cos(π /2) and hence we have for − 1 < t < 0

∞ cosu π du =− ∫ t+1 0 uttΓ+(1)sin(/2)π

∞ cos ky π k t dy =− ∫ t+1 0 yttΓ+(1)sin(/2)π

This then gives us

1(2)ππt+1 ∞∞1 cos2kx ς (1)tkdx+−=− t − 2 ∑∑∫ t+1 ttΓ+(1)sin(/2) tπ kk==110 x

Designating ts=− we obtain for 0 < s < 1

1(2)()πς1−s sk∞ 1 cos2πx ς (1−+=sd )− 2 x ∑∫ 1−s sssΓ−(1 ) sin(π / 2) k=1 0 x and using the Riemann functional equation this becomes

1cos2∞ 1 π kx (4.3.217c) ς (1−−=sdx ) 2 ∑∫ 1−s sxk =1 0

We note that

⎡⎤1cos2∞ 1 π kx limς (1−−sdx ) = 2 s→0 ⎢⎥∑∫ ⎣⎦sxk=1 0 and hence we have

∞ 1 cos 2π kx γ = 2∑∫ dx k =1 0 x in accordance with (4.3.217b).

Having regard to (4.3.217b) it is interesting to note that in their paper, An integral representation of the generalized Euler-Mascheroni constants, Ainsworth and Howell

17 logn [6y] used a contour integral of the form cotπ zdz in order to evaluate the Stieltjes ∫C z constants: this is reminiscent of the integral identity (6.5a)

1 bb∞ ∫∫p(x )cot(αα x / 2) dx= ∑ p ( x )sin nx dx 2 aan=0

With s =1/2 we get

⎛⎞11cos2 ⎛⎞ ∞ 1 π kx ςς⎜⎟+=22 ⎜⎟ − 2∑∫ dx ⎝⎠22 ⎝⎠ k=1 0 x and we see with a simple substitution that

12cos 2ππkx 2 π k ∫∫dx= cos( t2 ) dt 00xkπ

The Fresnel integral is defined by G&R [74, p.880]

2 x Cx()= ∫ cos()t2 dt 2π 0 and we then see that

1 cos 2π kx C ( 2π k ) ∫ dx = 0 xk

Hence we obtain

31⎛⎞ ∞ Ck (2)π (4.3.217d) ς ⎜⎟−=22∑ 22⎝⎠ k=1 k ⎛⎞1 It therefore appears that a double series could be established for ς ⎜⎟ using the known ⎝⎠2 expansion for the Fresnel integral

∞ (/2)(1)π 241kkk− x + Cx()= ∑ k =0 (4kk+ 1)(2 )!

Letting p = 0 in (4.3.217b) gives us

18 ∞∞∞ cos 2π kx γγ==22[()(2)]dx= Ci ∞− Ci π k 0 ∑∑∫ kk==111 x

∞ =−2(2)∑Ciπ k k =1 where Ci() x is the cosine integral defined in G&R [74, p.878] and [1, p.231] as

x costx−− 1 ∞ ( 1)nn2 Ci() x=+γγ log x +∫ dt =+log x +∑ 0 tnnn=1 2(2)!

(see also (6.94g) et seq. in Volume V for a more detailed exposition) and we can see from (6.94ga) that Ci (∞= ) 0 .

We will see in (6.117r) in Volume V that

1 ∞ γ =−2(2)∑Ci kπ 2 k =1 and we therefore have an unexplained difference of 1/2 which remains to be investigated. Using the Wolfram Integrator, we find that integrals of the form log p x cos 2π kx dx may be evaluated in terms of generalised hypergeometric series. ∫ x

We briefly reconsider the equation

∞∞∞(1)−−pp (1) ∞ log px ttkppγπ= 2cos2xdx ∑∑∑p ∫ ppk===001pp!!1 x

∞∞(1)−−pppp∞ logx ∞∞ (1) 1 log x =−2c∑∑tkxdxtkxdxpp∫∫os22cππ ∑∑os2 pk==01pxpx!!00 pk == 01

We now have regard to the second summation. Dilcher [54a] proved by induction that if log p x fx()= then p x

p!lp ogj x fx()n ()=++− snp ( 1, 1 j ) p n+1 ∑ x j=0 j!

19 where sn(++ 1, p 1) is a Stirling number of the first kind. We then have

()n fpsnpp (1)=++ ! ( 1, 1) which results in the Taylor series expansion

p ∞∞()n logxsnpf p (1) nn (++ 1, 1) fxp ()==∑∑ (1)! x −= p (1) x − xnnn==00!! n

If x →+1 z this may be written as

∞∞∞sn(1,1)++ p sn (1,1)++ p sn (1,1) ++ p log(1)!(1)pn+=zp +zz∑∑∑=pz !n+pz ! n+1 nnn===000nnn!!!

We have [126, p.56] sn (++= 1, p 1) snp ( , ) − nsnp ( , + 1) and thus

∞∞snp(, )−+ nsnp (, 1)sn ( ++ 1, p 1) logpn (1+=zp ) !∑∑zp+ ! zn+1 nn==00nn!!

∞∞snp(, ) nsnp (,+++ 1) ∞sn ( 1, p 1) =−p!!∑∑zpnn zp + ! ∑z n+1 nn==00nn!! n = 0 n !

∞∞snp(, ) snp (,+++ 1) ∞sn ( 1, p 1) =−p!!∑∑zpnn zp + ! ∑z n+1 nn==00nn!(1)!!− n = 0 n

∞∞snp(, ) sm (++ 1, p 1)∞sn (++ 1, p 1) =−p!!∑∑zpnmz+11 + p ! ∑z n+ nm==01nm!! n = 0 n ! and we therefore have obtained the expansion (3.105) from Volume I for the Stirling number of the first kind which is valid for z < 1

∞ snp(, ) ∞ snp(, ) logpn (1+=zp ) !∑ z or logpnxp= !∑ ( x− 1) n=0 n! n=0 n!

We then have

11log p xkcos 2π x∞ s (n ,p ) ∫∫cos 2π kx dx=− p! ∑ (x 1)n dx 00xxnn=0 ! or alternatively without x in the denominator

20 11log p xs∞ (n++ 1,p 1) ∫∫cos 2ππkx dx=− p! cos 2 kx∑ (x 1)n dx 00xnn=0 !

We have

11 ∫∫(x −= 1)cos2nnππkx dx (1)(1−−x )cos2n kx dx 00

and with the substitutiony =− 1 x this becomes

1 =−(1)nn∫ ykydy cos2π 0

Unfortunately, any further analysis in this direction will have to wait for another day.

EVALUATION OF SOME STIELTJES CONSTANTS

Let us consider the function Z (su , ) defined by

∞ n 1 ⎛⎞n ks Z(,su )=−+∑∑⎜⎟ (1)(k u ) nk==00n +1 ⎝⎠k

∞ n 1 ⎛⎞n k =−+∑∑⎜⎟( 1) exp[sku log( )] nk==00n +1 ⎝⎠k

∞∞n pp 1l⎛⎞n k skuog(+ ) =−∑∑⎜⎟(1) ∑ nk==00np+1!⎝⎠k p = 0

∞∞p n s 1 ⎛⎞n kp =−+∑∑ ∑⎜⎟(1)log(ku ) pn==00pn!1+ k = 0⎝⎠k and using ∞ n 11⎛⎞n kp+1 γ p ()uuk=−∑∑⎜⎟ (1)log( − + ) pn++11nk==00⎝⎠k this becomes ∞ s p =−1()∑ γ p u p=0 p!

Then, using (4.3.217a), we have

∞ n 1 ⎛⎞n ks Z(,su )=−+=−−∑∑⎜⎟ (1)(k u ) sς (1 su , ) nk==00n +1 ⎝⎠k

∞ n 1 ⎛⎞n ks The fact that −−ssuς (1 , ) =∑∑⎜⎟ ( − 1) ( ku + ) may in fact be immediately be nk==00n +1 ⎝⎠k seen from the Hasse formula (3.12a) from Volume I.

Differentiating Z(su , ) gives us

∞ n ∂ 1 ⎛⎞n ks Z(su , )=−++∑∑⎜⎟ ( 1) (ku ) log( ku ) ∂+snnk==001 ⎝⎠k

=−−−ssusuς ′ (1 , )ς (1 , )

Letting sm=+21 we obtain another derivation of (4.3.133b)

2m ∞ n mk2(2π ) 1 ⎛⎞n 21m+ (4.3.217e) ς (2mkk+=− 1) ( 1)∑∑⎜⎟ ( − 1) ( + 1) log( + 1) (2mn++ 1)!nk==00 1 ⎝⎠k and with sm= 2 we get

∞ n 1 ⎛⎞n km2 (4.3.217f) 2mmumuςς′ (1−−−= 2 , ) (1 2 , )∑∑⎜⎟ ( −+ 1) ( k 1) log( k + 1) nk==00n +1 ⎝⎠k

Also, using Lerch’s identity (4.3.116), with s =1 we obtain

∞ n 11⎛⎞n k logΓ− (u ) log(2πς ) − (0, u ) =∑∑⎜⎟ ( − 1) ( ku + )log( ku + ) 21nk==00n + ⎝⎠k which we have seen before in (4.3.119). Note that we previously we used (4.3.119) to derive Lerch’s identity.

With regard to the definition of Z(,su ) we may also note from (A.25) that

Nn 1 ⎛⎞n kN BN ()uk=−+∑∑⎜⎟ (1)(u ) nk==00n +1 ⎝⎠k

22 ∞ n 1 ⎛⎞n kN =−+∑∑⎜⎟(1)(ku ) nk==00n +1 ⎝⎠k

It may also be useful to consider a function of the type

∞ n 1 ⎛⎞n krs Z(,,rsuv ,)=−++∑∑⎜⎟ (1)(k u )( k v ) nk==00n +1 ⎝⎠k

We also note for reference that Kluyver showed in 1922 that

logpk 2∞ (− 1)⎛⎞ log k (4.3.217g) γ pp= ∑ B +1 ⎜⎟ pk+1log2k =1 ⎝⎠ 

We may also consider a more complex series

∞∞ kk∞∞ ∞ m nk(1)−− n(1)⎧ 1 1 ⎛⎞m rkk+1 ⎫ ∑∑znuzγ k ()=− ∑∑⎨ ∑ ∑⎜⎟(1) −+ nur log( )⎬ nk==10kkkm!!nk== 10⎩⎭++1m= 01 r = 0⎝⎠r

∞∞nkm ∞ k++11k znur1(⎛⎞m r −1)log(+ ) =−∑∑ ∑⎜⎟(1) ∑ nm==10nm++1( r = 0⎝⎠r k = 0 k1)!

∞∞n m z 1 ⎛⎞m r =−−+−∑∑ ∑⎜⎟( 1) exp{} [nur log( )] 1 nm==10nm+1 r = 0⎝⎠r

∞∞zn 11m ⎛⎞m ⎛ ⎞ (1)r 1 =−−∑∑ ∑⎜⎟ ⎜n ⎟ nm==10nm++1() r = 0⎝⎠r ⎝ ur ⎠

The Hasse identity tells us that

∞ 1(1)m ⎛⎞m − r nn(1,) u ∑∑⎜⎟ n =+ς mr==00mur++1()⎝⎠r and we therefore obtain

∞∞ kn∞ ∞ nk(1)− nz (4.3.217h) ∑∑znunuzγςk ()=+− ∑ ( 1,) ∑ nk==10kn! n=1 n=1

With u =1 we get

23 ∞∞ kn∞ ∞ nk(1)− nz ∑∑znγςk =+− ∑(1) nz ∑ nk==10kn! n=1 n=1 which may be written for z <1 as

∞∞ k nk(1)− (4.3.218) ∑∑znγγk =−−−−log(1 z )ψ (1 z ) nk==10k!

This result was obtained in a different way by Lopez [101ab] in 1998.

Coffey [45e] employed a different method in 2006 to show that for integers j ≥ 0 and z < 1

∞∞ k ∞ nkjjnj(1)− − () ∑∑znunuzjLizγςkj()=+−− ∑ ( 1,) (1)!+1 () nk==10()!kj− n= 1 and with j = 0 this becomes

∞∞ k ∞ nk(1)− n ∑∑znunuzzγςk ( )= ∑ (+−− 1, ) log(1 ) nk==10k! n=1

=ψ (uuz )−−−−ψ ( ) log(1 z )

Letting zz→− we obtain

∞∞k ∞ nn(1)− k n n−1 (4.3.219) ∑∑(−=−−++ 1) znunuzzγςk ( ) ∑ ( 1) ( , ) log(1 ) nk==10k! n= 2 and integrating this with respect to z produces

∞∞nk ∞ n (1)−−nk+1 (1) nx ∑∑x nuγςk ( )= −− ∑ ( 1) ( nu , ) ++ (1 x )log(1 +− xx ) nk==10nk+1! n=2 n

We have the well-known result [126, p.159] for x < u

∞ xn (4.3.220) ∑(1)(,)−=Γ+−Γ−n ςψnu log( u x ) log() u x () u n=2 n and hence we have

∞∞nk (1)−−nk+1 (1) ∑∑x nuγψk ()=− log( Γ ux + ) + log() Γ uxu + () + (1 + x )log(1 + xx ) − nk==10nk+1!

Differentiating (4.3.220) results in

∞ ∑(1)(,)−=+−nnςψψnux−1 ( u x ) () u n=2 and we obtain another proof of Coffey’s formula (4.3.219) above

∞∞k nn(1)− k ∑∑(−=−++++ 1) znuuzuzγψk ( ) ( ) ψ ( ) log(1 ) nk==10k!

Multiplying (4.3.219) by z and then integrating gives us

∞∞nk++21 ∞ n nkzz(1)− n1 22 ∑∑(−=−−+−+−+ 1) nuγςk ( ) ∑ ( 1) ( nu , ) [2( z 1)log(1 zzz ) 2 ] nk==10nk++2! n = 2 n14 and we know from [126, p.211] that for z < u (see also Coffey [45j, p.13])

∞ zzzn+12 ∑(−=− 1)n ςπ (nu , ) [2 u 1−+ log(2 )] [1−+ψ (u )] (1−Γ u )log ( u+ z ) n=2 n +122

+++−Γ− logGu ( z ) ( u 1)log ( u ) log Gu ( )

Hence we obtain

(4.3.221)

∞∞nk+2 2 nkzz(1)− z ∑∑(−=−−−−−−−Γ+ 1) nuγπk ( ) [2 u 1 log(2 )] [1ψ (u )] (1 u )log ( uz ) nk==10nk+ 2! 2 2

1 −+−−Γ++−+−+logGu ( z ) ( u 1)log ( u ) log Gu ( ) [2( z22 1)log(1 z ) z 2 z ] 4  We may also consider the following summation

∞∞ppp ∞n (1)−−−p (1)(s 1)⎧ 1 1 ⎛⎞n kpr++1 ⎫ ∑∑(1)()su−=γ pr+ ⎨ − ∑∑⎜⎟(1)log( −uk + )⎬ pp==00pp!!⎩⎭p++r1 n == 00n +1k⎝⎠k

(1)−−r ∞∞ 1 n ⎛⎞n (1)pr++111 (suk− 1) pr ++ log pr ++ (+ ) =−( 1)k (prpr+ )(+− 1)...( p+ 1) r+1 ∑∑⎜⎟ ∑ (1)sn−+nk==00 1⎝⎠k p = 0 (pr++ 1)!

With r =1 we get

∞∞∞pppn (1)−−p (1)(s − 1)⎧ 1 1 ⎛⎞n kp+2 ⎫ ∑∑∑∑(1)()su−=γ p+1 ⎨ −⎜⎟(1)log() −uk +⎬ ppn====0000pp!!⎩⎭p++2n1k⎝⎠k

11∞∞n ⎛⎞n (1)(1)log()− ppp+++222suk−+ (1)k (p 1) =−2 ∑∑⎜⎟ − ∑ + (1)sn−+nk==00 1⎝⎠k p = 0 (p + 2)!

We see that

∞ x p+2 ∑ =−−exx 1 p=0 (2)!p + and also that

xp∞ +2 2 de⎛⎞−−1 x x xp⎜⎟=+∑(1) dx⎝⎠ x p=0 (2)!p +

We therefore have

∞ x p+2 1(+−xe 1)x =∑ ( p + 1) p=0 (2)!p + and hence we get

∞∞(−+ 1) p 1 1 n ⎛⎞n ⎡ log(uk ) 1 ⎤ (1)()su−=−pkγ (1)1(1) −−−− s ∑∑p+1 21∑⎜⎟ ⎢ ss−−1⎥ pnk===00ps!(−+1)1(n0⎝⎠k ⎣ u +k)(u +k)⎦

1 1∞∞ 1 nn⎛⎞nnlog(uk+ ) 1 1 ⎛⎞ 1 =− + (1)− kk+ (1)− 21∑∑⎜⎟ ss− 21∑∑⎜⎟ − (1)ssn−−+ 1nk==00 1⎝⎠kk ( uks + ) (1) − nk ==00 n + 1⎝⎠ ( uk + )

1 =− −ς ′(,su ) (1)s − 2

Therefore we obtain

26 1(1)∞ − p (4.3.222) ′(,su ) (s 1)p () u −=ςγ2 +∑ −p+1 (1)sp− p=0 !

This then implies that lim[(ssu−=− 1)2ς ′ ( , )] 1 which concurs with (4.3.203). s→1

More generally, with a little more algebra, we may show that

r!(1)∞ − p (4.3.222a) (1)−=+−rrςγ() (,)su ( s 1) p () u r+1 ∑ pr+ (1)sp− p=0 !

In 1995, Choudhury [45ad] mentioned that, for r ≥1 and Re ()s > 1, Ramanujan determined that ∞∞logrppr ! (− 1) (4.3.222b) (1)−=rrς () ()ss =+ ( − 1) pγ ∑∑sr+1 pr+ pp==10ps(1)− p !

(see also Ramanujan’s Notebooks [21], Part I, p.224). This is a particular case of (4.3.222a) where u =1, and it should be noted that (4.3.222a) is valid for all s ≠ 1.

Letting s = 0 in (4.3.222) we see that

∞ γ ()u −=+ς ′(0,u ) 1 ∑ p+1 p=0 p! and applying Lerch’s identity (4.3.116) this may be written as

1 ∞ γ ()u (4.3.223) log(2π )−Γ=+ log (u ) 1 ∑ p+1 2!p=0 p which may be compared with (4.3.216)

∞ (1)− p ′ ψ ()uu=+ 1∑ γ p () p=0 p!

Differentiating (4.3.223) we have

∞ γ ′ ()u ψ ()u =−∑ p+1 p=0 p!

∞∞ n p+1 11⎧ ⎛⎞n k log()uk+ ⎫ =−∑∑∑⎨ ⎜⎟(1) ⎬ pn==00pn!1⎩⎭++ k = 0⎝⎠k uk

27 ∞∞p n log (uk++ )⎧ 1 ⎛⎞n k log( uk ) ⎫ =−∑∑∑⎨ ⎜⎟(1) ⎬ pnk===00pn!1⎩⎭++0⎝⎠k uk

∞ n 1 ⎛⎞n k = ∑∑⎜⎟(−+ 1) log(uk ) nk==00n +1 ⎝⎠k which we saw earlier in (4.3.74).

A further differentiation of (4.3.223) gives us

∞ γ ′′ ()u ψ ′()u =−∑ p+1 p=0 p!

∞∞11⎧⎫nn⎛⎞nn log()1lppuk++∞ ⎛⎞ og()+1 uk (1)p (1)kk(1) =+∑∑∑⎨⎬⎜⎟ −22 − ∑∑ ⎜⎟ − pnk===000pn!1()1()⎩⎭++++⎝⎠kk uknnk== 00 ⎝⎠ uk

∞∞11ln ⎛⎞n ogp (uk+ ) (1)k (p 1) =−∑∑⎜⎟ 2 ∑ + nk==00nu++1()!⎝⎠k k p= 0 p

∞∞1 n ⎛⎞n log(uk++ ) logp ( uk ) (1)k −−∑∑⎜⎟ 2 ∑ nk==00nu++1()!⎝⎠k kp p= 0

∞ 11n ⎛⎞n ( 1)k [1 log(uk )]( uk ) =−+++∑∑⎜⎟ 2 nk==00nu++1()⎝⎠k k

∞ 1 n ⎛⎞n log(uk+ ) (1)k (uk ) −−∑∑⎜⎟ 2 + nk==00nu++1()⎝⎠k k

∞ n 11⎛⎞n k =−∑∑⎜⎟(1) nk==00nuk++1 ⎝⎠k

=ψ ′()u (which is where we started from).

Letting s =−1in (4.3.222) gives us

12∞ p ′ −−ςγ(1,)uu =+∑ p+1 () 4!p=0 p

Differentiation gives us

28 d ∞ 2 p ′′ −−=ςγ(1,)uu∑ p+1 () dup=0 p! and using

∞ 1ln ⎛⎞n og()p+1 uk+ ′ k γ p+1()u =−∑∑⎜⎟(1)− nk==00nu++1 ⎝⎠k k we have

∞∞ppn +1 duk21⎛⎞n k log()+ ς ′(1,)−=u ∑∑ ∑⎜⎟ (1)− du pn==00p!1 n ++ k = 0⎝⎠k u k

∞∞pp n 2 log (uk+ ) 1 ⎛⎞n k log( uk+ ) =−∑∑∑⎜⎟(1) pnk===00p!1nu++0⎝⎠k k

∞∞n pp 1 ⎛⎞n k log(uk++ ) 2 log ( uk ) =−∑∑⎜⎟(1) ∑ nk==00nukp++1!⎝⎠k p =0

∞ n 1 ⎛⎞n k log(uk+ ) =−∑∑⎜⎟( 1) exp[2log(uk + )] nk==00nu++1 ⎝⎠k k

∞ n 1 ⎛⎞n k =−++∑∑⎜⎟( 1) (uk )log( uk ) nk==00n +1 ⎝⎠k

Then referring to (4.3.108) we see that

d ςςς′′(1,)−=uuu (0,) − (0,) du

We also see that

∞∞p (1)− γ ′p′+1()u 1()+=−∑∑γ p u pp==00p!!p

We have from (4.3.222a)

∞ γ ()u (1)−=−+rrς () (0,)ur (1) r+1 ! ∑ pr+ p=0 p! and hence with u =1

29 ∞ γ (1)−=−=−+rrςς() (0,1)(1) rr() (0)(1) r+1r ! ∑ pr+ p=0 p!

From (4.3.222b) we see that

⎡⎤(1)!−−rpr ∞ (1) rr() p (1)−−=−⎢⎥ςγ (,)su r+1 ∑ (s 1)pr+ () u ⎣⎦(1)sp− p=0 ! and hence as we have seen before

r rr⎡⎤() (1)!− r (1)lim−−=⎢⎥ςγ (,)su r+1 r ()u s→1 ⎣⎦(1)s −

It seems that a lot of the above analysis was superfluous: having regard to

∞ p 1(1)− p ςγ(,su )=+∑ p ()(u s − 1) sp−1!p=0 we immediately see that

(1)!−−rpr ∞ (1) ςγ()r (su , )=+ p ( p −−+− 1)...( p r 1) ( u )( s 1) pr− r+1 ∑ p (1)sp− p=0 ! which is in fact exactly equivalent to (4.3.222a)

(1)!−−rpr ∞ (1) ςγ()rrp(,su )=+−− (1) ( s 1) () u r+1 ∑ pr+ (1)sp− p=0 !

We have ∂−∞ (1)p ′ p ςγς(,su )=−∑ p ()(u s 1)=− s ( s+ 1,) u ∂upp=0 ! and hence we obtain

∞∞(1)−−pp(1) ′ pp+1 ∑∑γγpp()(us−=−− 1) 1 ()us pp==00pp!!

The left-hand side is equal to

∞∞ppn (1)−+ 1 ⎛⎞n kplog(uk ) −−−∑∑∑⎜⎟(1) (s 1) pnk===000pn!1++⎝⎠k uk

∞∞1(1)(1)(1)log()n ⎛⎞n −−−kpppsuk + =−∑∑⎜⎟ ∑ nk==00nukp++1!⎝⎠k p = 0

∞ 1(1)n ⎛⎞n − k =−∑∑⎜⎟ exp[ − (suk − 1)log( + )] nk==00nuk++1 ⎝⎠k

∞ 1(1)n ⎛⎞n − k

=−∑∑⎜⎟ s nk==00nuk++1()⎝⎠k

=−ssς ( + 1, u ) 

We see that d ∞ 1(1)n ⎛⎞n − k γ 0 ()u =−∑∑⎜⎟ du nk==00n+1 ⎝⎠k u+ k

∞ 1 (2,u ) =−ς =−∑ 2 n=0 ()nu+ where we have employed the Hasse formula (4.3.106a). Integration then results in the familiar formula for the digamma function [126, p.14]

∞∞⎡⎤11 1 (4.3.223a) γ 00()xxx−=γψ (1) ∑∑⎢⎥ − =−−( 1) =−−() γ nn==00⎣⎦nxn++1(1n ++)() nx

Similarly we find that

duk∞ 1n ⎛⎞n (− 1)k log(+ ) γ1()u =−∑∑⎜⎟ du nk==00n ++1 ⎝⎠k u k

The Hasse formula (4.3.106a) gives us

∂−∞ 1n ⎛⎞n ( 1)k log(uk+ ) [(s−=−+=− 1)ςςς ( su , )] ( s 1)′ ( su , ) ( su , ) ∑∑⎜⎟ s−1 ∂+snnk==001()⎝⎠k u+k and with s = 2 we have

∞ 1n ⎛⎞n (− 1)k log(uk+ ) ςς′(2,uu )+=− (2, ) ∑∑⎜⎟ nk==00nu++1 ⎝⎠k k

31 Therefore we get

(4.3.223b)

dn∞∞∞log(+ un ) 1 1−+ log(u ) ′ γς1(uuu )=+=−+ (2, ) ς (2, ) ∑∑∑22 = 2 du nnn===000()nu++ () nu () nu +

We may easily evaluate the following integral

x 1−+ log(nu ) log( nu + ) x du = ∫ 2 1 ()nu++ nu 1 and hence we have upon integrating (see also (4.3.228b))

∞ ⎡log(nx++ ) log( n 1) ⎤ (4.3.223c) γγ11()x −= (1) ∑ ⎢ − ⎥ n=0 ⎣ nx++ n1 ⎦

As mentioned previously, the Stieltjes constants γ n are the coefficients in the Laurent expansion of ς (s ) about s =1.

∞ n ∞ n 1(1)− n 1(1)− n ςγ()ss=+∑ n ( − 1) or ς (1)ss+=+∑ γ n sn−1!n=0 snn=0 !

⎡⎤1 Since lim⎢⎥ς (s ) −=γ (see also (4.4.99m)), it is clear that γ 0 = γ . It may be shown, s→1 ⎣⎦s −1 as in [81, p.4], that

⎡⎤NNlognnkN log +1 ⎡ logn ktN log n⎤ (4.3.224) γ n =−=−lim lim dt NN→∞ ⎢⎥∑∑→∞ ⎢ ∫ ⎥ ⎣⎦kk==11kn+1 ⎣ k1 t⎦

For example, Bohman and Fröberg [24a] noted that

∞ (1)s − ∞ ⎡ 11⎤ (1)()ss and 1 −=ς ∑ s ∑ ⎢ ss−−11− ⎥ = k=1 k k=1 ⎣kk(1)+ ⎦ and, assuming that s is real and greater than 1, the above two equations may be subtracted to give

∞ ⎡⎤11(1)s − (4.3.225) (1)()1ss −=+ς ∑ ⎢⎥ss−−11 −+ s k=1 ⎣⎦(1)kkk+

32 It will be noted from the above that

lim(ss−= 1)ς ( ) 1 s→1 which we shall also see in (4.4.100x).

Equation (4.3.225) may be written as

∞ −1 (ss−=+−−+−−−+−−− 1)ς ( ) 1∑ ⎣⎦⎡⎤ exp( ( s 1)log( k 1)) exp( ( sksksk 1)log ) ( 1) exp( ( 1)log ) k =1

∞∞ nn ∞ nn ⎡⎤(1)(−−ss 1) nn −1 (1)( −−s 1) n =+1 ∑∑⎢⎥()log (kk + 1) − log + ∑ log k kn==10⎣⎦nk!!n=0 n

Dividing by (s − 1) we get (4.3.224) where

∞∞⎡⎤lognnkk log (+− 1) log n k⎡ log n kk+1 log n t⎤ (4.3.225a) γ =− =−dt n ∑∑⎢⎥⎢ ∫ ⎥ kk==11⎣⎦kn+1 ⎣ ktk ⎦

Osler [105(v)] has provided an interesting derivation of the above formula using partial sums of the Riemann zeta function.

Differentiating (4.3.225) we get

∞ ⎡ log(kksk+− 1) log ( 1) log 1 ⎤ (1)()()sss′ −+=−ςς∑ ⎢ s−−11 +−sss +⎥ k =1 ⎣ (1)kkkk+ ⎦ and hence we have

∞ ⎡ log(kksk+− 1) log ( 1) log ⎤ (1)()ss′ −=−ς ∑ ⎢ ss−−11 +− s⎥ k=1 ⎣ (1)kkk+ ⎦ which implies that (as is easily seen by inspection)

∞∞log(kk+ 1) log ∑∑ss−−11= kk==11(1)kk+

Ivić [81, p.41] reports that the coefficients γ p (x ) for x ∈ (0,1] may be expressed as

⎛⎞N logpp (nx++ ) log+1 ( Nx ) (4.3.226) γ p ()x =− lim N →∞ ⎜⎟∑ ⎝⎠n=0 nx++ p1

33 and in particular we have

⎛⎞N 1 (4.3.226i) γ 0 ()x =−+ lim log(Nx ) N →∞ ⎜⎟∑ ⎝⎠n=0 nx+

N ⎛⎞log(nx+ ) 1 2 (4.3.226ii) γ1()x =−+ lim log(Nx ) N →∞ ⎜⎟∑ ⎝⎠n=0 nx+ 2

We also see that ⎛⎞N logpp (nN++ 1) log+1 ( 1) γ p (1)=− lim N →∞ ⎜⎟∑ ⎝⎠n=0 np++11

⎛⎞N logpp (nN++ 1) log+1 ( 1) =−lim N →∞ ⎜⎟∑ ⎝⎠n=1 np++11

⎛⎞N logppkN log (++ 1) log p+1 ( N 1) =+−lim N →∞ ⎜⎟∑ ⎝⎠k =1 kN++11 p

logp (N + 1) and since lim = 0 this becomes N →∞ N +1

⎛⎞N logppkN log+1 (+ 1) =−lim N →∞ ⎜⎟∑ ⎝⎠k=1 kp+1

⎛⎞N logppkNN log+++111 log p log p ( N+ 1) =−+−lim N →∞ ⎜⎟∑ ⎝⎠k =1 kp++11 p p + 1

Therefore, assuming that lim⎡⎤ logpp++11NN− log (+= 1) 0 , we have another proof of the N →∞ ⎣⎦ well-known result (see for example the 1955 paper, “The power series coefficients of ς ()s ”, by Briggs and Chowla [35a])

⎛⎞NNlogppkN log +1 ⎛⎞logp kxN log p γγpp==(1) lim − =lim ⎜⎟ − dx NN→∞⎜⎟∑∑ →∞ ∫ ⎝⎠kk==11kp+1 ⎝⎠ k1 x

As shown by Dilcher in [54a] we also have for k ≥ 0

∞−kkk 1 +1 kk+−1() nlog n ⎛⎞k j kj log 2 (4.3.226iii) (−=−=−− 1)ςγaj (1)∑∑ ( 1) ⎜⎟( 1) log 2 nj==10nk⎝⎠j +1

34 of which (C.61) in Volume VI is a particular case. It should be noted that I have inserted a factor of (− 1) j which appears to be required in (4.3.226iii). We then see that (see also (4.3.237) and (4.4.44j))

1 (4.3.226iv) ςγγ(2)(1)=−+ log3 2 log2 2 2 log 2 a 3 1

See also the paper by Collins [46aa] entitled “The role of Bell polynomials in integration” ∞ logn x where the integrals dx are evaluated in terms of the Stieltjes constants. It may of ∫ x 0 e +1 course be noted that

∞∞lognnsnxdx−1 d dx ==Γdx [()sς ()] s ∫∫edsedsxnxn++11 a 00s=1 s=1

From (4.3.226ii) we have

N ⎛⎞⎡⎤log(nx++ ) log( n 1) 1 221 γγ11()xN−= (1)lim − −log( ++x ) log(N + 1) N →∞ ⎜⎟∑ ⎢⎥ ⎝⎠n=0 ⎣⎦nx++ n12 2

We see that

log22 (NNxNNxNNx+− 1) log ( + ) = [log( ++ 1) log( + )][log( +− 1) log( + )]

(1)N + =++log[(NNx 1)( )]log ()Nx+

⎛⎞1 ⎜⎟1+ ⎡⎤2 ⎛⎞⎛⎞1 x ⎝⎠N =++log⎢⎥N ⎜⎟⎜⎟ 1 1 log ⎣⎦⎝⎠⎝⎠NN⎛⎞x ⎜⎟1+ ⎝⎠N

⎛⎞11 ⎛⎞ ⎛⎞ 1 ⎜⎟11++ ⎜⎟ ⎜⎟ 1 + ⎡⎤2 ⎛⎞⎛⎞11x ⎝⎠NN ⎝⎠⎛⎞ ⎝⎠ N =++log⎢⎥NN⎜⎟⎜⎟ 1 1 log = 2log log ++ log⎜⎟ 1 log ⎣⎦⎝⎠⎝⎠NN⎛⎞x ⎛⎞xx⎝⎠N ⎛⎞ ⎜⎟11++ ⎜⎟ ⎜⎟ 1 + ⎝⎠NN ⎝⎠ ⎝⎠ N

35 ⎛⎞1 ⎜⎟1+ ⎛⎞x ⎝⎠N ++log⎜⎟ 1 log ⎝⎠N ⎛⎞x ⎜⎟1+ ⎝⎠N

Therefore, using L’Hôpital’s rule we may prove that

lim[log22 (NNx+− 1) log ( + )] = 0 N →∞ and this limit is almost intuitively obvious.

Hence we see that (as originally obtained in (4.3.223b))

∞ ⎡⎤log(nx++ ) log( n 1) (4.3.226v) γγ11()x −= (1) ∑ ⎢⎥ − n=0 ⎣⎦nx++ n1

More generally, with the identity

abpp++11−=−+( abababab )( pp−122 + p − ++... p ) we may also prove that

lim[logpp++11 (NNx+− 1) log ( + )] = 0 N →∞

The following alternative proof is quite succinct. Define f ()xNx= log(p+1 + ) and, by the mean value theorem of calculus, we have

logp (N +α ) logpp++11 (Nx+− ) log ( N +=− 1) ( x 1)( p + 1) N +α where 1 < α < x . Successive applications of L’Hôpital’s rule easily show that

logp (N +α ) lim = 0 N →∞ N +α

Hence we have

∞ ⎡⎤logpp (nx++ ) log ( n 1) (4.3.227) γγpp()x −= (1) ∑ ⎢⎥ − n=0 ⎣⎦nx++ n1 and in particular we have

36 ∞∞⎡⎤11 1 (4.3.228) γγ00()x −= (1) ∑∑⎢⎥ − =−−(xx 1) =−−γψ() nn==00⎣⎦nxn++1(1 n ++)() nx

We therefore have

∞ 1 ′′()x ()x γψ0 =−∑ 2 =− n=0 ()nx+ and hence upon integration we see that

γ 0 ()x =−ψ ()xc +

We already know that the integration constant c = 0 but we may also determine this in the following way:

As x →∞ we see that

j ⎛⎞k k ⎡ k (1)− ⎜⎟⎤ ⎛⎞k j j ⎜⎟(−+=+ 1) log(xj ) log⎢ ( xj ) ⎝⎠⎥ ∑ j ∏ j=0 ⎝⎠ ⎣⎢ j=0 ⎦⎥

k ⎛⎞k j ⎡⎤⎜⎟(1)− ∑ j ∼ log ⎢⎥x j=0 ⎝⎠ = δ log x ⎢⎥k ,0 ⎣⎦

Therefore we have

∞ k 1 ⎛⎞k j ∑∑⎜⎟(−+→ 1) log(x jx ) log as x →∞ kj==00k +1 ⎝⎠j

and hence −→γ 0 ()x logx as x →∞. We have

lim[γ 0 (xx )+= log ] 0 x→∞ and therefore

lim[−++=ψ (xc ) log x ] 0 x→∞ which implies that c = 0 since we already know from (E.66a) in Volume VI that lim[−+ψ (xx ) log ] = 0 . x→∞

37 We easily see that

(4.3.228a) ∞ pp p ⎡⎤log (nx++ ) log ( n 1) p ∂ γγpp()x −= (1) − =− (1)lim [(,)ςsx − ς (,1)] s ∑ ⎢⎥s→1 p n=0 ⎣⎦nx++ n1 ∂ s

∂ p =−(1)limp [(,)ςςsx − ()] s s→1 ∂s p

From this we easily see that

p p ∂ (4.3.228b) γγpp()x −=− ()ys lim(1) [(,)ςx − ς (,sy )] s→1 ∂s p which we shall derive in a slightly different manner in (4.3.233e). It may also be readily derived from (4.3.207). This rather innocuous little equation actually turns out to be quite useful later in (4.3.233f).With p = 0 and y =1 we obtain

(4.3.228c) γ 0 ()x −=γς lim[(,)sx − ς ()] s s→1 which we saw in (4.3.204a). We see that

⎛⎞111∞ ⎡ ⎤ lim[ςς (sx , )−= ( s )] lim⎜⎟ + − ss→→11∑ ⎢ ss⎥ ⎝⎠xnxnn=1 ⎣()+ ⎦

111∞ ⎡ ⎤ =+∑ ⎢ −⎥ x n=1 ⎣nxn+ ⎦ and, using (E.14) in Volume VI, we have

111∞ ⎡⎤ +−=−−∑ ⎢⎥γψ()x xnxnn=1 ⎣⎦+ which gives us another derivation of (4.3.228c).

From (4.3.228b) we also see that

⎛⎞11 ⎛⎞ 2 (4.3.228d) ς ′′⎜⎟1,−=−+=+ςγγ() 1,111 ⎜⎟ log 2 2 γ ⎝⎠22 ⎝⎠ as reported by Adamchik in [2a]. See also (4.3.233b) et seq.

38 Differentiation of (4.3.228a) with respect to x gives us

p d p ∂∂ γςp ()x =− (1) lim [(,)sx −ς ()] s dx ∂∂xs→1 s p and, since the partial derivatives commute in the region where ς (sx , ) is analytic, we have ∂∂p =−(1)limp [(,)ςςsx − ()] s s→1 ∂∂sxp

∂ p =−(1)limp+1 [ssς ( + 1,)] x s→1 ∂s p

∂ since ςς(,sx )=− s ( s + 1,) x. With p = 1 we obtain ∂x

d γς(x )=+′ (2,xx ) ς (2, ) dx 1 which we have previously seen in (4.3.223b).

This may also be obtained directly as follows. Referring to (4.3.227) we see that

dn∞∞logpp−1 (+ xn ) log (+ x ) γ p ()xp=−∑∑22 dx nn==00()nx++ () nx and in particular

dn∞∞1 log(+ x ) γ1()x =−∑∑22 dxnn==00() n++ x () n x

=+ς (2,x )ς ′ (2,x )

We now recall (4.3.112b)

∞ 1 n ⎛⎞n log(x + k ) (1)(,)(,)ssxsxςς′ (1)k −+=−−∑∑⎜⎟ s−1 nk==00nxk++1()⎝⎠k and note that

∞ n 1 ⎛⎞n k log(x + k ) ςς′(2,xx )+=− (2, ) ∑∑⎜⎟( − 1) nk==00nxk++1 ⎝⎠k

Therefore we get as expected

∞ n dxk1 ⎛⎞n k log(+ ) γ1()x =−∑∑⎜⎟(1)− dx nk==00n ++1 ⎝⎠k x k

Integration results in

∞∞nn 11⎛⎞nnkk2211⎛⎞ γγ11(x )−=− (1) ∑∑⎜⎟( − 1) log (xk ++ ) ∑∑⎜⎟( − 1) log (1 +k ) 21nk==00nn++⎝⎠kk21 nk == 00⎝⎠ and by symmetry we see that

∞ n 11⎛⎞n k 2 γ1()x =−∑∑⎜⎟(1)log( −xk + ) + c 21nk==00n + ⎝⎠k

It remains for us to determine the integration constant. Employing the same method as before, as x →∞ we see that

kk ⎛⎞kkjj2 ⎛⎞ ∑∑⎜⎟(−+=−++ 1) log (x jx )⎜⎟ ( 1) log(j )log(xj ) jj==00⎝⎠jj⎝⎠

j ⎛⎞k k (1)− ⎜⎟ = ∑log(x ++jxj )⎝⎠j log( ) j=0

2 = δk ,0 log x

Therefore we have

∞ k 11⎛⎞k j 221 −−+→−∑∑⎜⎟(1)log(x jx ) log as x →∞ 21kj==00k + ⎝⎠j 2

It is seen that

∞ n 11122⎛⎞n k γ1(x )=− log xx −∑∑⎜⎟( − 1) log ( +k ) +c 221nk==10n + ⎝⎠k

However, where we go from here I do not know! When x =1 we do however find that c = 0 provided we already know that

40 ∞ n 11⎛⎞n k 2 γ1()uu=−∑∑⎜⎟(1)log( − +k ) 21nk==00n + ⎝⎠k 

We see that starting the summation of (4.3.227) at n =1 gives us

logppxnxn∞ ⎡ log (++ ) log p ( 1) ⎤ (4.3.228e) γγpp()x −− (1) =∑ ⎢ − ⎥ xnxnn=1 ⎣ ++1 ⎦ and thus, in view of the telescoping of the above series, as x → 0 we have

log pppxn∞ ⎡log (++xn ) log ( 1) ⎤ lim[γγpp (x )−− (1) ] = lim − xx→→00∑ ⎢ ⎥ xnn=1 ⎣ ++xn1 ⎦

∞ ⎡⎤logppnn log (+ 1) = ∑ ⎢⎥−=0 n=1 ⎣⎦nn+1 We then have

log p x (4.3.228f) lim[γ pp (x )−= ] γ x→0 x

Therefore we get

⎡⎤logp xxk 1∞ 1 n ⎛⎞n + lim +−=( 1)kp log+1 0 x→0 ⎢⎥∑∑⎜⎟ ⎣⎦xp++11nk==00 n⎝⎠k 1 + k

Using L’Hôpital’s rule we see that

p ∞ n ⎡⎤logx 1 1 ⎛⎞n kp+1 xk+ ⎢⎥+−⎜⎟(1)log x pn++11∑∑ k 1+ k lim ⎢⎥nk==00⎝⎠ x→0 ⎢⎥x ⎢⎥ ⎣⎦

⎡⎤p logpp−1 xx−+ log∞ 1 n ⎛⎞n logp (xk ) =+−lim ( 1)k x→0 ⎢⎥2 ∑∑⎜⎟ ⎣⎦xnnk==00++1 ⎝⎠k xk

⎡⎤pxlogpp−1 − log x ′ =−lim⎢⎥2 γ p (x ) x→0 ⎣⎦x

By differentiating (4.3.228e) we get

p logpp−−11x− log x∞∞ log p ( nx++ ) logp (nx ) ′ γ p ()xp−=−+222∑∑ xnnn==11()++xn ()x and hence

⎡⎤p logpp−−11xx− log∞∞ log p n log p n ′ lim⎢⎥γ p (xp ) −=222−+∑∑ x→0 xnn ⎣⎦nn==11

This gives us

⎡⎤pxlogpp−1 − log x ′ pp++−1() p 1 ( p 1) (4.3.228f) lim⎢⎥γςp (xp )−=2 (−+ 1) (2) (− 1)ς (2) x→0 ⎣⎦x

We also have

dn∞∞log(+ xn ) log2 (+ x ) γ 2 ()x =− 2∑∑22 dx nn==00()n++ x () n x

=− 2ς ′′′ (2,x ) −ς (2,x )

We have

∞ 1ln ⎛⎞n og2 ()x + k (1)(,)2(,)ssxsx′′ ′ (1)k −+=ςς∑∑⎜⎟ − s−1 nk==00nxk++1()⎝⎠k and therefore we see that

∞ n 2 1l⎛⎞n k og()x + k ςς′′(2,xx )+= 2 ′ (2, ) ∑∑⎜⎟( − 1) nk==00nxk++1 ⎝⎠k

Hence we get

∞ n 2 dxk1l⎛⎞n k og()+ γ 2 ()x =−∑∑⎜⎟(1)− dx nk==00n ++1 ⎝⎠k x k and upon integration we obtain

∞∞nn 11⎛⎞nnkk3311⎛⎞ γγ22(x )−=− (1) ∑∑⎜⎟( − 1) log (xk ++ ) ∑∑⎜⎟( − 1) log (1 +k ) 31nk==00nn++⎝⎠kk31 nk == 00⎝⎠

42 and thus we have

∞ n 11⎛⎞n k 3 γ 2 ()x =−∑∑⎜⎟(1)log( −xk + ) + c 31nk==00n + ⎝⎠k

Using (4.3.227) we see that

∞∞(1)−−++pp(1)∞⎡ log(pnx ) log(p n 1)⎤ ∑∑∑[()γγppx −= (1)] ⎢ − ⎥ pp==00ppnxn!!n=0⎣ ++ 1⎦

∞∞1(1)log()−+ppnx ∞∞ 1(1)log(1) −+ppn =−∑∑ ∑∑ np==00nx++ p!1!np==00 n p

∞∞11 =−+−−+∑∑exp[ log(nx )] exp[ log(n 1)] nn==00nx++n1

∞∞11 =−∑∑22 nn==00()nx++ (1) n

=−ψ ′′()x ψ (1)

This concurs with (4.3.216) above.

We see from (4.3.215) that

xx∞ 1 n ⎛⎞n γ ()udu=−( − 1)k log(u + kdu ) ∫∫0 ∑∑⎜⎟ 11nk==00n +1 ⎝⎠k

x ∞ n 1 ⎛⎞n k =−⎜⎟( − 1) [(uk + )log( uk + ) − u ] ∑∑n +1 k nk==00⎝⎠ 1

∞∞nn 11⎛⎞nnkk⎛⎞ =−−x 1 ∑∑⎜⎟(− 1) (xk + )log( xk + ) +∑∑⎜⎟(− 1) (1 +k )log(1 + k ) nk==00nn++11⎝⎠kk nk ==00⎝⎠ and using (4.3.74a) we obtain for x > 0

x (4.3.229) γς(udu )=− log Γ ( x ) =− [′ (0, x ) −ς′ (0)] ∫ 0 1

as expected since γ 0 ()uu=−ψ (). Similarly, we have

xx11∞ n ⎛⎞n γ ()udu=−(1)log( −k 2 u + kdu ) ∫∫1 ∑∑⎜⎟ 1121nk==00n + ⎝⎠k

x ∞ n 11⎛⎞n k 2 =−⎜⎟( − 1) (uk + )[log ( uk +− ) 2log( uk ++ ) 2] 21∑∑n + k nk==00⎝⎠ 1

We recall (4.3.107a)

∞ 1 n ⎛⎞n log(uk+ ) (1)(,)(,)ssusu′ (1)k −+=−−ςς ∑∑⎜⎟ s−1 nk==00nu++1()⎝⎠k k and differentiation results in

∞ 1ln ⎛⎞n og2 ()uk+ (1)(,)2(,)ssusu′′ ′ (1)k −+=ςς∑∑⎜⎟ − s−1 nk==00nu++1()⎝⎠k k

∞ 1ln ⎛⎞n og3 ()uk+ (ssusu 1)(,)3(3) (2) (,) (1)k −+=−−ςς ∑∑⎜⎟ s−1 nk==00nu++1()⎝⎠k k

With s = 0 we get

∞ n 1 ⎛⎞n k −+=−ςς′(0,u ) (0, u ) ∑∑⎜⎟( −++ 1) (uk )log( uk ) nk==00n +1 ⎝⎠k

∞ n 1 ⎛⎞n k 2 −+ςς′′(0,u ) 2 ′ (0, u ) =∑∑⎜⎟( −++ 1) (uk )log ( uk ) nk==00n +1 ⎝⎠k

∞ n (3) (2) 1 ⎛⎞n k 3 −+ςς(0,uu ) 3 (0, ) =−−++∑∑⎜⎟( 1) (ukuk )log ( ) nk==00n +1 ⎝⎠k

Using the Leibniz rule for differentiation we note that

∂∂∂nnn−1 (4.3.229a) [(ssus−=−+ 1)ςςς ( , )] ( 1) ( sunsu , ) ( , ) ∂∂∂sssnnn−1 and using the above we obtain

(4.3.230)

44 ∞ n ()mm(− 1) m1 ⎛⎞n km −+ςς(0,um ) (0, u ) =− ( 1) ∑∑⎜⎟( −++ 1) (uk )log ( uk ) nk==00n +1 ⎝⎠k

With u =1 we get

(4.3.230i) ∞ n ()mm(− 1) m1 ⎛⎞n k m −+ςς(0)mk (0) =− ( 1) ∑∑⎜⎟( −++ 1) (1 )log (1k ) nk==00n +1 ⎝⎠k and therefore we see that

(4.3.230ii) ∞ n ()mm(− 1) m m1 ⎛⎞n km −+ςς(0)mm (0) +−=− ( 1) γm−1 ( 1) ∑∑⎜⎟( − 1)kk log (1 + ) nk==00n +1 ⎝⎠k

∞ n 1 ⎛⎞n km Hence, using Apostol’s paper [14aa], we may express ∑∑⎜⎟(1)log(1− kk+ ) in nk==00n +1 ⎝⎠k terms of Stieltjes constants and other known constants.

For example we have with m =1

∞ n (1) (0) 1 ⎛⎞n k −+ςςγ(0) (0) −=−∑∑⎜⎟( − 1)kk log(1 + ) nk==00n +1 ⎝⎠k which gives us

∞ n 11⎛⎞n k 1 (4.3.230iii) ⎜⎟(−+=−−+ 1)kk log(1 ) log(2π ) γ ∑∑k nk==00n +12⎝⎠ 2

When m = 2 we get

∞ n (2) (1) 1 ⎛⎞n k 2 −+ςςγ(0)2(0)2 +=1 ∑∑⎜⎟(1)log(1) −kk + nk==00n +1 ⎝⎠k and hence we have

(4.3.230iv)

∞ n 11122 1⎛⎞n k 2 log (2πς )+−+−= (2) γγ1 log(2 π ) ∑∑⎜⎟( − 1)kk log (1 + ) 242 nk==00n +1⎝⎠k

In passing, we note that

1 1∞ 1 n ⎛⎞n log(uk+ ) ′(,su ) (, su ) (1)k ςς=− − ∑∑⎜⎟− s−1 ssnuk−−++111()nk==00⎝⎠k

2 2∞ 1 n ⎛⎞n log(uk+ ) ′′(,su ) (, su ) (1)k ςς=+22∑∑⎜⎟ −s−1 (1)ssnu−−++ (1)nk==00 1⎝⎠k (k )

11∞ n ⎛⎞n log()2 uk+ (1)k +−∑∑⎜⎟ s−1 sn−+11nk==00⎝⎠k () uk +

3! 3!∞ 1 n ⎛⎞n log(uk+ ) (3) (,su ) (,su ) (1)k ςς=− 33− ∑∑⎜⎟− s− 1 (1)ssnu−−++ (1)nk==00 1⎝⎠k (k )

31∞ n ⎛⎞n log()2 uk+ (1)k −−21∑∑⎜⎟ s− (1)sn−+nk==00 1⎝⎠k ( uk + )

11∞ n ⎛⎞n log()3 uk+ (1)k −−∑∑⎜⎟ s−1 sn−+11nk==00⎝⎠k () uk +

We then obtain

x 11x γςςςς()udu=−+−−+′′ (0,) u ′ (0,) u ′ (0,) u (0,) u u ∫ 1 1 221

Hence we get

x 1 (4.3.231) γςς()udu=− [′′′′ (0,) x (0)] ∫ 1 1 2

Moving on we have

xx11∞ n ⎛⎞n γ ()udu=−(1)log( −k 3 u + kdu ) ∫∫2 ∑∑⎜⎟ 1131nk==00n + ⎝⎠k

x x log33 (ukduuk+=+ ) ( )[log ( uk +−+++− ) 3log2 ( uk ) 6log( uk ) 6] ∫ 1 1

Therefore we obtain

x ⎛⎞1 γςςςςςς()udu=−(3) (0,) u +(2) (0,) u +−(2) (0,)2 u +(1) (0,) u +− 2(1) (0,)2(0,)u + u ∫ 2 ⎜⎟()() 1 ⎝⎠3

x ⎛⎞1 +−2⎜⎟u ⎝⎠2 1 and thus we have

x 1 (4.3.231i) γςς()udu=− [(3) (0,) x − (3) (0)] ∫ 2 1 3

More generally we have

xx11∞ n ⎛⎞n γ ()udu=−(1)log( −kp+1 u + kdu ) ∫∫p ∑∑⎜⎟ 11pn++11nk==00⎝⎠k

With the substitution y =+ log(uk ) we get

p+1 ym ∫∫logppypym+++111 (u+= k ) du y e dy =−+( 1) ( p 1)! e ∑ ( − 1) m=0 m! and therefore

p+1 (1)− m ∫ logpp++11 (ukdu+=−+ ) ( 1) ( p 1)!∑ ( uk + )log m ( uk + ) m=0 m!

Hence we obtain

x p+1 (1)− m ∞ 1 n ⎛⎞n γ ()udu=− (1)pkm+1 p ! (1)(1− +k )log( x + k ) ∫ p ∑∑∑⎜⎟ 1 mnk===000mn!1+ ⎝⎠k

p+1 m ∞ n pkm+1 (1)− 1 ⎛⎞n −−(1)p !∑∑∑⎜⎟(1)(1− +kk )log(1 + ) mnk===000mn!1+ ⎝⎠k and using (4.3.107a) this may be written as

x p+1 (1)− m γςς()udu=− p ! [()mm (0,)x m(− 1) (0,)] x ∫ p ∑ 1 m=1 m!

47 p+1 (1)− m −−pm![(0,1)(0,1)]∑ ςς()mm(− 1) m=1 m!

Hence we obtain the general result

x (1)− p+1 (4.3.231ii) γςς()udu=− [(1)pp++ (0,)x (1) (0)] ∫ p 1 p +1

We saw above that

x 1 γςς()udu=− [′′′′ (0,) x (0)] ∫ 1 1 2 and differentiation gives us

11d ∂∂2 ′′ γς1()xx== (0,)2 ς (,) sx 22dx∂∂ x s s=0

We have

∂∂2 ςς(,sx )=− s(2) ( s + 1,) x − 2 ς (1) ( s + 1,) x ∂∂xs2 and hence we get

1 (2) (1) γςς1()x =− lim[ssx (+ 1,) + 2 ( sx + 1,)] 2 s→0

We have therefore come around full circle since

lim[sςς(2) ( s++ 1,)2 x(1) ( s + 1,)]lim[(1) x = s − ςς(2) (,) sx + 2(1) (,)] sx ss→→01

∞ n 11⎛⎞n k 2 and thus, as we have already seen, γ1()x = −−+∑∑⎜⎟(1)log(uk ). 21nk==00n + ⎝⎠k 1 By considering ∫ς (,su ) du= 0 for Re ()s < 1 Coffey [45c] has shown that 0

∞ (1)− p 1 1 (4.3.231iii) (1)()sudu−=p γ ∑ ∫ p p=0 p!10 − s

48 but it should be noted that my expression for γ p (u ) is not defined at x = 0 .

We see from (4.3.214) that

∞∞ppp ∞n (1)−−−p (1)(s 1)⎧ 1 1 ⎛⎞n kp+1 ⎫ ∑∑∑∑(1)()su−=γ p ⎨ −⎜⎟(1)log( − uk + )⎬ pp==00pppn!!⎩⎭++1 n == 001k⎝⎠k

∞∞n ppp+++111 11⎛⎞n k (1)(1)log()− suk−+ =−∑∑⎜⎟(1) ∑ sn−+11nk==00⎝⎠k p = 0 (1)! p+

∞ n 11⎛⎞n k =−−−+−∑∑⎜⎟( 1)() exp[ (suk 1)log( )] 1 sn−+11nk==00⎝⎠k

11∞ n ⎛⎞n ⎡1 ⎤ =−−(1)k 1 ∑∑⎜⎟ ⎢ s−1 ⎥ sn−+11nk==00⎝⎠k ⎣ () uk + ⎦

1 =−ς (,su ) s −1

We have thus rediscovered that

∞ p 1(1)− p ςγ(,su )=+∑ (s − 1)p () u sp−1!p=0 

Coffey [45c] has shown that for integers q ≥ 2

qp−1 p+1 ⎛⎞rqpjjlog ⎛⎞p (4.3.232) ∑∑γγpp⎜⎟=− +qq(1) − +⎜⎟(1) − γ pj− log q rj==10⎝⎠qp+1 ⎝⎠j and, taking the simplest case q = 2 , we get

p+1 p ⎛⎞1lpjjog2⎛⎞p (4.3.232a) γγpp⎜⎟=− +2( − 1) + 2∑⎜⎟ ( − 1)γ pj− log 2 ⎝⎠21p + j=0 ⎝⎠j

With p = 0 we get

0 ⎛⎞1 ⎛⎞1 j j γγ00⎜⎟=− +2log 2 + 2∑⎜⎟ ( − 1)γ 0− j log 2 ⎝⎠2 j=0 ⎝⎠j

49 =+γ 2log 2 =−ψ (1/ 2) in agreement with (4.3.215).

With p =1 we see that

1 ⎛⎞1 2 ⎛⎞1 j j γγ11⎜⎟=− −log 2 + 2∑⎜⎟ ( − 1)γ 1− j log 2 ⎝⎠2 j=0 ⎝⎠j

2 =−γγγ11 −log 2 + 2 − 2 log 2 and thus we have

⎛⎞1 2 (4.3.233) γγ11⎜⎟=−log 2 − 2 γ log 2 ⎝⎠2

A different proof of this is given in (4.3.263).

Using (4.3.233) in conjunction with (4.3.226iv)

∞ ⎡log(nx++ ) log( n 1) ⎤ γγ11()x −= (1) ∑ ⎢ − ⎥ n=0 ⎣ nx++ n1 ⎦ we obtain with x =1/2

∞ 2 ⎡log(nn++ 1) log(2 1) log 2 ⎤ log 2+= 2γ log 2 ∑ ⎢ −2 +2 ⎥ n=0 ⎣ nnn+12121++⎦ and this corrects a misprint in Dilcher’s paper [54a].

We also have

1 ⎛⎞123 ⎛⎞1 j j (4.3.233a) γγ22⎜⎟=− +log 2 + 2∑⎜⎟ ( − 1)γ 2− j log 2 ⎝⎠23j=0 ⎝⎠j

2 =−γγ2 log 2 + log3 2 21 3  Integrating (4.3.226iv) gives us

u ∞ ⎡1 1 log(n + 1) ⎤ γγ()(1)xdx−− u = log()22n +− u log(1)(1) n +−− u ∫ 11∑ ⎢ ⎥ 1 n=0 ⎣22n + 1⎦

∞ ⎡⎤1 221 ⎛⎞log(nu++ ) log( n 1) log(nu + ) =+−++−−−−∑ ⎢⎥log (nu ) log ( n 1) ( u 1)⎜⎟(u 1) n=0 ⎣⎦22 ⎝⎠nu++ n1 nu + and using (4.3.226iv) this becomes

∞ ⎡⎤1 221 log(nu+ ) =+−+−−+−−∑ ⎢⎥log (nu ) log ( n 1) ( u 1) (u 1)[γ11 ( u )γ ] n=0 ⎣⎦22 nu+

Then referring to (4.3.231) we may write this as

11 ς ′′(0,uu )− ςγ ′′ (0)−− ( 1) = 22 1

∞ ⎡⎤1 221 log(nu+ ) ∑ ⎢⎥log (nu+− ) log ( n +−− 1) ( u 1) +−(u 1)[γ11 ( u ) −γ ] n=0 ⎣⎦22 nu+

We have

⎛⎞1 s ς ⎜⎟ss,(21)()=−ς ⎝⎠2

⎛⎞1 ss ςςς′′⎜⎟sss,(21)()2()log2=− + ⎝⎠2

⎛⎞1 sss2 ςςςς′′⎜⎟ssss,=− (21)()2.2()log22()log2 ′′ + ′ + ⎝⎠2 and with s = 0 we have

⎛⎞1 2 ςςς′′⎜⎟0,=+ 2 ′ (0)log 2 (0)log 2 ⎝⎠2

1 =−log(2π ) log 2− log2 2 2

Then with u = 1/2 and using (4.3.233) we obtain

1111 −−−+=log(2π ) log 2 log2 2ςγ′′ (0) 24221

51 ∞ ⎡⎤1122⎛⎞11 1⎛⎞ 12 ∑ ⎢⎥log ⎜⎟nn+−log ( ++ 1) log ⎜⎟ n + +log 2 +γ log 2 n=0 ⎣⎦22212⎝⎠22n + ⎝⎠

We note from (4.3.247) that

111 ς ′′(0) =+γγ22 − π −log 2 (2 π ) 1 2242 and we obtain (on the assumption that integrating (4.3.226iv) term by term is valid)

1311 log(2ππ )log( / 2)− log22 2−−+= γ log 2 γ π2 44248

∞ ⎡⎤1122⎛⎞11 1⎛⎞ ∑ ⎢⎥log ⎜⎟nn+−log ( ++ 1) log⎜⎟ n + n=0 ⎣⎦2221⎝⎠22n + ⎝⎠



Adamchik [2a], using Rademacher’s formula [14, p.261] for the Hurwitz zeta function, showed that

(4.3.233b)

⎛⎞⎛p ppj ⎞ ⎛⎞ππq−1 ⎛⎞⎛2 jp ⎞ ςς′′⎜⎟⎜1,−−= 1,1 ⎟ ππγπ [log(2q ) + ]cot⎜⎟ − 2∑ log Γ ⎜⎟⎜ sin ⎟ ⎝⎠⎝qq ⎠ ⎝⎠qj=1 ⎝⎠⎝ qq ⎠ where p and q are positive integers and p < q . In particular we have

⎛⎞13 ⎛⎞ ⎡ ⎛⎞1⎤ (4.3.233c) ςςπγπ′′⎜⎟1,−=++−Γ ⎜⎟ 1,⎢ 4log 2 3log 4log ⎜⎟⎥ ⎝⎠44 ⎝⎠⎣ ⎝⎠4⎦

⎛⎞12 ⎛⎞π ⎡ ⎛⎞1⎤ (4.3.233d) ςς′′⎜⎟1,−= ⎜⎟ 1,⎢ 2 γ −+−Γ log3 8log(2 π ) 12log ⎜⎟⎥ ⎝⎠33 ⎝⎠23⎣ ⎝⎠3⎦

We have from (4.3.206) that

∞ p 1(1)− p ςγ(,su )=+∑ p ()(u s − 1) sp−1!p=0 and therefore

52 ∞ p (1)− p ςς(,su )−−= (,1 s u )∑ [γγpp ()u −−− (1 u )]( s 1) p=0 p!

Differentiation with respect to s results in

∞ (1)− p ′′ p−1 (4.3.233di) ςς(,su )−−= (,1 s u )∑ p [γγpp () u −−− (1 u )]( s 1) p=0 p! and in the limit as s →1 we have

(4.3.233e) ς ′′(1,uuuu )−−=−−−ςγγ (1,1 ) [11 ( ) (1 )]

In fact, as seen in (4.3.228b), more generally we have

p p ∂ γγpp()x −=− ()ys lim(1) [(,)ςx − ς (,sy )] s→1 ∂s p

Hence we obtain ⎛⎞13 ⎛⎞ ⎡ ⎛⎞1⎤ (4.3.233f) γγ11⎜⎟−=−++−Γ ⎜⎟ πγ⎢ 4log 2 3log π 4log ⎜⎟⎥ ⎝⎠44 ⎝⎠⎣ ⎝⎠4⎦ and equivalently

⎛⎞31 ⎛⎞ ⎡ ⎛⎞1⎤ (4.3.233g) ςςπγπ′′⎜⎟1,−=++−Γ ⎜⎟ 1,⎢ 4log 2 3log 4log ⎜⎟⎥ ⎝⎠44 ⎝⎠⎣ ⎝⎠4⎦

As mentioned above in (4.3.232), Coffey [45c] has shown that for integers q ≥ 2

qp−1 p+1 ⎛⎞rqpjjlog ⎛⎞p ∑∑γγpp⎜⎟=− +qq(1) − +⎜⎟(1) − γ pj− log q rj==10⎝⎠qp+1 ⎝⎠j and from (4.3.233) we saw that

⎛⎞1 2 γγ11⎜⎟=−log 2 − 2 γ log 2 ⎝⎠2

We see that

31 ⎛⎞r 2 ⎛⎞1 j j ∑∑γγ11⎜⎟=− −4log 4 + 4⎜⎟ ( − 1)γ 1− j log 4 rj==10⎝⎠4 ⎝⎠j

53 2 =−316log28log2γγ1 − and we also have

3 ⎛⎞r ⎛⎞113 ⎛⎞ ⎛⎞ ∑γγγγ1111⎜⎟=++ ⎜⎟ ⎜⎟ ⎜⎟ r=1 ⎝⎠4424 ⎝⎠ ⎝⎠ ⎝⎠

⎛⎞13 ⎛⎞ 2 =++−−γγγ111⎜⎟ ⎜⎟ log 2 2 γ log 2 ⎝⎠44 ⎝⎠

Therefore we get

⎛⎞13 ⎛⎞ 2 γγ11⎜⎟+=−− ⎜⎟215log26log2 γ 1 γ ⎝⎠44 ⎝⎠ and using (4.3.233a) we easily see that

(4.3.233h)

⎛⎞11 2 1⎡ ⎛⎞1⎤ γγ11⎜⎟=−[2 15log 2 − 6 γπγπ log 2] −++−Γ⎢ 4log 2 3log 4log ⎜⎟⎥ ⎝⎠42 2⎣ ⎝⎠4⎦

⎛⎞31 2 1⎡ ⎛⎞1⎤ γγ11⎜⎟=−[2 15log 2 − 6 γπγπ log 2] +++⎢ 4log 2 3log −Γ 4log ⎜⎟⎥ ⎝⎠42 2⎣ ⎝⎠4⎦

⎛⎞1 Assuming that each of the terms π ,,log2,logγπ and log Γ⎜⎟ has weight equal to one ⎝⎠4 and γ1 has weight 2, then the above represents a homogenous equation of weight 2.

⎛⎞11 An alternative proof is contained in (4.3.268). Since ψ ⎜⎟=−γπ − −3log2 we may ⎝⎠42 write this as

⎛⎞111 ⎛⎞ 22 1⎡ ⎛⎞1⎤ γγψγγ11⎜⎟=++−− ⎜⎟ [2 15log 2] π⎢ 4log 2 +−Γ 3log π 4log ⎜⎟⎥ ⎝⎠442 ⎝⎠ 2⎣ ⎝⎠4⎦ or

⎛⎞11 ⎛⎞ 22 151⎡ ⎛⎞1⎤ γγγγγ11⎜⎟=− ⎜⎟ + + −log 2 − π⎢ 4log 2 + 3log π − 4log Γ⎜⎟⎥ ⎝⎠44 ⎝⎠ 22⎣ ⎝⎠4⎦

54 Similarly we have

12⎡⎤ 1 2 ⎛⎞ ⎛⎞′′ ⎛ ⎞ ⎛ ⎞ γγ11⎜⎟−=−− ⎜⎟⎢⎥ ς ⎜1, ⎟ ς ⎜ 1, ⎟ ⎝⎠33 ⎝⎠⎣⎦ ⎝ 3 ⎠ ⎝ 3 ⎠ and using (4.3.233d) this becomes

π ⎡ ⎛⎞1 ⎤ =−⎢2γπ − log3 + 8log(2 ) − 12log Γ⎜⎟⎥ 23⎣ ⎝⎠3 ⎦

Equation (4.3.232) gives us

⎛⎞12 ⎛⎞ 32 γγ11⎜⎟+=−− ⎜⎟ 23log3log3 γγ 1 ⎝⎠33 ⎝⎠ 2 and this then gives us

⎛⎞13 32 π ⎡ ⎛⎞1⎤ (4.3.233i) γγγ11⎜⎟=−log3 − log 3 −⎢ 2γ − log3 + 8log(2 π ) − 12log Γ⎜⎟⎥ ⎝⎠32 4 43⎣ ⎝⎠3⎦

⎛⎞2 with a similar expression for γ1 ⎜⎟. ⎝⎠3

In [54a] Dilcher defined a generalised polygamma function by

logppx ∞ ⎡ log (nx+ ) log p n⎤ ψ pp()x =−γ − −∑ ⎢ − ⎥ xnxnn=1 ⎣ + ⎦

logppxnxnnn∞ ⎡ log (++ ) log p ( 1) log pp log ( + 1) ⎤ =−γ p − −∑ ⎢ − − + ⎥ xnxnnnn=1 ⎣ ++11 +⎦

logppxnxn∞ ⎡ log (++ ) log p ( 1) ⎤ =−γ p − −∑ ⎢ − ⎥ xnxnn=1 ⎣ ++1 ⎦

Hence, comparing this with (4.3.228e), we see that

(4.3.233j) ψ pp (x )=−γ (x )

This relationship was also noted by Coffey in [45c].

Dilcher [54a] also showed that

⎛⎞11⎡⎤π 33 ⎡ 2 1⎛⎞ 1⎤ ψ11⎜⎟=−γγππ +⎢⎥3log3 + + log 3 + 3⎢ log(2 )− log3 − log Γ⎜⎟⎥ ⎝⎠32⎣⎦3 4⎣ 3 123⎝⎠⎦ and a little algebra shows that this is the same as (4.3.233i) above. It was also shown by Dilcher [54a] that

k +1 k ⎛⎞11 ⎛⎞ kjj+1 log2 ⎛⎞k ψ kk⎜⎟=−γγγ ⎜⎟ =(1)2 − − k −2∑⎜⎟ (1) − kj− log2 ⎝⎠22 ⎝⎠ k + 1 j=0 ⎝⎠j which is the same as Coffey’s formula (4.3.232a).

With s = 2 in (4.3.233di) we obtain

∞ (1)− p ′′ (4.3.233k) ςς(2,uu )−−= (2,1 ) ∑ puu[γγpp ( ) −− (1 )] p=0 p! We see that

∞∞p p ∞n (1)− puγ p () (1)− p ⎧ 1 1 ⎛⎞n kp+1 ⎫ ∑∑=−⎨ ∑∑⎜⎟(1)log( −+uk )⎬ pp==00pppn!!11⎩⎭++ nk == 00⎝⎠k

∞∞n pp++11 1(⎛⎞n k p −1)log(uk+ ) =−∑∑⎜⎟(1) ∑ nk==00npp++1!⎝⎠k p = 0 (1)

We see that

∞ (1)− pp++11x ∑ =−e−x 1 p=0 (1)!p + and also that

∞ pp++11 −x (1)−−px2 d⎛⎞ e 1 ∑ = x ⎜⎟ p=0 (1)!pdxx+ ⎝⎠

We therefore have

∞ (1)− pp++11px ∑ =−1(x + 1)e−x p=0 (1)!p + and hence we get

56 ∞∞p+1 n (1)− puγ p () 1⎛⎞n k ⎧ [1++ log(uk )]⎫ ∑∑∑=−−⎜⎟(1)1⎨ ⎬ pn===000pn!1++k⎝⎠k ⎩⎭ uk

∞∞nnk ∞ n 1 ⎛⎞nnkk1⎛⎞ (− 1) 1 ⎛⎞ nlog(uk+ ) =−∑∑⎜⎟(1)− ∑∑⎜⎟ −− ∑∑⎜⎟(1) nk==00nnuknuk+++++11⎝⎠kk nk == 00⎝⎠ nk == 00 1⎝⎠ k

∞ n 1 ⎛⎞n k log(uk+ ) =−1(2,)ς u −∑∑⎜⎟(1)− nk==00nu++1 ⎝⎠k k

Using (4.3.107a)

∞ 1 n ⎛⎞n log(uk+ ) (1)(,)(,)ssusu′ (1)k −+=−−ςς ∑∑⎜⎟ s−1 nk==00nu++1()⎝⎠k k we see that

∞ n 1 ⎛⎞n k log(uk+ ) ςς′(2,uu )+=− (2, ) ∑∑⎜⎟( − 1) nk==00nu++1 ⎝⎠k k

This gives us

∞ (1)− p+1 puγ () ∑ p =+1(2,)ς ′ u p=0 p! and this verifies (4.3.233k).

Similarly, with s = 3 in (4.3.233di) we obtain

∞ (1)2− pp ′′ ςς(3,uu )−−= (3,1 ) ∑ puu[γγpp ( ) −− (1 )] p=0 p!

Differentiating (4.3.217a)

∞ p (1)− p 1 ∑ tuγςp ()= ( tu+− 1,) p=0 p! t we see that

∞ (1)−∂p 1 ptp−1 () u ( t 1,) u ∑ γςp = ++2 p=0 p! ∂tt

57 and with t =1 we obtain the same result as before.



We recall from (4.3.210) that

∞ n 11⎛⎞n k 2 γ1()uk=−∑∑⎜⎟(1)log( − +u ) 21nk==00n + ⎝⎠k and therefore

∞ n ⎛⎞111⎛⎞n k 2 (21)k + γ1 ⎜⎟=− ∑∑⎜⎟(1)log− ⎝⎠22nk==00n + 1⎝⎠k 2

∞ n 11⎛⎞n k 22 =−∑∑⎜⎟( − 1) [log (2kk + 1) − 2log 2log(2 + 1) + log 2] 21nk==00n + ⎝⎠k

∞∞nn 11⎛⎞nnkk2 1⎛⎞ 12 =− ∑∑⎜⎟(− 1) log (2kk + 1) + log 2∑∑⎜⎟(− 1) log(2 + 1) − log 2 21nk==00nn++⎝⎠kk nk == 001⎝⎠ 2

From (4.3.74a) we have previously seen that

∞ n 1 ⎛⎞n k ∑∑⎜⎟(−+=−− 1) log(2k 1)γ log 2 nk==00n +1 ⎝⎠k and accordingly we obtain

∞ n ⎛⎞111⎛⎞n k 223 γγ1 ⎜⎟=− ∑∑⎜⎟(− 1) log (2k + 1) − log 2 − log 2 ⎝⎠22nk==00n + 1⎝⎠k 2

Hence we have using (4.3.233)

∞ n 11⎛⎞n k 221 (4.3.233l) ∑∑⎜⎟(−+=+− 1) log (2k 1)γγ1 log 2 log 2 21nk==00n + ⎝⎠k 2



Using (4.3.212) we see that

uu11∞ n ⎛⎞n γ ()x dx =− (1)log( −k 2 x + k ) dx ∫∫1 ∑∑⎜⎟ 0021nk==00n + ⎝⎠k

u u log22 (x +=+kdxxkxkxkxkxk ) ( )log ( +−+ ) 2( )log( +++ ) 2( ) ∫ 0 0

=+()log()2()log()2()uk2 uk +− uk + uk ++ uk +

−+klog2 ( kkkk ) 2 log( ) − 2 and we shall abbreviate this as I (uI )− (0) . Therefore we have

u 11∞ n ⎛⎞n γ ()xdx=− (1)[() −k Iu − I (0)] ∫ 1 ∑∑⎜⎟ 0 21nk==00n + ⎝⎠k where

Iu( )=+ ( u k ){ [log( u +−+ k ) 1]2 1}

With reference to

∞ 1(1)n ⎛⎞n − k (1)(,)ssu −=ς ∑∑⎜⎟ s−1 nk==00nuk++1()⎝⎠k we see that

∞∞11nn⎛⎞nn∂−⎛⎞ (1)k (−+ 1)k (uk )log( uk +=− ) lim ∑∑⎜⎟ s→0 ∑∑⎜⎟ s−1 nk==00ns+∂11⎝⎠kk nk == 00n+⎝⎠(u+k)

∂ =−lim [(ssu − 1)ς ( , )] s→0 ∂s

= ς ′ (0,uu )−ς (0, ) which we have in fact already seen from (4.3.108). Then using Lerch’s identity (4.3.116) we see that

∞ n 11⎛⎞n k (4.3.233m) ∑∑⎜⎟(−+ 1) (uk )log( uk +=Γ− ) log ( u ) log(2πς ) − (0,u ) nk==00n +12⎝⎠k

We note that it is not possible to substitute u = 0 in the above equation. With u =1 we get

59 ∞ n 11⎛⎞n k 1 ∑∑⎜⎟(−+ 1) (kk 1)log( +=− 1) log(2π ) nk==00n +12⎝⎠k 2 which we have seen before in (4.3.115).



We see from (4.3.110) that

∞ n 1 ⎛⎞n k ς (0,uu ) =−∑∑⎜⎟( − 1) ( +k ) nk==00n +1 ⎝⎠k and therefore using (4.3.108) we have

∞ n 1 ⎛⎞n k ς ′(0,uu ) =−++−∑∑⎜⎟( 1) (k )[log(uk ) 1] nk==00n +1 ⎝⎠k

Furthermore we have

∞∞11n ⎛⎞nn∂−2 n ⎛⎞ (1)k (4.3.233n) (1)(−+k uk )log(2 uk += ) lim ∑∑⎜⎟ s→0 21∑∑⎜⎟ s− nk==00ns+∂11⎝⎠kk nk== 00n+⎝⎠(u+k)

∂2 =−lim [(ssu 1)ς ( , )] s→0 ∂s2

= −+ς ′′′ (0,uu ) 2ς (0, )

= −+Γ−ς ′′(0,uu ) 2log ( ) log(2π )

We now differentiate (4.3.233n) to obtain

∞∞nn 11⎛⎞nnkk⎛⎞ 2 d 2∑∑⎜⎟(−++ 1) log(uk ) ∑∑⎜⎟( − 1) log (uk +=−+ )ςψ′′ (0, u ) 2 ( u ) nk==00nnd++11⎝⎠kk nk == 00⎝⎠ u

We therefore see that

d ∞ 1 n ⎛⎞n ′′ k 2 ςγ(0,uu ) =−∑∑⎜⎟(1)log( − +k ) = 21 ()u du nk==00n +1 ⎝⎠k

It is readily seen that

60 ∂∂22 ∂ ∂ −=ςς(,su )− (,su ) ∂∂us22 ∂ s ∂ u

∂2 =−[(1,)] −ssς + u ∂s2

=+++ssς ′′ ( 1, u ) 2ς ′ ( s 1, u )

We also have

∞ n 1 ⎛⎞n k 2 ∑∑⎜⎟(1)(−+uk )log( uk += ) nk==00n +1 ⎝⎠k

∞ n 1 ⎛⎞n k −+ςς′′ (0,uu ) 2∑∑⎜⎟ ( −+++ 1) (k )log(uk ) 2 (0,u ) nk==00n +1 ⎝⎠k and therefore we get

∞ n 1 ⎛⎞n k 2 ς ′′(0,uu ) =−∑∑⎜⎟( − 1) ( +k )[log(u +k ) − 1] nk==00n +1 ⎝⎠k

We also note for reference from Apostol’s paper [14aa] that

111 (4.3.234) ς ′′(0) =+γγ22 − π −log 2 (2 π ) 1 2242 and we may write this as

(4.3.234i)

2 ∞∞nn 11⎛⎞nnkk2 1⎡ 1 ⎛⎞ ⎤ ς ′′(0)=− ∑∑⎜⎟(1)log(1)− +kk + ⎢∑∑⎜⎟(1)log(1)− + ⎥ 21nk==00nn++⎝⎠kk2⎣ nk == 00 1⎝⎠ ⎦

2 ∞ n 11⎡⎤⎛⎞n k −−ς (2)2⎢⎥⎜⎟(1)(1)[log(1)1] −+++kk 41∑∑n + k ⎣⎦nk==00⎝⎠

Upon differentiating (4.3.107a) we obtain

∞ 1n ⎛⎞n (− 1)k log2 (k + 1) (sss 1)()2()′′ ′ −+=ςς∑∑⎜⎟ s−1 nk==00nk++1(1)⎝⎠k

61 and as s → 0 we get

∞ n 1 ⎛⎞n k 2 2ςς′′′ (0)−= (0) ⎜⎟( −+ 1) (kk 1)log ( + 1) ∑∑k nk==00n +1 ⎝⎠

∞∞nn 11⎛⎞nnkk22⎛⎞ =−+∑∑⎜⎟(1)log(kk 1)+−+ ∑∑⎜⎟(1)log(k 1) nk==00nn++11⎝⎠kknk == 00⎝⎠ and we therefore see that

∞ n ∞ n 11⎛⎞nnk 2 ⎛⎞ k 2 ∑∑⎜⎟(−+=−−−+ 1)kk log ( 1) 2ςς′′′ (0) (0) ∑∑⎜⎟( 1) log (k 1) nk==00nn++11⎝⎠kknk == 00⎝⎠

= 2(0)ς ′ −+ςγ′′ (0)21 and using (4.3.234) we then obtain

∞ n 11⎛⎞n k 221212 ∑∑⎜⎟(−+=−−++ 1)kk log ( 1)γ1 log(2πγ ) πlog (2 π ) nk==00n +12⎝⎠k 242

In Coffey’s recent paper [45c], “New results on the Stieltjes constants: Asymptotic and exact evaluation”, it was shown that the Stieltjes constant could be written in the form

11∞ 1 n ⎛⎞n log(k + 1) (4.3.235) log 2 (1)k γγ0 == − ∑∑n+1 ⎜⎟− 2log2nk==0121⎝⎠k k +

Since log1= 0 we may also write this as

11∞ 1 n ⎛⎞n log(k + 1) log 2 (1)k γγ0 == − ∑∑n+1 ⎜⎟− 2log2nk==0021⎝⎠k k + and using (c.61) in Volume VI

∞ 1n ⎛⎞n (−+ 1)k log(k 1) ⎡ log 2⎤ =−ςγ′ (1) =− log 2 − ∑∑n+1 ⎜⎟ a ⎢ ⎥ nk==0021⎝⎠k k + ⎣ 2⎦ it is seen that Coffey’s result becomes evident. Coffey also reported that

(4.3.236)

1 1∞∞ 1 nn⎛⎞nnlog(kk+ 1) 1 1 ⎛⎞ log2 (+ 1) =−log2 2 ( − 1)kk + (− 1) γ1 ∑∑nn++11⎜⎟ ∑∑⎜⎟ 12 2nk==01 2 ⎝⎠kkkk++1 2log 2 nk == 01 2 ⎝⎠ 1 and, using (C.61), this may be written as

112 ⎡⎤ log21 γγς1 =+log 2 log 2 −+a′′(1) 12 2⎣⎦⎢⎥ 2 2log 2 which, in turn, is equivalent to (see also Dilcher’s result (4.3.226iv))

1 (4.3.237) ςγ′′ (1)=−+ 2 log 2 γ log22 2 log 2 a 1 3

The above is in fact a particular case of the general formula reported by Briggs and Chowla [35a] in 1955 where they showed that (see also Dilcher’s result (4.3.226iv))

k +1 rr+1 ()k (1)log2− (4.3.238) ς akr(1)= kA !∑ − r=1 r!

(1)− n where A = γ and A =1. nnn! −1

In [45e] Coffey extended his results to show that

mnB logml− 2∞ 1 ⎛⎞n logl (k + 1) γ m!(ml−+1 1)k m =−∑∑∑n+1 ⎜⎟ − ln===111(1)!!2ml−+ l k⎝⎠k k +1 (4.3.239)

∞ n m+1 11⎛⎞n km log(1)k + Bm+1 +1 −−−∑∑n+1 ⎜⎟( 1) log 2 (1)log22mk+++nk==11⎝⎠k 1m 1 or alternatively

(4.3.239i)

m ml− BBml−+1 log 2 ll() 1 m++1 ( m 1) m+1 m +1 γςςmaa=−m!∑ (− 1) (1) − (− 1) (1) − log 2 l=1 (1)!!ml−+ l (1)log2m+ m+1



Differentiating the Hasse formula (3.12) we obtain

63 1∞∞ 1nn⎛⎞nn (1)log(−+kkk 1) 1 1⎛⎞ (1)− ς ′()s =− − ∑∑⎜⎟ s−12∑∑⎜⎟ s−1 sn−+1nk==00 1⎝⎠kk ( k + 1) ( s − 1) nk ==00 n + 1⎝⎠ ( k + 1)

1∞ 1n ⎛⎞n (−+ 1)k log(k 1) 1 (4.3.239ii) =− − ς ()s ∑∑⎜⎟ s−1 sn−+11nk==00⎝⎠k (1)1 k + s − and with s = 0 we have using (F.6)

∞ n 11⎛⎞n k 1 ςπ′(0)=− log(2 ) =∑∑⎜⎟( − 1) (kk + 1)log( + 1) − 21nk==00n + ⎝⎠k 2

Hence we see that

∞ n 11⎛⎞n k 1 ∑∑⎜⎟(−+ 1) (kk 1)log( +=− 1) log(2π ) nk==00n +12⎝⎠k 2 and therefore we see that

∞ n 11⎛⎞n k 1 (4.3.240) ∑∑⎜⎟(−+=+− 1)kk log( 1)γ log(2π ) nk==00n +12⎝⎠k 2

With s = 2 we have

∞ 1n ⎛⎞n (−+ 1)k log(k 1) ς ′(2) =−∑∑⎜⎟ −ς (2) nk==00nk++11⎝⎠k and using

∞ n 11⎛⎞n kp+1 γ p ()uu=−∑∑⎜⎟(1)log( − +k ) pn++11nk==00⎝⎠k we see that

∞ 1ln ⎛⎞n ogp ()uk+ ′ k (4.3.241) γ p ()u =−∑∑⎜⎟(1) − nk==00nu++1 ⎝⎠k k and

∞ 1ln ⎛⎞n ogp (1)+ k ′ k (4.3.242) γ p (1) =−∑∑⎜⎟( − 1) nk==00nk++11⎝⎠k

Hence we have shown that

(4.3.243) ς ′′(2)=−γς1 (1) (2) which we have already seen in a more general form in (4.3.223b). Using (F.7)

11 ςγπς′′(1)−= (1 −− log2) + (2) 12 2π 2 we see that

ς ′′(2)=−−−− 2πς2 ( 1) ς (2)(1 γ log2 π ) and hence we have

2 (4.3.244) γ1′′(1)=−++ 2πς ( 1) ς (2)( γ log 2 π )

With sm=−2 in (4.3.239ii) we have

∞ n 11⎛⎞n km21+ 1 ςς′(2−=mk ) ∑∑⎜⎟(1)( − + 1) log(k ++ 1) (2 −m ) 21mn++nk==00 1⎝⎠k 21m +

∞ n 11⎛⎞n km21+ =−++∑∑⎜⎟( 1) (kk 1) log( 1) 21mn++nk==00 1⎝⎠k

Therefore, using (F.8a)

(2m )! ςς′(−=− 2mm ) ( 1)m (2 + 1) 2(2π )2m we obtain (which we also saw in Volume II(a))

2m ∞ n mkm2(2π ) 1 ⎛⎞n 21+ (4.3.245) ς (2mk+=− 1) ( 1) ∑∑⎜⎟( − 1) ( + 1) log(k + 1) (2mn++ 1)!nk==00 1 ⎝⎠k and in particular with m = 1 we have

2 ∞ n 41π ⎛⎞n k 3 (4.3.246) ς (3) =−∑∑⎜⎟( − 1) (kk + 1) log( + 1) 31nk==00n + ⎝⎠k

65 ∞ n 1 ⎛⎞n k 3 We may therefore deduce a formula for ∑∑⎜⎟(− 1)kk log(+ 1) involving ς (3) . nk==00n +1 ⎝⎠k

From (4.3.107a) we have

∞ k 1 ⎛⎞k j 3 3ςς′ (−−−= 2,x ) ( 2,xx ) ∑∑⎜⎟( − 1) ( +j ) log(x +j ) kj==00k +1 ⎝⎠j and letting x =1 this becomes

∞ k 1 ⎛⎞k j 3 3ς ′ (−= 2) ∑∑⎜⎟( − 1) (1 +j ) log(1 +j ) kj==00k +1 ⎝⎠j

ς (3) From (F.8b) we have ς ′(2)−=− and hence we recover (4.3.246). 4π 2

With sm=−12 in (4.3.239ii) we have

∞ n 11⎛⎞n km2 1 ςς′(1−= 2mk ) ∑∑⎜⎟( − 1) ( + 1) log(km ++ 1) (1 − 2 ) 21mnnk==00+ ⎝⎠k 2m

B and using (F.12b)ς (1−=− 2m ) 2m this becomes 2m

11∞ n ⎛⎞n B ς ′(1 2mk ) ( 1)km ( 1)2 log(k 1) 2m −=∑∑⎜⎟ − + +−2 21mnnk==00+ ⎝⎠k 4m

In this regard, we also recall (F.8)

ς ′(2m ) 2(12)mmHςγπ′ −−⎡⎤(1) −− log2 B = B ⎣⎦21mmm− 2ς (2m ) 2 and we therefore obtain

∞ 1(n ⎛⎞n ς ′ 2m)B km2(⎡⎤1) 2m ∑∑⎜⎟(−+ 1) (kkH 1) log( +=−− 1) ⎣⎦21mmm− γπlog 2 B2 + B2 − nk==00nmm+1(⎝⎠k ς 2)2

A further differentiation results in

1∞∞ 1nn⎛⎞nn (− 1)kk log2 (kk+− 1) 1 1⎛⎞ ( 1) log(+ 1) ς ′′()s =+ ∑∑⎜⎟ ss−−12∑∑⎜⎟ 1 sn−+1nk==00 1⎝⎠kk ( k + 1) ( s − 1) nk == 00 n + 1⎝⎠ ( k + 1)

11 −+ς ′()ssς () ss−−1(1)2

11∞ n ⎛⎞n (1)log(1)2−+k 2 k =−ς ′()s ∑∑⎜⎟ s−1 sn−+11nk==00⎝⎠k (1)1 k + s − and thus we get

(4.3.246a) ∞ n 1 ⎛⎞n k 2 ς ′′(0) =−∑∑⎜⎟( − 1) (kk + 1)log ( + 1) − log(2π ) nk==00n +1 ⎝⎠k

∞ n 1 ⎛⎞n k 2 =−∑∑⎜⎟( − 1)kk log ( + 1) + 2γ1 − log(2π ) nk==00n +1 ⎝⎠k

We also note from Apostol’s paper [14aa] that

111 (4.3.247) ς ′′(0) =+γγ22 − π −log 2 (2 π ) 1 2242 and we then obtain

∞ n 11⎛⎞n k 221212 (4.3.248) ∑∑⎜⎟(−+=−−++ 1)kk log ( 1)γ1 log(2πγ ) πlog (2 π ) nk==00n +12⎝⎠k 242

With s = 2 we have

∞ 1n ⎛⎞n (−+ 1)k log2 (k 1) ςς′′(2) =−∑∑⎜⎟ 2 ′ (2) nk==00nk++11⎝⎠k

=−γ 2′′(1) − 2ς (2) and therefore we have

(4.3.249) ς ′′(2)=−γγς21 ′ (1) − 2 ′ (1) + 2 (2)

A CONNECTION WITH LOGARITHMIC INTEGRALS

Using the integral definition of the gamma function for s > 0

67 ∞ Γ=()sexdt∫ −−xs1 0 and making the substitution x = ut (where u > 0) we obtain for us, > 0

∞ Γ()s et−−ut s 1 dt= ∫ s 0 u and using the binomial theorem we get for n ≥ 0

∞ k n n ⎛⎞n (1)− (4.3.250) eetdts−−−ut1(−=Γ t s 1 ) ∫ () ∑⎜⎟ s 0 k =0 ⎝⎠k ()ku+

Making the summation gives us

∞∞∞ ∞∞1 nn ∑∑∫∫∫eetdtedxeetdt−−−ut()11−= ts1( −+−−− nx1) ut() − ts1 nn==00n +1 000 and this may be expressed as

∞∞ ∞∞ ∞∞n n −+(1)nx − ut − t s −1 − utx −⎡⎤ − x − t s −1 ∑∑∫∫edxeetdteeeetdtdx()11−= ∫∫⎣⎦() − nn==0000 00

∞∞ eet−−−ut x s 1 = dt dx ∫∫ −−xt 001[1]−−ee

We have ∞ et−x 1 ∞ dx =−−=log(1e−−xt [1 e ]) ∫ −−x tt − −t 0 1[1]−−ee 1 − e 0 1−e and therefore we see that

∞∞ eet−−−ut x s 1(∞ et−ut s ∞ e−−u1) t t s dt dx== dt dt ∫∫ −−xt ∫−t ∫ t 001[1]−−ee0 1 − e0 e − 1

We note from [126, p.92] that for Re (s ) > 0 and Re (u ) > 0

∞ et−−(1)uts (4.3.251) dt=Γ(1)(1,) s +ς s + u ∫ t 0 e −1

and with u =1 we obtain the familiar formula

∞ t s−1 (4.3.252) dtss=Γ()(,1)ς =Γ ()() ssς ∫ t 0 e −1

We therefore have

∞ et−−(1)uts ∞ 1(1)n ⎛⎞n − k (4.3.253) dt=Γ() s ∫ ts∑∑⎜⎟ 0 enk−++11()nk==00⎝⎠k u

This gives us another proof of the Hasse identity for Re ()s > 0

11∞ n ⎛⎞n (1)− k (1,)su ς +=∑∑⎜⎟ s snnk==00++1()⎝⎠k ku

Integrating (4.3.251) with respect to u we see that

x ∞∞et−−(1)uts[1− e−−(1) xts ] t − 1 du dt ==Γ−()[()sςς s (, s x )] ∫∫tt ∫ 10ee−−110

We also note that

∞ et−−(1)uts ∂ dt=Γ( s + 1)ςς ( s + 1, u ) =Γ ( s ) s ( s + 1, u ) =−Γ ( s ) ς ( s , u ) ∫ t 0 eu−∂1

∞ 1(1)n ⎛⎞n − k ()s =Γ ∑∑⎜⎟ s nk==00nku++1()⎝⎠k

We have

∞ et−−(1)uts ∞ 1(1)n ⎛⎞n − k dt = ∫ ts∑∑⎜⎟ 0 Γ−()[se 1]nk==00 n + 1⎝⎠k ( k + u )

and differentiating this with respect to s results in

∞ et−−(1)uts[log ts−−+ψ ( )]∞ 1n ⎛⎞n ( 1)k log(ku ) dt =− ∫ ts∑∑⎜⎟ 0 Γ−()[se 1]nk==00n + 1⎝⎠k (k + u )

Hence we have

(4.3.254)

∞ ett−−(1)utslog ∞ 1n ⎛⎞n (−+ 1)k log( ku ) dt=Γ++−Γψς() s ( s 1)( s 1,) u () s ∫ t ∑∑⎜⎟ s 0 en−++11nk==00⎝⎠k (ku)

With u =1 in (4.3.254) we get

∞ tts log ∞ 1n ⎛⎞n (− 1)k log( k+ 1) t dt=Γ++−Γψς() s ( s 1)( s 1) () s ⎜⎟ s ∫ en−+11∑∑k (k+1) 0 nk==00⎝⎠ which is equivalent to

∞ tts log d dt=Γ++[( s 1)(ς s 1)] ∫ t 0 eds−1

With s =1 in (4.3.254) we get

∞ ett−−(1)utlog ∞ 1n ⎛⎞n (− 1)k log( ku+ ) t dt=−γς (2, u ) − ⎜⎟ ∫ enk−++11∑∑k u 0 nk==00⎝⎠ and using (4.3.70) this becomes

∞ ett−−(1)utlog ∞ 1n ⎛⎞n (− 1)k log( ku+ ) ′ t dt=−γψ () u − ⎜⎟ ∫ enk−++11∑∑k u 0 nk==00⎝⎠

Reference to (4.3.210) then shows that

∞ ett−−(1)utlog (4.3.255) dt=−γγψγγγ′′′′() u () u =+ () u () u ∫ et −1 110 0

Integrating this with respect to u we get

x ∞∞ett−−(1)utlog [1− e−−(1) xt ]log t du dt ==dtγ () x−γγψγ− () x −2 ∫∫eett−−11 ∫ 11 10 0

The substitution y = e−t gives us

1 1− y x−1 (4.3.256) log log(1/ydy )= γ ( x )−−γγψγ ( x ) −2 ∫ 11 0 1− y

and with x =1/2 this becomes

(4.3.257)

1111/− yy1/ ⎛⎞11 ⎛⎞ log log(1/ydy ) = −=log log(1/ydy ) γ −γγψ−− γ2 ∫∫11⎜⎟ ⎜⎟ 0012− y 1+ y ⎝⎠ ⎝⎠2

We now recall Adamchik’s result (C.57) in Volume VI (see reference [2a] and also Roach’s paper [110b])

(4.3.258)

1 x p−1 ⎛⎞1γ ++ log(2np ) ⎡ ⎛ ⎞ ⎛ np ⎞⎤⎡1 ⎛ p ⎞ ⎛ np + ⎞⎤ log log dx =−+−ψψ ς′′1, ς 1, ∫ n ⎜⎟ ⎢ ⎜ ⎟ ⎜ ⎟⎥⎢ ⎜ ⎟ ⎜ ⎟⎥ 0 12+ xx⎝⎠ n⎣ ⎝2 nnnnn ⎠ ⎝2 ⎠⎦⎣2 ⎝2 ⎠ ⎝2 ⎠⎦ where, notwithstanding the suggestive notation, neither p nor n need be integers.

With p = n we get

1 xnn−1 ⎛⎞1γ + log(2 )⎡ ⎛⎞ 1 ⎤⎡1⎛ 1 ⎞ ⎤ (4.3.259) log log dx =−ψψ1+− ς′′ 1, ς 1,1 ∫ n ⎜⎟ ⎢ ⎜⎟ ()⎥⎢⎜ ⎟ () ⎥ 0 12+ xx⎝⎠ n⎣ ⎝⎠2⎦⎣2n ⎝2 ⎠ ⎦

From (4.3.228c), and also from Adamchik’s paper [2a], we see that

⎛⎞1 2 (4.3.260) ςς′′⎜⎟1,−=+() 1,1 log 2 2 γ log 2 ⎝⎠2

⎛⎞1 and hence, using ψ ⎜⎟=−γ −2log2, we have ⎝⎠2

1 xn−1 ⎛⎞11 (4.3.261) log log dx =−⎡log2 2 + 2log 2log n⎤ ∫ n ⎜⎟ ⎣ ⎦ 0 12+ xxn⎝⎠

With pn==1/2 we get

1 1/ x ⎛⎞1 2 (4.3.262) ∫ log log⎜⎟dx = log 2 0 1+ x ⎝⎠x

We then see from (4.3.257) and (4.3.262) that

⎛⎞1 2 (4.3.263) γγ11⎜⎟=−log 2 − 2 γ log 2 ⎝⎠2 and this is the same as the result obtained by Coffey [45c] in (4.3.233).

From (4.3.256) we have

−3 1 11− x 4 ⎛⎞ ⎛⎞1 (4.3.264) log log(1/xdx ) = γ −−γγψ − γ2 ∫ 11⎜⎟ ⎜⎟ 0 14− x ⎝⎠ ⎝⎠4 and using [126, p.22]

⎛⎞11 (4.3.265) ψ ⎜⎟=−γπ − −3log2 ⎝⎠42 we get

−3 1 11− x 4 ⎛⎞ 1 (4.3.266) log log(1/xdx ) =−++γγγπγ3 log 2 ∫ 11⎜⎟ 0 14− x ⎝⎠ 2

We easily see that

−−33 33 ⎡ ⎤ −3 xx44−−11 xx44 −3 1 11−−xxx4 ( ) ( ) 44⎢ 1 ⎥ =− =− =− x 4 + 11 1⎢ 1⎥ 11−−xx 22 2 2 xx−+11 x + 1⎢ x − 1⎥ ( )( ) ( ) ⎣ ( )⎦

−−33⎡⎤1 − xxx42⎢⎥1 1 4 =− x 4 + =− − 11111⎢⎥ xxxxx24224+++++11111⎢⎥ ( ) ⎣⎦( ) ( ) ( )( )

Using partial fractions we see that

−−−−331 3 xxxx4 424 =−+ 11 1 1 1 (xx24++1)( 1212121) ( x 4 +) ( x 2 +) ( x 2 +) and we therefore obtain

72 −−33 −1 −3 1− xx44 x 2 x 4 =− − − 111 1− x 212121(xxx422+ ) ( ++) ( )

1 There is a more straightforward derivation using yx= 4 but, for posterity, I have left my slightly more complex struggle intact!

Letting p =1/4 and n =1/2 in (4.3.258) we get

−3 1 x 4 11313⎡ ⎤⎡ ⎤ ⎛⎞ ⎛⎞ ⎛⎞′′ ⎛ ⎞ ⎛ ⎞ 1 log log⎜⎟dx =−+−γψ⎢ ⎜⎟ ψ ⎜⎟⎥⎢ ς ⎜ 1, ⎟ ς ⎜ 1, ⎟⎥ ∫ 2 0 1+ x ⎝⎠x ⎣ ⎝⎠44 ⎝⎠⎦⎣ ⎝ 4 ⎠ ⎝ 4 ⎠⎦ and with p = 1/ 2 and n =1/2 in (4.3.258) we have already seen that

− 1 1 2 x ⎛⎞1 2 1 log log⎜⎟dx = log 2 ∫ 2 0 1+ x ⎝⎠x

With pn==1/ 4 we get

−3 1 4 x ⎛⎞1 2 1 log log⎜⎟dx = 6log 2 ∫ 4 0 1+ x ⎝⎠x

We therefore obtain

−3 1 4 17− x 2 1⎡ ⎛⎞1 ⎛⎞3⎤⎡1 ⎛1 ⎞ ⎛3 ⎞⎤ ∫ log log(1/xdx )=− log 2 −γψ⎢ ⎜⎟ − ψ ⎜⎟⎥⎢ − ς′′ ⎜1, ⎟ − ς ⎜ 1, ⎟⎥ 0 12− x 2⎣ ⎝⎠4 ⎝⎠4⎦⎣2 ⎝4 ⎠ ⎝4 ⎠⎦

We then refer to (4.3.228b) and see that

13⎡⎤ 13 ′′⎛⎞ ⎛⎞ ⎛⎞⎛⎞ ςς⎜⎟1,−=−− ⎜⎟ 1, ⎢⎥ γγ11 ⎜⎟⎜⎟ ⎝⎠44 ⎝⎠⎣⎦ ⎝⎠⎝⎠ 44 and using [126, p.22]

⎛⎞11 ⎛⎞31 ψ ⎜⎟=−γπ − −3log2 ψ ⎜⎟=−γπ + −3log2 ⎝⎠42 ⎝⎠42 we get

73 −3 1 17− x 4 11⎡ ⎛⎞1 ⎛⎞3⎤ (4.3.267) log log(1/xdx )=− log2 2 +γπ + γ − γ ∫ ⎢ 11⎜⎟ ⎜⎟⎥ 0 12− x 22⎣ ⎝⎠4 ⎝⎠4⎦

Using (4.3.233a)

⎛⎞13 ⎛⎞⎡⎤ ⎛⎞1 γγ11⎜⎟−=−++−Γ ⎜⎟ πγ⎢⎥4log 2 3log π 4log ⎜⎟ ⎝⎠44 ⎝⎠ ⎣⎦⎝⎠4 this then becomes

−3 1 4 17− x 2 11⎡ ⎛⎞ 1⎤ ∫ log log(1/xdx )=− log 2 +γπ − π⎢ γ +4log 2 + 3logπ − 4log Γ⎜⎟⎥ 0 12− x 22⎣ ⎝⎠ 4⎦ and equating this with (4.3.266) we get

(4.3.268)

⎛⎞11 72 11⎡ ⎛⎞1⎤ γγγπγ11⎜⎟−+ +3 log 2 =− log 2 +γππγ −⎢ +4log 2 + 3log π − 4log Γ⎜⎟⎥ ⎝⎠42 2 22⎣ ⎝⎠4⎦

This is in fact another proof of equation (4.3.233h). 

Differentiating (4.3.251) results in

∞ ett−−(1)utslog (4.3.269) dt=Γ(1)(1,)(1)(1,) s +ςς′′ s + u +Γ s + s + u ∫ t 0 e −1 and with s =1 we obtain

∞ ett−−(1)utlog (4.3.270) dt=+Γςς′′(2, u ) (2) (2,u ) ∫ t 0 e −1

Hence we have using the derivative of (4.3.253) combined with (4.3.254)

(4.3.271) ς ′′(2,uuuuuu )+Γ=− (2)ςγγψγγγ (2, )11 ′′′′ ( ) ( ) =+ ( )0 ( ) and integration over the interval [1,x ] produces

x x ∞∞1−+ψ (2) log(nu ) ∫ [(2,)ςψς′ uudu+=+ (2)(2,)] ∑∑ 1 nn==00nu++ nu 1

∞∞⎡ 1 1⎤⎡ log(nx++ ) log( n 1) ⎤ =−[1ψ (2)]∑∑⎢ −⎥⎢ + − ⎥ nn==00⎣nxn++11⎦⎣ nx + n +⎦

=−[1 −ψ (2)][γψ + (xx )] + γ11 ( ) − γ

2 (4.3.272) =−−−γ11()xxγγψγ () which we have seen previously.

Letting u =1in (4.3.271) gives us (see also (4.4.42bi))

∞ ttlog (4.3.273) dt =+−ς ′(2) (1γς ) (2) ∫ t 0 e −1 which may also be represented differently using (F.7)

11 ςγπς′′(1)−= (1 −− log2) + (2) 12 2π 2

Differentiating (4.3.253) with respect to u gives us

∞ et−−(1)1uts + ∞ 1(1)n ⎛⎞n − k dt= s ∫ ts∑∑⎜⎟ +1 0 Γ−()[se 1]nk==00 n + 1⎝⎠k ( k + u ) which is of course equivalent to letting ss→+1 in (4.3.253).

Alternatively, differentiating (4.3.253) with respect to s results in

∞ et−−(1)uts(log ts−−+ψ ( )) ∞ 1n ⎛⎞n ( 1)k log(ku ) (4.3.274) dt=−Γ() s ∫ t ∑∑⎜⎟ s 0 enk−++11nk==00⎝⎠k (u) and a further differentiation gives us

−−(1)uts 2 ∞ et{[log ts−−ψψ ( )]′ ( s )} ∞ 1(1)log()n ⎛⎞n −+k 2 ku (4.3.275) dt=Γ() s ∫ t ∑∑⎜⎟ s 0 en−++11()nk==00⎝⎠k ku

Completing the summation of (4.3.250) starting at n =1 results in

∞∞∞ ∞∞1 nn ∑∑∫∫∫eetdtedxeetdt−−−ut()11−= ts1( −+−−− nx1) ut() − ts1 nn==11n +1 000

∞∞ ∞ n −−ut x⎡⎤ − x − t s −1 =−∑∫∫ee⎣⎦ e()1 e t dtdx n=1 00

∞∞ ee−−ut21 x(1− e − t) t s − = dt dx ∫∫ −−xt 00 1[1]−−ee

We have with the obvious substitution y = e−x

1 ∞ eyye−2x 1 log(1−− [1−t ]y ) dx ==−−dy ∫∫−−xt − t − t − t2 001[1]1[1]1−−ee −− ey − e [1] − e 0

1 t =− + 1[1]−−ee−−tt2 and we therefore see that

−−ut21 x − t s − ∞∞ee()1− e t ∞et−−−ut s ∞ et ut s 1 dt dx =−dt dt ∫∫ −−xt∫ − t2 ∫ − t 001[1]−−ee 0 [1] −e 0 1 − e

∞∞et−−(2)uts et−−(1)1 uts − =−dt dt ∫∫tt2 00[1]ee−− 1

We now let tt→ α and obtain

∞∞et−−(2)1uts − e −− (2)uα tssα t − 1 I ==dt dt ∫∫ttα 00ee−−11

Then we have

s−−1 αα t( ut−− 2) s 1 s−( utsst− 2)αα dI ∞ te⎡⎤(1)−−−− e{ sαα (2) ut} e α te ==⎣⎦dt 0 ∫ αt 2 deα 0 [1]− and with α =1 this becomes

∞ tees−−1(⎡⎤(1) t−−−− ut−2){ sutete (2)} −(utst−2) ⎣⎦dt = 0 ∫ t 2 0 [1]e − and we therefore see that

∞∞etete−−(2)utst s−1 −− (2) ut ∞tes−−(2) ut dt=− s dt(2) u −dt ∫∫tt2 ∫ t 00[1]ee−− [1] 0 e −1

With u = 2 this becomes for Re (s ) > 1

∞∞test ts−1 (4.3.276) dt==Γ s dt s()() sς s ∫∫tt2 00[1]ee−− 1 which is reported in [126, p.103].

This may also be shown using integration by parts: we have

∞ ∞∞etts ts ts−1 dt =− +s dt ∫∫ttt2 00[1]eee−−− 10 1

Application of L’Hôpital’s rule shows us that

t s t s lim= 0 if s > 0 and lim= 0 t→0 et −1 t→∞ et −1 and therefore, as noted above, for s > 0 we have

∞∞etts ts−1 dt= s dt ∫∫tt2 00[1]ee−− 1

We have therefore shown that for us , > 0

−−ut21 x − t s − ∞∞ee()1− e t ∞tesut−−−1(2)∞ te sut−− (2)∞ e−− (1)1 uts t − dt dx=−−− s dt(2) u dt dt ∫∫ −−xt∫ t∫ t ∫ t 001[1]−−ee 0 e−1 0e−1 0 e − 1

∞ 1(1)n ⎛⎞n − k ()s =Γ ∑∑⎜⎟ s nk==10nku++1()⎝⎠k and, using (4.3.251), for u > 1 , s > 1 we have

∞∞ee−−ut21 x()1− e − t t s − dt dx=Γ( s + 1)(,ςςς s u − 1)( − u − 2)( Γ s + 1)(, s u − 1) −Γ ()(,) s s u ∫∫ −−xt 00 1[1]−−ee

77 ∞ 1(1)n ⎛⎞n − k ()s =Γ ∑∑⎜⎟ s nk==10nku++1()⎝⎠k

Hence we have for u > 1 , s > 1

∞ 1(1)n ⎛⎞n − k ssu(,1)(2)(,1)(,) u ssu su ∑∑⎜⎟ s =−−−−−ςςς nk==10nku++1()⎝⎠k and with u = 2 we obtain

∞ 1(1)n ⎛⎞n − k ss(,1) (,2) s ss () (,2) s ∑∑⎜⎟ s =−=−ςς ςς nk==10nk++1(2)⎝⎠k

Differentiating (4.3.251) results in

∞ et−−(1)utslog p t∂−+ p ∞ 1n ⎛⎞n ( 1)kp log ( ku ) =Γ++=−Γ(1)(1,)(1)()ssuς p s ∫ tp[]∑∑⎜⎟ s 0 es−∂11nk==00n +⎝⎠k ( ku +) and for convenience we may let ss→−1 and write the above equation as

∞ −−(1)1uts − p p ∞ n k p 1 etlog t 1 ∂ p 1⎛⎞n (−+ 1) log (ku ) tp=Γ=−[]()(,ssuς ) (1) ⎜⎟ s−1 Γ−(1)se∫ − 1 Γ−∂ (1) ss ∑∑n+1k ( ku+ ) 0 nk==00⎝⎠



In the book “Concrete Mathematics” [75, p.252] it is an exercise to prove that

∞ ⎛⎞x (4.3.280) Γ+=(1)xs∑ n ⎜⎟ n=0 ⎝⎠n

where ss01==0 and for n ≥ 2 we have

∞ n−1 seetdt=−(1)nt−−− 1 − t 1 n ∫ () 0

n nk⎛⎞n −1 +1 =−( 1)∑⎜⎟ ( − 1) log(1 +k ) k =0 ⎝⎠k and we note the connection with (4.3.250) which is valid for s > 0.

78 ∞ k n n ⎛⎞n (1)− eetdts−−−ut1(−=Γ t s 1 ) ∫ () ∑⎜⎟ s 0 k=0 ⎝⎠k ()ku+

⎛⎞n Therefore, since ⎜⎟= 0 , for n ≥1 we have ⎝⎠n +1

∞ n −−−ttn 11⎛⎞n k+ ∫ eetdt()1−=−+∑⎜⎟( 1) log(1 k ) 0 k =0 ⎝⎠k 

In 1999 Coppo [46b] showed that there exists a polynomial pn (z ) such that

∞ (4.3.290) xnz=−px(log) zedz− ∫ n 0 and the following related analysis is extracted from Brede’s dissertation [33b].

n k With pnk()zaz= ∑ we have k=0

∞∞n p (log)xzedzaxzedz−=−−−zk (log)z ∫∫nk∑ 00k =0

nk∞ ⎛⎞k =−axzedzlk(log)−−lz ∑∑k ∫ ⎜⎟ kl==000 ⎝⎠l

nk ⎛⎞k kl−−() kl l =−Γ∑∑axk ⎜⎟(1) (1) kl==00⎝⎠l

Hence we have

∞ nnk⎡ − ⎛⎞kl+⎤ p (log)xzedz−=−Γ− zllk (1)() (1)ax ∫ nk∑∑⎢ ⎜⎟ +1 ⎥ 0 kl==00⎣ ⎝⎠l ⎦

and we then select the coefficients ak such that

nk− ll⎛⎞kl+=() ⎧1for k n ∑(1)−Γ=⎜⎟ (1)ak +1 ⎨ l=0 ⎝⎠lk ⎩0 for less than n

79 resulting in

∞ xnz=−px(log) zedz− ∫ n 0

For example we have

(4.3.291) pz0 ()= 1

pz1()=− z γ

22 pz2 ()=− z 2γ z +−γς (2)

Letting x → log x we obtain

∞ lognzx =−px (log log zedz ) − ∫ n 0 and, with the substitution zx /=− log t, we have

logn x 1 =−p ( log log(1/tt ) x−1 dt ∫ n x 0

Referring to (4.3.225a)

∞∞⎡⎤lognnkk log (+− 1) log n k⎡ log n kk +1 log n t⎤ γ =− =−dt n ∑∑⎢⎥⎢ ∫ ⎥ kk==11⎣⎦kn+1 ⎣ ktk ⎦

⎡⎤N lognnktN +1 log =−lim dt N →∞ ⎢⎥∑ ∫ ⎣⎦k =1 kt1 we then see that

N 11N + 1 ⎡⎤kx−−11 γ nn=−limp ( log log(1/tt ) dt −− p n( log log(1/ tt ) dtdx N →∞ ⎢⎥∑∫∫∫ ⎣⎦k =1 010

11N −1 N + ⎡ kx⎤ =−limpn ( log log(1/tt ) − tdxdt N →∞ ∫∫⎢∑ ⎥ 01⎣ k =0 ⎦

80 1 ⎡ttNN−−11⎤ =−limpn ( log log(1/td ) −t N →∞ ∫ ⎢ ⎥ 0 ⎣ tt−1log⎦

Hence, as shown more completely in Brede’s dissertation, we obtain as N →∞

1 ⎡ 11⎤ (4.3.292) γ =−p ( log log(1/td ) − t nn∫ ⎢ ⎥ 0 ⎣logtt− 1⎦

1111 − ut Since −=∫ du we may write this as a double integral logtt−− 10 1 u

11 1− ut γ =−p ( log log(1/t ) dudt nn∫∫ 00 1− u

From (4.3.68d) we saw that for n ≥1

1 yyun−1(1− ) 1 n ⎛⎞n (4.3.293) dy= (−+ 1)jr (uj )−1 log( uj + ) ∫ r ∑⎜⎟ 0 log yrΓ() j=0 ⎝⎠j and we have the summation

∞∞1(1)111 yyun−1 − n ⎛⎞n dy= (−+ 1)jr (uj )−1 log( uj + ) ∑∑∫ r ∑⎜⎟ nnj===11nyrn+Γ+1log0 ()10⎝⎠j

We have

∞ (1− y )n 1 ∑ =−[logy + 1 −y ] n=1 ny+−11

and therefore we obtain

1 ⎡⎤yyuu−−1111∞ n ⎛⎞n (4.3.294) +=−−++dy ( 1)jr (uj )−1 log( uj ) ∫ ⎢⎥rr−1 ∑∑⎜⎟ 0 ⎣⎦(1−Γ+yyy ) log log ( rn )nj==10 1 ⎝⎠j

Starting the summation at n = 0 gives us

1 ⎡⎤yyuu−−11111∞ n ⎛⎞n +=dy urjr−−11log u −( −++ 1) (uj ) log( uj ) ∫ ⎢⎥rr−1 ∑∑⎜⎟ 0 ⎣⎦(1−ΓΓ+yyy ) log log ( r ) ( rn )nj==00 1 ⎝⎠j

and reference to (4.3.107a) shows that this is related to ς ′ (2− ru , ) .

With r =1 we get

1 uu−−11 ∞ n ⎡⎤yy 1 ⎛⎞n j (4.3.295) ∫ ⎢⎥+=−dylog u ∑∑⎜⎟( −+ 1) log(u j ) 0 ⎣⎦(1−+yy ) log nj==00 n1 ⎝⎠j and using (4.3.74) this becomes

1 ⎡⎤yyuu−−11 (4.3.296) ∫ ⎢⎥+=−dylog uψ ( u ) 0 ⎣⎦(1− yy ) log

With the substitution y = 1/ t in (4.3.293) we get

1 yyun−1(1−− )∞ (1 1 / t ) n dy = dt ∫∫ru+1 r 01logy tt log (1/ ) and a summation gives us (starting at n = 1)

∞ 1∞∞ (1− 1/ttt )n ⎡⎤ log(1/ ) 1 dt =−1 dt ∑ ∫∫ur++11⎢⎥ ur n=1 nt+−111 log (1/ t )⎣⎦ 1 t t log (1/ t )

∞ ⎡ 11⎤ =−dt ∫ ⎢ ur−+11 ur⎥ 1 ⎣tt(1− ) log (1 / tt ) log (1 / t ) ⎦

Differentiation of this with respect to r results in

(4.3.297) ∞ ⎡⎤⎛log log(1/ttr ) log log(1/ )ψ ( )∞ 1 n n⎞ −−=dt−−( 1)jr (uj++ )−1 log( uj ) ∫ ⎢⎥ur−+11 ur ∑∑⎜⎟ 1 ⎣⎦⎝tt(1−Γ ) log (1 / tt ) log (1 / t ) ( rn )nj==10+ 1 j ⎠

∞ n 11⎛⎞n jr−12 + ∑∑⎜⎟(1)(−+uj ) log( uj + ) Γ+()rnnj==10 1 ⎝⎠j and for r = 1 we get

∞ ⎡⎤⎛log log(1/tt ) log log(1/ )∞ 1 n n⎞ (4.3.297a) −−dt =γ ( −+ 1)j log(u j ) ∫ ⎢⎥uu+1 ∑∑⎜⎟ 1 ⎣⎦⎝tt(1−+ ) t log(1/ t )nj==10 n 1 j ⎠

∞ n 1 ⎛⎞n j 2 +∑∑⎜⎟(1)log(−+uj ) nj==10n +1 ⎝⎠j

∞ n 1 ⎛⎞n j = γ ∑∑⎜⎟(−+ 1) log(uj ) nj==00n +1 ⎝⎠j

∞ n 1 ⎛⎞n j 2 +∑∑⎜⎟(1)log(−+uj ) nj==00n +1 ⎝⎠j

−−γ loguu log2

Using (4.3.210) we see that this may be written in terms of the Stieltjes constants

∞ ⎡⎤1 1 log log(1/t ) (4.3.298) −=dtγγ() u+ 2 γ () u+ γ log u+ log2 u ∫ ⎢⎥u 01 1 ⎣⎦1− tt log(1/ t ) t

2 =−++2()γγψγ1 uu () loglog uu and with u =1 we obtain

∞ ⎡⎤log log(1/tt ) log log(1/ ) (4.3.299) −=+dt 2γ γ 2 ∫ ⎢⎥2 1 1 ⎣⎦tt(1− ) t log(1/ t )

∞ ⎡lognn log(1/tt ) log log(1/ ) ⎤ Further differentiations will result in integrals − dt ∫ ⎢ uu+1 ⎥ 1 ⎣ tt(1− ) t log(1/ t ) ⎦ involving the Stieltjes constants γ n (u ) .

Differentiation of (4.3.298) with respect to u results in

∞ ⎡⎤logtt log log(1/ ) log log(1/ t ) (4.3.299a) −+=−dt2()γγψ′ u′ () u ∫ ⎢⎥uu+1 1 1 ⎣⎦tt(1− ) t and with u =1 we get

∞ ⎡⎤logtt log log(1/ ) log log(1/ t ) (4.3.299b) −+=dt 2(1)γ ′ −γς (2) ∫ ⎢⎥2 1 1 ⎣⎦tt(1− ) t

Using (4.3.244)

2 γ1′′(1)=−++ 2πς ( 1) ς (2)( γ log 2 π )

we may write this as

83 ∞ ⎡⎤logtt 1 log log(1/ ) 2 (4.3.299c) −+∫ ⎢⎥ dt =−++4(1)(2)(2log2)π ςςγπ′ 1 ⎣⎦1− tt t

A HITCHHIKER’S GUIDE TO THE RIEMANN HYPOTHESIS

I was vaguely aware that the Stieltjes constants were relevant to the Riemann Hypothesis but I never really paid much attention to that aspect of analysis because I assumed that the mathematics was fiendishly difficult. However, having derived an interesting formula for γ p ()u , I decided to dip into this exotic topic via Coffey’s papers entitled “Toward verification of the Riemann Hypothesis: Application of the Li criterion” [45f] and “Polygamma theory, the Li/Keiper constants, and validity of the Riemann Hypothesis” [45g]. The following is a brief synopsis of some of the results contained in those papers, with some new material added.

Defining the Riemann xi function ξ (s ) as

1 − s (4.3.300) ξ ()sss=− ( 1)πς2 Γ ( s /2)() s 2 we see from the functional equation for ς (s ) that (see Appendix F in Volume VI)

(4.3.301) ξ (ss )=−ξ (1 )

In 1996, Li [101i] defined the sequence of numbers ()λ n by

n 1 d n−1 (4.3.302) λξn = n [log()]ss (1)!nds− s=1 and proved that a necessary and sufficient condition for the non-trivial zeros ρ of the 1 Riemann zeta function to lie on the critical line si= + τ is that λ is non-negative for 2 n every positive integer n . Earlier in 1992, Keiper [83a] showed that if the Riemann hypothesis is true, then λn > 0 for all n ≥1.

Li also showed that

n ⎡⎤⎛⎞1 (4.3.303) λn =−−∑ ⎢⎥11⎜⎟ ρ ⎣⎦⎢⎥⎝⎠ρ

84 Taking logarithms of (4.3.300) gives us

s (4.3.304) logξπ (ss )=− log 2 + log[] Γ (s / 2) − log + log[( ss − 1)ς ( )] 2 and we also have the Maclaurin expansion about s =1

∞ σ (4.3.305) logξ (ss )=− log 2 −∑ ( − 1)kkk ( − 1) k =1 k which implies that

∞ ξ′()s kk−1 =−∑(1) −σ k (s − 1) ξ ()s k=1

The constant term in (4.3.305) arises because

11− 1 limξπ (ss )=Γ2 (1/ 2) lim[( − 1) ς (s )] = ss→→1122

and we note that the coefficients σ k are defined by

(1)− kk+1 d σξk = k log (s ) (1)!kds− s=1

Comparing this with (4.3.302) we immediately see that λ11= σ .

In passing, we may also note that

11 limξς (ss )=− (0)lim[ Γ (s / 2)] =− ς (0)lim[ ss / 2 Γ ( / 2)] = ss→→0022 s → 0 in accordance with the functional equation for ξ()s .

Eliminating logξ (s ) from (4.3.304) and (4.3.305) results in

s ∞ σ (4.3.306) log[]ssΓ− ( / 2) logπς +− log[(s 1) ( s )] =−−−∑ ( 1)kkk (s 1) 2 k =1 k

Using the following identity (see [1, p.256] which was also employed by Keiper [83a] and Coffey [45f])

85 ∞ 1 1 [()ς k − 1] k (4.3.307) log[]ssΓ=+−−−−+ ( / 2) log 2 (γγ 1) ( 1)(s 1) ∑ k [1 −− (s 1)] 22 k =2 k2 we get s ∞ σ log[(1)()]logss−=ςπ −−−∑ (1)(1)kkk s 2 k =1 k (4.3.308) 1 1∞ [()ς k − 1] −−−+−−− log 2 (γγ 1) ( 1)(ss 1) [1 −− ( 1)]k ∑ k 22k =2 k 2

From (4.3.112a) we have for the left-hand side of (4.3.308)

lim log[(ss−= 1)ς ( )] log lim[( ss −== 1)ς ( )] log1 0 ss→→11 and hence we get

∞ [()ς k − 1] 1 ⎛⎞π (4.3.309) =−+(1γ ) log ∑ k ⎜⎟ k=2 k22⎝⎠ 2 in agreement with the formula reported in the book by Srivastava and Choi [126, p.174] and also Keiper’s paper [83a] . In fact, Srivastava and Choi [126, p.173] also give the general result valid for s < 2

∞ [()ς k − 1] (4.3.310) ∑ sssk =Γ−+−log() 2 (1γ ) k=2 k

Differentiating this gives us

∞ (4.3.311) ∑[()ς ks−=−−− 1]k −1 ψγ() 2 s k=2 and with s =1/2 we see that

∞ [()ς k − 1] 1 (4.3.312) ∑ k = −−[3/2]ψ ()γ k=2 22

We have [126, p.20]

⎛⎞11n−1 (4.3.313) ψ ⎜⎟n +=−−γ 2log2 + 2∑ ⎝⎠22k =0 k +1

86 and hence

∞ [()ς k − 1] 1 (4.3.314) ∑ k = log 2 − k=2 22

Similarly, letting t =1 in (4.3.310) we obtain the expansion

∞ [()ς k − 1] (4.3.315) ∑ =1−γ k=2 k

This may also be obtained by letting s = 0 in (4.3.307).

In (4.3.106a) we let gs( )=− log[( s 1)ς ( s )] and we note that limgs ( )= 0 . We have s→1

⎡⎤∞ 1(1)n ⎛⎞n − k gs() log log() hs ==⎢⎥∑∑⎜⎟ s−1 ⎣⎦nk==00nk++1(1)⎝⎠k and we also see that g (1)= 0 . Therefore we have the limit

⎡⎤ddkk lim⎢⎥kk log[(ss−= 1)ς ( )] lim gs ( ) ss→→11⎣⎦ds ds

From (4.3.308) we see that

(4.3.316)

∞∞ dk1 kk−11[()ς − 1] k− log[(ss−= 1)ςπγσ ( )] (log +−−−−+ 1)∑∑ ( 1)k (s 1) k [1 −− (s 1)] ds 22kk==12 and hence

dk1 ∞ [()ς − 1] gss′(1)=− log[( 1)ςπ ( )] =+−++ (logγσ 1) 1 ∑ k ds s=1 22k =2

11 =+−++−(logπγ 1) σ log 2 221

With reference to gs ( )= log hs ( ) and the fact that h (1)= 1, elementary calculus results in

∞ 1(1)log(1)n ⎛⎞n − kp+ k hs()pp()=− (1) ∑∑⎜⎟ s−1 nk==00nk++1(1)⎝⎠k

∞ n ()pp1 ⎛⎞n kpp+1 hkp(1)=− ( 1)∑∑⎜⎟ (− 1) log (1 + ) =− ( 1) γ p−1 for p ≥1 nk==00n +1 ⎝⎠k

hs′() gs′()= gh′ (1)= ′ (1) = γ hs()

Therefore we have obtained

11 (4.3.317) σπγ=−log + + 1 − log 2 1 22 as reported, inter alia, by Coffey in [45f]. We have

(4.3.317a) σ11= λ 0.023...

Differentiating (4.3.316) results in for n ≥ 1

d n+1 ∞ log[(ss−=−−−−−− 1)ςσ ( )] ( 1)kkn ( kkkns 1)( 2)...( ) ( 1) −−1 n+1 ∑ k ds k =1

∞ [()ς k − 1] +−( 1)nk (kk − 1)( − 2)...( kns − )[1 − ( − 1)] −−n1 ∑ k k=2 2

Therefore we have

(4.3.318) n+1 ∞ dknn[()ς − 1] lim log[(ss−=−+− 1)ςσ ( )] ( 1) n !n+1 ( 1) (kkkn −−− 1)( 2)...( ) s→1 n+1 ∑ k ds k=2 2

By differentiating (4.3.311) n times we obtain

∞ (4.3.319) ∑[ςψ (kkk )−−− 1]( 1)( 2)...( kns − )kn−−11( =− ( 1)n+ n)() 2 −s k=2 and, multiplying across by sn+1 and letting s = 1/2, we get

∞ [()ς k −− 1] (1)n+1 (4.3.320) (kk 1)( 2)...( kn ) ()n 3/ 2 ∑ kn−− −=+1 ψ () k=2 22

Since ψ ()nnn()1(+=x ψ () ()xnx +−1)!−n−1 we have

88 (4.3.321) ψ ()nn()3/2=+−ψ ()( 1/2) ( 1) nn !2n+1

We have

(4.3.322) ψς()nn()1/2= (−−+ 1)++1nn ![2n 1 1] ( 1) and hence we get

(4.3.323) ψ ()nn()3/2=− ( 1)+1nnn ![2n++ 1 − 1]ς ( + 1) +− ( 1) n !2n1

Accordingly we see that

∞ [()ς k − 1] ⎛⎞1 (4.3.324) (kk 1)( 2)...( knn ) ! 1 ( n 1) n ! ∑ kn− −−=−⎜⎟+1 ς +− k=2 22⎝⎠

Hence we see that for n ≥1

(4.3.324a) n+1 d nn⎛⎞1 n limn+1 log[(ss−=−+−− 1)()](1)!ςσ nn+1 (1)!1 n⎜⎟n+1 ς ( n +−− 1)(1)! n s→1 ds ⎝⎠2 and with n =1 we get

d 2 3 gss′′(1)=−=−−+ lim log[( 1)ςσς ( )]2 (2) 1 s→1 ds2 4

We have 2 hsh()′′ () s−[ h ′ () s] 2 gs′′()= gh′′(1)=− ′′ (1)[ h ′ (1)] []hs()2

Therefore we obtain

2 g′′(1)=− 2γ1 −γ

3 =−σς −(2) + 1 2 4 and hence we see that

3 (4.3.325) σ =−ςγγ(2) + 1 + 2 + 2 214

89 We have from Keiper’s paper [83a]

1 (4.3.326) σ = n ∑ n ρ ρ and since n k ⎡⎤⎛⎞1 n ⎛⎞n ()−1 λn =−−=−⎢⎥11 ⎜⎟ k ∑∑∑⎜⎟ k ρρ⎣⎦⎢⎥⎝⎠⎝⎠ρ k=0 ρ we see that

n ⎛⎞n k (4.3.327) λnk=−∑⎜⎟(1) − σ k=1 ⎝⎠k

which implies that λ11= σ .

This gives us in accordance with [45f]

3 (4.3.328) λ =−=−++−+−−2log12log2(2)2σσ πγ ς γγ2 2124 1

We see from (4.3.324a) that

n+1 nn11d ⎛⎞ n (4.3.329) (1)−=σςςn+1 limnn++11 log[(ss −−−− 1)()](1)1⎜⎟ (n ++− 1)(1) s→1 nds!2⎝⎠ and we have

m ⎛⎞m n λmn=−∑⎜⎟(1) − σ n=1 ⎝⎠n

mm ⎛⎞mmnn ⎛⎞ =−=−+∑∑⎜⎟(1)σ nn++11 ⎜⎟(1)σσm 1 nn==01⎝⎠nn++11 ⎝⎠

⎛⎞m since ⎜⎟= 0 . Therefore we obtain ⎝⎠m +1

m ⎛⎞m 1 d n+1 λςm =−lim log[(ss 1) ( )] s→1 ∑⎜⎟ n+1 n=1 ⎝⎠n +1 nds!

90 mm⎛⎞mm⎛⎞1 ⎛⎞ −−−++−+(1)1nnς (nm 1) (1) σ ∑∑⎜⎟ ⎜⎟n+1 ⎜⎟ 1 nn==11⎝⎠nn++11⎝⎠2 ⎝⎠

This may be written as

mmn+1 ⎛⎞mm11d ⎛⎞r ⎛⎞ λm =−lim log[(ss 1)ςς ( )]+ (− 1) 1− (rmm )+ 1−+σ1 s→1 ∑∑⎜⎟ nr+1 ⎜⎟ ⎜⎟ nr==12⎝⎠nr+1 nds!2⎝⎠ ⎝⎠

m ⎛⎞m n+1 since ∑⎜⎟(1)−=− 1 m . Employing Coffey’s notation [45g] we have n=1 ⎝⎠n +1

(4.3.330)

mdm ⎛⎞m 1 n+1 λπγm =−1 [log − + 2log 2] +Sm1 ( ) + lim log[(s− 1)ς ( s )] s→1 ∑⎜⎟ n+1 2!n=1 ⎝⎠n +1 nds where m ⎛⎞m ⎛⎞1 (4.3.331) Sm()=−− (1)1r ς ()r 1 ∑⎜⎟ ⎜⎟r r=2 ⎝⎠r ⎝⎠2

Using the Leibniz rule we see that

11ddmnm ⎛⎞m lim⎡⎤sssm−1 log[(−= 1)ςς ( )] lim log[(ss − 1) ( )] ss→→11mn⎣⎦∑⎜⎟ (mds−− 1)!n=1 ⎝⎠n (nds 1)!

m ⎛⎞m 1 d n+1 =−lim log[(ss 1)ς ( )] s→1 ∑⎜⎟ n+1 n=0 ⎝⎠n +1 nds!

m ⎛⎞mm1 ddn+1 ⎛⎞ =−lim log[(ss 1)ςς ( )]+− lim log[(ss 1) ( )] ss→→11∑⎜⎟ n+1 ⎜⎟ n=1 ⎝⎠n +11nds! ⎝⎠ds

Therefore we obtain

11ddmnm ⎛⎞m +1 lim⎡⎤sssm−1 log[(−= 1)ς ( )] lim log[(ssm −+ 1)ςγ ( )] ss→→11mn⎣⎦∑⎜⎟ +1 (1)!mds− n=1 ⎝⎠n +1 nds! and hence we derive Coffey’s expression

91 m md1 m−1 λm =−1 [logπγ − + 2log 2] +Sm1 ( ) + lim⎡⎤s log[( s− 1)ς ( s )] − m γ 2(s→1 mds−1)!m ⎣⎦ or equivalently

m (4.3.332) λπγ=−1 [log + + 2log 2] +Sm ( ) + Sm ( ) m 2 12 where m ⎛⎞m 1 d n (4.3.333) Sm2 ( )=− lim log[(s 1)ς ( s )] s→1 ∑⎜⎟ n n=1 ⎝⎠n (1)!nds−

We have

(4.3.334) ςς(,3/2)rr=−() 2rr 1 () − 2 and therefore we see that

ς (,3/2)r ⎛⎞ 1 11(+=−rr⎜⎟ς r) 22⎝⎠

Hence we may write

m ⎛⎞m ⎛⎞ς (,3/2)r Sm()=−+ (1)1r 1 ∑⎜⎟ ⎜⎟r r=2 ⎝⎠r ⎝⎠2

mm⎛⎞mmmm ⎛⎞⎛⎞ ⎛⎞ ς (,3/2)r =−−++−(1)rr (1) ∑∑⎜⎟ ⎜⎟⎜⎟ ⎜⎟ r rr==02⎝⎠rr ⎝⎠⎝⎠01 ⎝⎠ 2

m ⎛⎞m ς (,3/2)r =−+m 1(1) −r ∑⎜⎟ r r=2 ⎝⎠r 2

Coffey [45f] has shown that

mmmm11 logmS+− (γγ 1) +≤ (mm ) ≤ log ++ ( 1) − 2222221 and hence we have

mmm1 m 1−+++≥+−++−++ [logπγ 2log 2]Sm ( ) log m (γ 1) 1 [logπγ 2log 2] 222221

92 mm⎡⎤3 = log−+ 1 24⎣⎦⎢⎥π 2

Referring to Kölbig’s paper [91aa] (and see also Coffey [45h]) we may write

ddnn (4.3.335) ⎡⎤⎡⎤sssnn−−11log[(−= 1)ςς ( )] explog sss log[(− 1) ( )] dsnn⎣⎦⎣⎦ds and then express this in terms of the complete Bell polynomials (albeit this operation does not really simplify the output).

The constants ηn are defined by reference to the logarithmic derivative of the Riemann zeta function

∞ dsς ′() 1 k (4.3.336) [logςη (ss )]==−−∑ k ( − 1) s −1 < 3 ds ς ()s s − 1 k =0 and we may also note that

∞ dsς ′() 1 k log[(1)()]ss−=+=−−ςη∑ k (1)s ds ς ()s s − 1 k =0

We then see that

d n+1 ∞ log[(ss 1) ( )] kkkns ( 1)...( 1)( 1)kn− n+1 −=−−−+−ςη∑ k ds k =0 and hence we get

d n+1 (4.3.337) lim log[(ssn−=− 1)ς ( )] !ηn s→1 dsn+1

Referring to (4.3.329)

n+1 nn11d ⎛⎞ n (1)−=σςςn+1 limnn++11 log[(ss −−−− 1)()](1)1⎜⎟ (n ++− 1)(1) s→1 nds!2⎝⎠ we then obtain

nn⎛⎞1 n (−=−−−− 1)σηnn+1 ( 1)⎜⎟ 1n+1 ς (n ++− 1) ( 1) ⎝⎠2

93 n+1 ⎛⎞1 (4.3.338) σηnn+1 =−(1) −⎜⎟ 1 −n+1 ς (n + 1)1 + ⎝⎠2 in accordance with Theorem 3 in Coffey’s paper [45f].

In 2004 Maślanka [101c] reports that the following expression was also implicitly given in the 1999 paper by Bombieri and Lagarias [24b]

m λπγ=−1 [log + + 2log 2] +Sm ( ) + Sm ( ) m 2 12 and Maślanka makes the decomposition

− ∼

λmmm=+λλ

− ∼ where the “trend” λ m is strictly increasing and the “oscillations” λ m are given by

− m λπγ=−1 [log + + 2log 2] +Sm ( ) m 2 1

∼ m ⎛⎞m λ mn=−∑⎜⎟η −1 n=1 ⎝⎠n

From (4.3.337) we see that

1 d n lim log[(ss−=− 1)ς ( )] ηn−1 s→1 (1)!nds− n and therefore we have

mm⎛⎞mm ⎛⎞ 1 d n −=ηςn−1 lim log[(ss− 1) ( )] ∑∑⎜⎟s→1 ⎜⎟ n nn==11⎝⎠nn ⎝⎠(1)!nds−

This then shows that

∼ m ⎛⎞m (4.3.339) λ mn==−Sm21() ∑⎜⎟η − n=1 ⎝⎠n and we have

m ⎛⎞m 1 d n Sm2 ( )=− lim log[(s 1)ς ( s )] s→1 ∑⎜⎟ n n=1 ⎝⎠n (1)!nds−

ddnn∞ 1(1)m ⎛⎞m − p log[(ss−= 1)ς ( )] log nn∑∑⎜⎟ s−1 ds dsmp==00 m ++1(1)⎝⎠p p

In 1968 Gould [123c] reminded the mathematical community of the “not well-known” formula for the n th derivative of a composite function f (z ) where z is a function of x , namely nkk ()nk() (1)− jkj⎛⎞k − (4.3.340) Dfzxz( )= ∑∑ Dfz ( ) (−≥ 1)⎜⎟ z , for n 1 kj==11k! ⎝⎠j

This differentiation algorithm was also reported by Gould [73a] in 1972 in a somewhat different form in the case where z is a function of s

nkk ()nk() (1)− jkjnj⎛⎞k − () (4.3.341) Dfzsz( )=−∑∑ Dfz ( ) ( 1)⎜⎟ zDz s , for n ≥ 1 kj==00k! ⎝⎠j

This expression is frequently easier to handle than the di Bruno algorithm and it may be worthwhile using it in an attempt to simplify, and shed further light on, the structure of ∼

λ m = Sm2 () . ∞ 1(1)q ⎛⎞q − r Letting f (zzs )= log[ ( )] and zs()= . ∑∑⎜⎟ s−1 qr==00qr++1(1)⎝⎠r we see that

()kk() k−−1 k z(1)= 1 and fz(1)== log[ (1)] 0 DfzDzz( )== log[ zs ( )] (−− 1) ( k 1)! z and we have

nk(1)− k ⎛⎞k Dfz()nkkjkjnj( )=− ( 1)−−1 ( k − 1)! z ( − 1) z− Dz() s s=1 ∑∑⎜⎟ s s=1 kj==10k! ⎝⎠j

nk1 ⎛⎞k =−(1) − jjnjzDz− () ∑∑ ⎜⎟ s s=1 kj==10k ⎝⎠j

1 z ⎛⎞1 Letting s = we have (1)()ss−→ςς⎜⎟ and note that the function 1− z 11− zz⎝⎠−

95 ⎡⎤z ⎛⎞1 Fz()= log⎢⎥ς ⎜⎟ has been considered by Smith [120ai] and Coffey [45g] in ⎣⎦11−−zz⎝⎠ connection with the Riemann Hypothesis. In [45g] Coffey reports that

j+1 n ∞∞nj⎡⎤ ∞ (1)−− (1) jp+1 ⎛⎞ (4.3.341) Fz()=− ⎢⎥γ j z⎜⎟ z ∑∑nj! ∑ nj==11⎣⎦⎢⎥⎝⎠ p = 0

and an alternative proof is shown below using the formula for the Stieltjes constants γ j (it should however be noted that in the expression used by Coffey the summation starts at j = 0 )

j+1 n ∞∞nj⎡⎤ ∞m ∞ (1)−− (1)⎧⎫ 1 1 ⎛⎞m kj++11 j⎛ p⎞ =− ⎢⎥⎨⎬− ⎜⎟(1)log(1− +kz ) ⎜⎟ z ∑∑njjm!11++ ∑∑k ∑ nj==11⎢⎥⎣⎦⎩⎭ mk == 00⎝⎠ ⎝⎠p = 0

j+1 n ⎡ ⎛⎞∞ ⎤ ⎢ (1)log(1−+jj++11kzz ) p j + 1⎥ ∞∞n m ∞ ⎜⎟∑ (1)− ⎢ 1 ⎛⎞m k ⎝⎠p=0 ⎥ =−⎜⎟(1) − ∑∑∑⎢ k ∑ ⎥ nmk===100nm⎢ ++1(1)!⎝⎠ j = 1 j ⎥ ⎢ ⎥ ⎣ ⎦

∞ 1 Since ∑ z p = for z < 1 this becomes p=0 1− z

j+1 n ⎡⎤⎛⎞z (1)log(1−+jj++11k ) ∞∞n ⎢⎥m m ∞ ⎜⎟ (1)− ⎢⎥ 1 ⎛⎞ k ⎝⎠1− z =−∑∑∑⎜⎟(1)− ∑ nmk===100nm⎢⎥++1(1)!⎝⎠k j = 1 j ⎢⎥ ⎣⎦

n ∞∞n m (1)− ⎡⎤ 1 ⎛⎞m k ⎧ ⎛⎞z ⎫ =−∑∑∑⎢⎥⎜⎟( − 1) exp⎨ −⎜⎟ log(1 +k )⎬ nmk===100nm⎣⎦+−11⎝⎠k ⎩⎭⎝⎠z

n ⎡⎤ ∞∞(1)−−nk 1m ⎛⎞m (1) =− ⎢⎥ ∑∑∑⎢⎥⎜⎟ z nmk===100nm+1 ⎝⎠k ⎣⎦⎢⎥(1+ k ) 1−z and using the Hasse formula for the Riemann zeta function this becomes

n ∞ (1)− n ⎡⎤z ⎛⎞ 1 =−∑ ⎢⎥ς ⎜⎟ n=1 nzz⎣⎦11−−⎝⎠

⎡⎤z ⎛⎞1 = log ⎢⎥ς ⎜⎟ ⎣⎦11−−zz⎝⎠ 

With reference to Maślanka’s paper [101c] we consider the sum

∞∞∞m nn11⎛⎞m kn+1 ∑∑∑∑γ nss=−⎜⎟(1)log(1 − + k ) nnmk====0000nm++11⎝⎠k

∞∞m nn++11 11⎛⎞m k sklog(1)+ =− ∑∑⎜⎟(1)− ∑ smmk==00++11⎝⎠k n = 0 n

∞ m 11⎛⎞m k =−−+∑∑⎜⎟( 1) log[1sk log(1 )] smmk==00+1 ⎝⎠k 

Another approach to the Riemann Hypothesis is outlined below. We slightly rearrange the definition of the Riemann xi function as below

1 −−ssss⎛⎞ −s⎛s ⎞ ξ ()sss=− ( 1)πςπ22 Γ ( s /2)() s = Γ⎜⎟ ( s − 1)() ςπ s =Γ+−2⎜ 1 ⎟ ( s 1)() ς s 2222⎝⎠ ⎝ ⎠ and upon taking logarithms we have

ss⎛⎞ logξπ (sss )=− log + log Γ⎜⎟ 1 + + log[( − 1)ς ( )] 22⎝⎠

(the presence of the (1)()ss− ς term indicates a source for the Stieltjes constants)

Using the Hadamard representation of the Riemann zeta function

cs es⎛⎞s/ρ (4.3.343) ς ()se=−∏⎜⎟1 ⎛⎞s ρ ⎝⎠ρ 2(s −Γ 1)⎜⎟ 1 + ⎝⎠2 where c =−−log(2π ) 1γ / 2 , Snowden [120aa] showed in 2001 that

97 ∞ ⎡(1)− kk+1 ⎤ s (4.3.344) log[(ss−=−+ 1)ςπς ( )] log 2 s [log(2 ) −+ 1]∑ ⎢ k (k ) −σk ⎥ k =2 ⎣ 2 ⎦ k

We also have the Maclaurin expansion for the gamma function (E.22n) in Volume VI

∞ (1)− k logΓ+ (1x ) =−+γςxkx∑ ( ) k k=2 k

Therefore we have for s ≤ 2

⎛⎞s 1(1)∞ − k (4.3.345) log 1sks ( ) k Γ+⎜⎟ =−+γς∑ k ⎝⎠22k =2 k 2 and this gives us the expansion

ss1(1)(1)∞∞−−kkk⎡⎤ log (ss ) log 2 [log(2 ) 1] logs ( ksk )k ( ) ξππ=− + − − −γςς +∑∑kk− ⎢⎥+σk 22kk==22kk 2⎣⎦ 2 which becomes

1 ∞ σ (4.3.346) logξπγ (ssss )=−−−−− [log ] log 2 [1 log 2] ∑ k k 2 k=2 k

In view of (4.3.317) this may be written as

∞∞ σσkkkk (4.3.347) logξσ (ss )=−1 − log 2 −∑∑ s =−log 2 − s kk==21kk and, having regard to the functional equation (4.3.301) ξ (ss )= ξ (1− ) , we immediately see that (4.3.347) is simply another representation of (4.3.305). Indeed we have

∞∞σσ (4.3.348) ∑∑kksskkk=−(1) ( − 1) kk==11kk from which we see that

∞ σ (4.3.349) ∑ k = 0 k =1 k

Letting ss→−1 we see that

98 ∞∞σσ ∑∑kk(1)ss−=kk (1)(2) − − k kk==11kk and with s = 2 (valid?) we obtain

∞ σ ∑(1)−=k k 0 k =1 k and hence we have

∞∞σσ ∑∑22kk==+10 kk==11kk21+

Differentiating (4.3.348) results in (we need to prove that term by term differentiation is valid)

∞∞ kkk−−11 (4.3.350) ∑∑σσkkss=−(1) ( − 1) kk==11 and with s =1 we obtain

∞ (4.3.351) ∑σ k =−σ1 k=1 as reported by Keiper [83a].

Let us now multiply (4.3.350) by s

∞∞ kkk−1 ∑∑σσkkss=−(1) (s − 1) kk==11 and differentiate to obtain

∞∞ kk−−12⎡ kk−1⎤ (4.3.352) ∑∑ksσσkk=−( 1)⎣sk ( −− 1)( s 1) +− ( s 1) ⎦ kk==11

Letting s =1 we have

∞ ∑kσ k =−σσ21 k=1

We now repeat the same operation with (4.3.352)

99 ∞∞ kk⎡⎤221k−− k ∑∑ksσσkk=−(1)⎣⎦sk ( −− 1)( s 1) +− ss ( 1) kk==11

∞∞ 21kk−−⎡⎤2 k3 kk−2− 1 ∑∑ksσσkk=−( 1)⎣⎦skk ( −− 1)( 2)( s − 1) + 3 sks ( −− 1)( 1) +− ( s 1) kk==11

Letting s =1 we have

∞ 2 ∑k σ k =−σσσ321 +3 − k=1

Having regard to the definition of the Li/Keiper constants

n 1 d n−1 λξn = n [log()]ss (1)!nds− s=1 we consider ∞∞σσ sssnn−−11logξ ( )=− log 2 −∑∑kk ss k +n− 1 =−n−1log 2 − ( − 1) kssn − 1 ( − 1)k kk==11kk whereupon we obtain

d n ∞ (kn+− 1)( kn +− 2).... kσ [log()]ssnk−−11k s n ξ =−∑ ds k=1 k

ddnn∞ σ [log()]ssnknk−−11 (1)k ss (1) nnξ =−∑ − − ds dsk =1 k

For n =1 we get

dd∞ σ [logξ (ss )]=−∑ ( − 1)kkk ( − 1) ds dsk =1 k

∞ kσ =−∑(1) −kkk (s − 1)−1 k =1 k

and therefore we see that λ11= σ . For n = 2 we have

dd22∞ σ [log()]ss (1)kkk ss ( 1) 22ξ =−∑ − − ds dsk =1 k

100

We have

d ∞∞σσ ∞ σ ∑∑(1)−−=−−+−−kkkkkss ( 1) (1) sks ( 1) k−1 ∑ (1) kk k ( s 1) dskk==11 k k k = 1 k

d 2 ∞∞σσ∞ σ (1)kkkkkss ( 1) (1)skk ( 1)(1) s k−21 2 (1) kkks ( 1) k− 2 ∑∑−−=−−−+−− ∑ dskk==11 k k k =1k

which evaluated at s =1 becomes σ 21− 2σ . Hence we obtain

λ 221=−σσ +2

More generally, using the Leibniz differentiation formula we obtain (4.3.327)

n ⎛⎞n k λnk=−∑⎜⎟(1) − σ k =1 ⎝⎠k

A MULTITUDE OF IDENTITIES AND THEOREMS

This part contains some ancillary theorems and miscellaneous comments.

(i) An alternative proof of (4.2.1) is set out below.

Theorem 4.1: n! 1 n ⎛n⎞ − )1( k (4.4.1) x−1 −= )1( n dttt = ⎜ ⎟ x(1++xnx )...( ) ∫ ∑⎜k ⎟ + xk 0 k =0 ⎝ ⎠ where x > 0.

Proof:

Consider the function (xg ) defined by

n! (4.4.2) xg )( = ++ xnxx ))...(1(

Using (4.3.3) we have

101 Γ xn )(! (4.4.3) = nx ++Γ )1(

Γ + Γ xn )()1( (4.4.4) = nx ++Γ )1(

(4.4.5) = + ,1( xnB )

(4.4.6) = nxB +1,( )

The function(xg ) is therefore related to the beta function defined by the integral

1 (4.4.7) ∫ x−1 −= )1(),( y−1dtttyxB , (Re x)( > 0, Re y)( > 0) 0 which, in turn, is related to the gamma function via the identity [115, p.193]

Γ Γ yx )()( (4.4.8) yxB ),( = +Γ yx )(

However, because this paper is deliberately pitched at an elementary level (well, it started out that way!), we shall not assume any prior knowledge of the fundamental identity (4.4.8). Instead, let us consider a function xh )( defined by

1 (4.4.9) ∫ x−1 −= )1()( n dtttxh 0

We can use integration by parts to evaluate this integral

1 − )1( tt xn n 1 = ∫ −+ )1( xn dttt x 0 x 0

n 1 ∫ −= )1( −1 xn dttt x 0

nn − )1( 1 = ∫ − )1( xn +− 12 dttt xx + )1( 0

nn − 1)...1( 1 = ∫ nx −+ 1dtt nxxx −++ )1)...(1( 0

102 and therefore we obtain

n! (4.4.10) xh )( = ++ nxxx ))...(1(

From (4.4.2) and (4.4.10) we therefore see that = ()( xhxg ) .

Let us now evaluate the integral (4.4.9) in a different way using the binomial theorem (4.1.2). We have

1 1 n x−1 n xkk−1 ⎛⎞n (4.4.11) ∫ − )1( dttt =−∫ttdt∑⎜⎟(1) 0 0 k =0 ⎝⎠k

n ⎛n⎞ − )1( k (4.4.11a) = ∑⎜ ⎟ k =0 ⎝k ⎠ + xk

Hence we deduce

1 n k n! x−1 n ⎛n⎞ − )1( (4.4.11b) xg )( = ∫ −= )1( dttt = ∑⎜ ⎟ ++ xnxx ))...(1( 0 k =0 ⎝k ⎠ + xk

Γ + Γ xn )()1( (4.4.11c) = nxBxnB +=+= )1,(),1( nx ++Γ )1(

From (4.4.11) we also note that

11n xn−+11⎛⎞n kkkx− ∫∫tytdt(1−= ) ∑⎜⎟( − 1) ytdt 00k =0 ⎝⎠k and hence

1 n k xn−1 ⎛⎞n ky (4.4.11d) ∫tytdt(1−= ) ∑⎜⎟( − 1) 0 k=0 ⎝⎠k kx+

Differentiation with respect to x results in

1 n ⎛⎞n yk (4.4.12) tyttdtsxnss−1(1−=−− )log (1)! (1) k ∫ ∑⎜⎟ s+1 0 k=0 ⎝⎠k ()kx+

103

Completing the summation we find

1 ∞ 1 n ⎛⎞n yk (4.4.13a) tLiytxssk−1 (1−=−− )log tdt (1)! s (1) ∫ p ∑∑ps⎜⎟ +1 0 nk==10nkx⎝⎠k ()+

1 ttxs−1 log ∞ n ⎛⎞n yk (4.4.13b) dt=−(1)!sn s u (1) − k ∫ ∑∑⎜⎟ s+1 0 1(1)−−uyt nk==00⎝⎠k ()kx +

Letting u =1/2 in (4.4.13b) results in

1 ttxs−1 log ∞ 1 n ⎛⎞n yk (4.4.13c) dt=−(1)!sk s (1) − ∫ ∑∑ns+11⎜⎟ + 0 12(++yt nk==00⎝⎠k k x)

Letting y =1 gives us

1 ttxs−1 log ∞ 1n ⎛⎞n (− 1)k (4.4.13d) dt=−(1)!s s ∫ ∑∑ns+11⎜⎟ + 0 12(++tknk==00⎝⎠k x)

With x =1 this becomes

1 log skt ∞ 1n ⎛⎞n (− 1) (4.4.13e) dt=−(1)!s s ∫ ∑∑ns+11⎜⎟ + 0 12(++tknk==00⎝⎠k 1)

We have from (3.86ii) in Volume I

1 log s t dt=−(1)!s sς ( s + 1) ∫ a 0 1+ t and we have therefore another derivation of the Hasse/Sondow identity (3.11)

With y =−1 we have

1 ttxs−1 log ∞ 1 n ⎛⎞n 1 (4.4.13f) dt=−(1)!s s ∫ ∑∑ns+11⎜⎟ + 0 12(−+tknk==00⎝⎠k x)

We have from (4.4.231)

11logsstt log (1− ) ∫∫dt ==dt(1)(−+s ς s 1)(Γ s + 1) ,s ≥ 1 001− tt

104 and we therefore obtain

∞ 11n ⎛⎞n (4.4.14) (1)s ς +=∑∑ns+11⎜⎟ + nk==002()⎝⎠k kx+

Further integrals may be obtained by multiplying (4.4.13b) by log p y and integrating using (3.237i) ff.

ttxs−1 log The Wolfram Integrator gives us dt in terms of generalised hypergeometric ∫ 1+ t functions.

Let us make a bold step and assume that (4.4.13b) is valid where s is a real number (and is not simply restricted to positive integers): we then conjecture that

1 ttxs−1 log ∞ n ⎛⎞n yk dt=Γ+cos(π s ) ( s 1) unk ( − 1) ∫ ∑∑⎜⎟ s+1 0 1(1)−−uyt nk==00⎝⎠k ()kx + and differentiating with respect to s gives us

1 ttxs−1 log log(log t ) ∞ n ⎛⎞n ykxk log(+ ) dt=Γ+cos(π s ) ( s 1) unk ( − 1) ∫ ∑∑⎜⎟ s+1 0 1(1)−−uyt nk==00⎝⎠k ()kx+

∞∞nn⎛⎞nnyykk⎛⎞ cos(ss )′ ( 1) unk ( 1) sin( ss )( 1) u nk (1) +Γ+−−Γππ∑∑⎜⎟ ss+11π+−∑∑⎜⎟ + nk==00⎝⎠kk()kx++ nk == 00⎝⎠ () kx

With s = 0 we get

1 x−1 ∞∞nnkk ttlog(log ) nk⎛⎞nn ykxlog(+ ) nk⎛⎞ y ∫ dt=−∑∑ u ⎜⎟(1)− −Γ′(1)∑∑u ⎜⎟ (1)− 0 1(1)−−uyt nk==00⎝⎠kkkx+ nk == 00⎝⎠ kx+

With s =1 we get

1 ttxkk−1 log log(log t ) ∞∞nn⎛⎞nn ykxlog(+ ) ⎛⎞ y dt=− unk(1)− −Γ′(2)u nk (1)− ∫ ∑∑⎜⎟ 22∑∑⎜⎟ 0 1(1)−−uyt nk==00⎝⎠kk()kx+ nk == 00⎝⎠ ()kx+

Integration with respect to u over the interval [0,1] gives us

105 1 ttyttxs−1 log log(1−+ )log(log ) ∞ 1 n ⎛⎞n ykxk log( ) −=dtcos(π s )Γ ( s + 1) (− 1)k ∫ ∑∑⎜⎟ s+1 0 11−+yt nk==00n ⎝⎠k (k+ x)

∞ 1 n ⎛⎞n yk [cos()(1)sin()(1)]ss′ ss (1)k +Γ+−Γ+πππ∑∑⎜⎟ − s+1 nk==00nkx++1()⎝⎠k and with y =1 this becomes

1 tttxs−1 log log(1−+ )log(log t ) ∞ 1 n ⎛⎞n log( kx ) −=dtcos(π s )Γ ( s + 1) (− 1)k ∫ ∑∑⎜⎟ s+1 0 11−+tnnk==00⎝⎠k (k+x)

∞ 1(1)n ⎛⎞n − k [cos()(1)sin()(1)]ss′ ss +Γ+−Γ+πππ∑∑⎜⎟ s+1 nk==00nkx++1()⎝⎠k

Referring to (4.3.107a) we see that

∞ 1 n ⎛⎞n log(kx+ ) (1)k (ssxsx 1)(′ 2,) ( 2,) ∑∑⎜⎟−=−++−+s+1 ςς nk==00nk++1()⎝⎠k x and we therefore conjecture that

1 tttxs−1 log log(1− )log(log t ) ∫ dt=Γ+++++cos()(πςς s s 1)[(1)( s′ s 2,) x ( s 2,)] x 0 1− t

−Γ+−Γ+++ [cos(π ss )′ ( 1)ππ sin( ss ) ( 1)]( s 1) ς ( s 2, x )

With x =1 we get

1 logs tt log(1− )log(log t ) ∫ dt=Γ+++++cos(πςς s ) ( s 1)[( s 1)′ ( s 2) ( s 2)] 0 1− t

−Γ+−Γ+++[cos()(1)sin()(1)](1)(2)π ss′ ππ ss s ς s and with s = 0 we might have

1 log(1− tt )log(log ) ∫ dt = ς ′(2)++ςγ (2) 0 1− t

Alternatively, we may write (4.4.13b) as

106 1 ttxs−1(log)− ∞ n ⎛⎞n yk dt=Γ(1) s + unk (1)− ∫ ∑∑⎜⎟ s+1 0 1(1)−−uyt nk==00⎝⎠k ()kx+ and differentiating with respect to s gives us

1 ttxs−1 log log(−+ log t ) ∞ n ⎛⎞n ykxklog( ) dt=Γ(1) s + unk (1)− ∫ ∑∑⎜⎟ s+1 0 1(1)−−uyt nk==00⎝⎠k ()kx+

∞ n ⎛⎞n yk ′(1)sunk (1) +Γ + ∑∑⎜⎟− s+1 nk==00⎝⎠k ()kx+ 

Let us now perform an alternative integration by parts of (4.4.9). We have for x > 1

1 11(1−−tx )n+1 ( 1) ∫∫ttdttxnx−−11(1−=− ) + ttdt x −+ 21 (1 − ) n 00nn++110

(1)x − 1 =−∫ttdtxn−+21(1 ) n +1 0

Continuing the integration by parts we get for x > k

11(xx−− 1)( 2)...( xk − ) ∫∫txn−−11(1−= t ) dt txknk− (1− t ) + dt 00(nn++ 1)( 2)...( nk + )

1 nk+ (xx−− 1)( 2)...( xk − ) xk−−1 ⎛⎞nk+ j j =−∫tt∑⎜⎟(1) dt (nn++ 1)( 2)...( nk + ) 0 j=0 ⎝⎠j

(xx−− 1)( 2)...( xk − ) nk+ ⎛⎞nk+ ( − 1) j = ∑⎜⎟ (nn+ 1)(++ 2)...( nk ) j=0 ⎝⎠j xkj −+

Therefore, for x > n , we obtain

nxxxn! (−− 1)( 2)...( − )2n ⎛⎞2n ( − 1) j = ∑⎜⎟ x(1++xnx )...( ) ( n ++ 1)( n 2)...(2 n ) j=0 ⎝⎠j xnj −+ and this gives us

107 xx(222−− 1 )( x 2 2 )...( x 2 − n 2 )2n ⎛⎞2n ( − 1) j (4.4.14a) 1 = ∑⎜⎟ (2nxnj )! j=0 ⎝⎠j − +

An alternative derivation is shown below.

2n 2njj⎛⎞2n (1−=y )∑⎜⎟ ( − 1) y j=0 ⎝⎠j

2n x−−nn12⎛⎞2n jj+−−xn 1 yy(1−= )∑⎜⎟ ( − 1) y j=0 ⎝⎠j

1 2n j xn−−12 n ⎛⎞2n (1)− ∫ yydy(1−= ) ∑⎜⎟ 0 j=0 ⎝⎠j x − nj+ and we see that

1 (2n )! ∫ yydyxn−−12(1−= ) n 0 (x −−+−+nx )( n 1)...( x n 2 n )

(2n )! = x(xx222−− 1 )( 2 2 )...( xn 2 − 2 )

Let us now generalise (4.4.9) by letting ny→ . We have for 0 < x < 1 and y > −1

xx∞ aya−−11⎛⎞y kk ∫∫u(1−= u ) du u∑⎜⎟( − 1) u du 00k =0 ⎝⎠k

∞ k ⎛⎞y (1)− ak+ = ∑⎜⎟ x k=0 ⎝⎠k ka+

Letting x →1 and employing Abel’s theorem [13, p.245] we obtain

1 ∞ k ay−1 ⎛⎞y (1)− ΓΓ+ ()(ay 1) (4.4.14ai) ∫uudu(1−= ) ∑⎜⎟ = 0 k=0 ⎝⎠k ka+ Γ++(1) ay

It may also be noted that

∞∞k ⎛⎞y (1)− ()()−yakk 1 (4.4.14aii) ∑∑⎜⎟ ==−+21Fyaa(,;1;1) kk==00⎝⎠k ka++(1)! ak k a

108 and then employ the Gauss identity for the hypergeometric function.

Differentiation of (4.4.14ai) with respect to a results in

1 ∞ ⎛⎞y (1)−ΓΓ+k ()(ay 1) (4.4.14aiv) uuuduay−1(1−=−= ) logψψ ( aay ) −++ ( 1) ∫ ∑⎜⎟ 2 [] 0 k =0 ⎝⎠k ()ka+Γ++ ( ay 1)

1 ∞ ⎛⎞y (1)− k (4.4.14av) uuayss−1(1−=− ) log udus ( 1) ! ∫ ∑⎜⎟ s+1 0 k=0 ⎝⎠k ()ka+

Integration gives us

1 ty−1 ∞ (1)(1)uu−− ⎛⎞y k kt+ (4.4.14avi) ∫ du =−∑⎜⎟(1)log 0 logukk =0 ⎝⎠k + 1

We have previously mentioned the Flajolet and Sedgewick paper “Mellin Transforms and Asymptotics: Finite Differences and Rice’s Integrals” [68] in (3.13): their paper employed the following lemma:

Let ϕ()z be analytic in a domain that contains the half line [,)n0 ∞ . Then the differences of the sequence {ϕ()k } admit the integral representation

n ⎛⎞n (1)− n n! (4.4.14b) (1)()−=k ϕϕkz () dz ∑⎜⎟ ∫C k =0 ⎝⎠k 2πizzzn (−− 1)...( ) where C is a positively oriented closed curve that lies in the domain of analyticity of

ϕ()z , encircles [,)nn0 and does not include any of the integers 0,1,...,n0 − 1.

The similarity between the structure of gx ( ) and the integrand of (4.4.14b) clearly indicates why the Flajolet and Sedgewick paper was relevant in my earlier analysis.

Upon differentiation of the first part of (4.4.13) we have

d ⎡ 1 ⎤ n ⎛⎞n (1)− k gx′() n ! = ⎢ ⎥ =−∑⎜⎟ 2 dx⎣ x(1++ x )...( n x )⎦ k=0 ⎝⎠k ()kx+

The Pochhamer symbol is defined by

109

Γ()x + n (xxxnx )=+ (1 )...( −+= 1 ) n Γ()x and differentiation results in

dΓΓ+−Γ+Γ() x′′ ( xn ) ( xn ) () x ()x nn==+− ()xxnx[]ψψ ( ) () dxΓ2 () x

n! We see that gx()= and hence ()x n+1

dn!!()x ′ n gx′()==−=−++− n !n+1 []ψψ ( x n 1) () x dx() x2 () x nn++11[]()x n+1

Accordingly we get

nx!()Γ (4.4.14c) gx′()=−[][]ψψ ( xn + + 1) − () x = Bxn (, + 1)( ψψ xn + + 1) − () x Γ++(1)xn

n ⎛⎞n (1)− k

=−∑⎜⎟ 2 k=0 ⎝⎠k ()kx+ and in particular we have

n!(1)Γ 1 gn′(1) =−[]ψψ( + 2) − (1) =− Hn+1 Γ+(2)nn +1

Therefore we obtain

n ⎛⎞n (1)− k H n+1 ∑⎜⎟ 2 = k=0 ⎝⎠k (1)kn++ 1

which we also saw in (4.2.16)

Upon differentiation of (4.4.13) we see that

n ⎛n⎞ − )1( k (4.4.15) s− )1( xg )( = s−1 s −− )!1()1( ⎜ ⎟ ∑⎜ ⎟ s k=0 ⎝k ⎠ + xk )(

110 1 (4.4.16) ∫ x−1 −= log)1( sn −1 tdttt 0

Now let us consider the series

∞ gx(1)s− () (4.4.17) H = ∑ n n=1 2

∞ 1 1 (4.4.18) =−ttxns−−11(1 ) log tdt ∑ n ∫ n=1 2 0

1 ttxs−−11(1− ) log t (4.4.19) = ∫ dt 0 1+ t where we completed the summation by employing the geometric series. Using the binomial theorem to expand the denominator of (4.4.19) we obtain

∞ 1 (4.4.20) Httd=−∑∫ (1)kkx+−11 (1)log −s − t k=0 0

∞ 1 ∞ 1 =−∑∫ (1)kkxtdt+−11 log s − −∑∫ (1)− kkxtdt+− log s1 k=0 0 k=0 0

We have the elementary integral

1 1 ∫ α dxx = 0 α +1 and by parametric differentiation we deduce

∂s−1 1 1 (1)(− s−1 s − 1)! (4.4.21) α dxx = xxdxα logs−1 = s−1 ∫ ∫ s ∂α 0 0 (1)α +

Using (4.4.21) we can write (4.4.20) as

⎧ ∞∞(1)−−kk (1) ⎫ (4.4.22) s−1 Hs=−(1)( − 1)!⎨∑∑s − s ⎬ ⎩⎭kk==00()kx+++ ( kx 1)

Using (4.4.15) we may also express H as

111 ∞ 1(1)n ⎛⎞n − k (4.4.23) Hs(1)(s−1 1)! =− − ∑∑ns⎜⎟ nk==102()⎝⎠k kx+ and we therefore have an expression involving the alternating Hurwitz-Lerch zeta function (see (4.4.82))

∞ 1(1)n ⎛⎞n − k ⎧ ∞∞(1)−−kk (1) ⎫ (4.4.24) ∑∑ns⎜⎟ =−⎨∑∑s s ⎬ nk==102()⎝⎠k kx+ ⎩⎭kk==00()kx+++ ( kx 1)

This simplifies to (where the series now starts at n = 0 )

∞ 1(1)n ⎛⎞n − k ∞ (1)− k (4.4.24a) = ∑∑ns+1 ⎜⎟ ∑ s nk==002()⎝⎠k kx+ k =0 ()kx+

This identity is also employed in (4.4.112w).

Differentiating (4.4.24a) with respect to s we obtain (it should be noted that we have only proved (4.4.24a) for the case where s is an integer).

∞ 1n ⎛⎞n (− 1)k log(kx+ ) ∞ (− 1)k log(kx+ ) (4.4.24aa) = ∑∑ns+1 ⎜⎟ ∑ s nk==002()⎝⎠k kx+ k =0 ()kx+

Completing the summation of (4.4.15) we obtain

∞∞111 n ⎛⎞n (−1)k ttxns−−−111(1−=−− ) log tdts ( 1) s ( 1)! ∑∑pss∫ ∑⎜⎟ nnk===11nnkx0 0⎝⎠k ()+ and hence we get

1 ∞ 1(1)n ⎛⎞n − k (4.4.24aaa) tLitxss−−−111(1−=−− ) log tdt ( 1) ( s 1)! ∫ p ∑∑ps⎜⎟ 0 nk==10nkx⎝⎠k ()+

We may note that

d s−1 11 tLitdttLitxxs−−−111(1−= ) (1 − ) log tdt s−1 ∫∫pp dx 00

The Wolfram Integrator evaluates tLitdtx−1 (1− ) in terms of generalised ∫ 2 hypergeometric functions. See also (4.4.100qiv).

112

Let us now integrate equation (4.4.1), noting that it is not defined for x = 0

β β 1 β n k n! xn−1 ⎛⎞n (1)− ∫ dx =−∫∫dx t(1 t ) dt = ∫∑⎜⎟ dx α xxnx(1++ )...( ) α 0 α k=0 ⎝⎠k kx+

Therefore we obtain for β ,α > 0

ββ11 1 ()(1)ttβα−−11−− tn (4.4.24b) dx txn−−11(1−=− t ) dt (1 t ) nx dt t dx = dt ∫∫ ∫ ∫ ∫ log t αα00 0 and β n k n ⎛⎞n (1)− ⎛⎞n k k + β ∫∑⎜⎟ dx =−∑⎜⎟(1)log α k =0 ⎝⎠k kx+ k=0 ⎝⎠k k +α

Therefore we have

1 βα−−11 n n ()(1)tt−− t ⎛⎞n k k + β (4.4.24ba) ∫ dt =−∑⎜⎟(1)log 0 log t k=0 ⎝⎠k k +α

With n = 0 we have

1 ()ttβα−−11− β (4.4.24c) ∫ dt = log 0 log t α and a version of the above formula is employed in (4.4.112c).

1− t We also have the series for < 1 p

∞ 1(1 ttβα−−11−− )(1) tn ∞ 1 n ⎛⎞n k + β dt =−(1)logk ∑ n ∫ ∑∑n ⎜⎟ n=0 pt0 log nk==00pk⎝⎠k +α and hence after completing the summation we obtain

1 ()ttβα−−11− ∞ 1 n ⎛⎞n k + β (4.4.24d) dt =−(1)logk ∫ ∑∑n ⎜⎟ 0 []1(1)/log−−tp t nk==00pk⎝⎠k +α

With p = 2 we have

113 1 ()ttβα−−11− ∞ 1 n ⎛⎞n k + β (4.4.24e) dt =−(1)logk ∫ ∑∑n+1 ⎜⎟ 0 (1+ tt ) log nk==002 ⎝⎠k k +α

As regards the above equation, reference should also be made to (3.36h), (4.4.99c), (4.4.112g) and (4.4.112ga). See also (4.4.112aiii) in Volume III.

Letting up=1/ in (4.4.24d) we get

1 βα−−11 ∞ n ()tt−+nk⎛⎞n kβ (4.4.24f) ∫ dt=−∑∑ u ⎜⎟(1)log 0 []1(1)−−tu log t nk==00⎝⎠k k+α and integrating with respect to u results in

x 1 βα−−11∞ n+ 1n ()tt−+ x ⎛⎞n k kβ ∫∫du dt =−∑∑⎜⎟(1)log 00[]1(1)−−tu log tnk==00 n+ 1 ⎝⎠k k+α

Reversing the order of integration gives us

xx11()()1ttβα−−11−− ttβα−− 11 ∫∫du dt = ∫dt ∫ du 00[]1(1)−−tu log t 0 logt 0[] 1(1)−−tu

1 ()log1(1)ttβα−−11−−−[ tx] =−∫ dt 0 (1− tt ) log and hence we have

1 βα−−11 ∞ n+1 n ()log1(1)tt−−−[ tx] xk⎛⎞n k +1 + β (4.4.24g) ∫ dt =−∑∑⎜⎟(1)log 0 (1−++tt ) log nk==00 n1 ⎝⎠k kα

Note the close connection between (4.4.24g) and the series given in (4.3.72a) for the digamma function (and refer also to (4.4.24j)).

With x =1/2 this becomes

1 ()log(1)/2ttβα−−11−+[ t] ∞ 1 n ⎛⎞n k + β dt =−(1)logk+1 ∫ ∑∑n+1 ⎜⎟ 0 (1−+tt ) log nk==00( n 1)2 ⎝⎠k k +α and with β == 2, α 1 this becomes

114 1 log[ (1+ t ) / 2] ∞ 12n ⎛⎞n k + (4.4.24h) dt =−(1)logk ∫ ∑∑n+1 ⎜⎟ 0 log tnnk==00(+ 1)2 ⎝⎠k k+1

Letting x =1 in (4.4.24g) we obtain

1 βα−−11 ∞ n ()tt−+ 1⎛⎞n k+1 kβ (4.4.24hi) ∫ dt =−∑∑⎜⎟(1)log 0 (1−+tn ) nk==001 ⎝⎠k k +α and with β ==2, α 1 this becomes

∞ n 12⎛⎞n k +1 k + −=1(1)log∑∑⎜⎟− nk==00nk++11⎝⎠k

We will see in (4.4.92a) that

∞ n 1 ⎛⎞n k+1 γ =−+∑∑⎜⎟( 1) log(k 1) nk==00n +1 ⎝⎠k and hence we have

∞ n 1 ⎛⎞n k+1 ∑∑⎜⎟(−+=− 1) log(k 2)γ 1 nk==00n +1 ⎝⎠k

t N −1 N −1 1 t N −1 Since =− t j , we have dt=− H and therefore ∑ ∫ N 1− t j=1 0 1− t

1 tkNN −++11∞ n ⎛⎞n 1 dt=− H = (1)log− k+1 ∫ N ∑∑⎜⎟ 0 11−+tnnk==00⎝⎠k k + 1

This then gives us

∞ n 1 ⎛⎞n k+1 (4.4.24i) ∑∑⎜⎟(−++=− 1) log(kN 1) γ HN nk==00n +1 ⎝⎠k

We may generalise this by noting that

111()11ttβα−−11−−− tα− 1 tβ− 1 ∫∫∫dt=−=− dt dt ψ ()αψβ () 000111−−−ttt and hence from (4.4.24hi) we have again derived (4.3.72a)

115

∞ n 1 ⎛⎞n k+1 k + β (4.4.24j) ∑∑⎜⎟(1)log−=−ψ ()αψβ ( ) nk==00nk++1 ⎝⎠k α

Using (4.4.92a) this gives us another derivation of (4.3.74)

∞ n 1 ⎛⎞n k ∑∑⎜⎟(−+= 1) log(k α )ψα ( ) nk==00n +1 ⎝⎠k

Integrating (4.4.24g) gives us

βα−−11 11()log1(1)tt−−−[ tu] ⎛⎞1()ttβα−−11− ∫∫⎜⎟udtudt− −= 00(1−−−tt ) log⎝⎠ 1 t (1 tt ) log

∞ n+2 n uk⎛⎞n k+1 + β ∑∑⎜⎟(1)log− nk==00(1)(2)nn++⎝⎠k k +α and, using (4.4.24g), this may be written as

1 ()log1(1)ttβα−−11−−−[ tu] −=dt ∫ 2 0 (1− tt ) log

∞∞nn++−2211nn1 βα− ukuktt⎛⎞nnkk++11++ββ⎛⎞ ()− ∑∑⎜⎟(1)log−− ∑∑⎜⎟ (1)log −+∫ dt nk==00(1)(2)nn++ ⎝⎠kkk+αα nk= 00 n +1=⎝⎠ k+0 (1)log − tt

Differentiating (4.4.24g) with respect to β we get

1 ttxβ −1 log[ 1−− (1 ) ] ∞ xnk+11n ⎛⎞n (1)− + (4.4.24k) ∫ dt = ∑∑⎜⎟ 0 11−++tnknk==00⎝⎠k β

With β =+N 1 and x =1 we have

1 ttNklog∞ 1n ⎛⎞n (− 1) +1 ∫ dt = ∑∑⎜⎟ 0 (1−+++tn ) nk==001 ⎝⎠k kN1

We also have from (4.4.238aa)

116 ttNjlog NN t t dt=−log t +− Li (1 t ) ∫ ∑∑2 2 (1− tj ) jj==11 j

and hence 1 ttN logN 1 dt=−=− Li(1) H (2) ς (2) ∫ ∑ 2 2 N 0 (1− tj ) j=1

Therefore we have shown that

∞ n k+1 (2) 1(1)⎛⎞n − (4.4.24l) H N −=ς (2) ∑∑⎜⎟ nk==00nkN+++11⎝⎠k and, for example, with N =1 we have

∞ 1(1)n ⎛⎞n − k+1 (4.4.24m) 1(2)−=ς ∑∑⎜⎟ nk==00nk++12⎝⎠k

Differentiating (4.4.24k) with respect to β we get

1 ttβ −1 log log[ 1−− (1 tx ) ] ∞ xnk+1 n ⎛⎞n (1)− (4.4.24n) dt = ∫ ∑∑⎜⎟ 2 0 (1−++tnk )nk==00 1⎝⎠k (β )

With β =+N 1 and x =1 we get

1 ttNklog2 ∞ 1n ⎛⎞n (− 1) dt = ∫ ∑∑⎜⎟ 2 0 (1−+++tn )nk==00 1⎝⎠k ( kN 1)

As will be seen in equation (4.4.235) of Volume IV we have

xxkjjjlog2 kkk x x x dx=−2 + 2log x − log22 x − log(1 − x )log x − 2log x Li ( x )+ 2 Li ( x ) ∫ ∑∑∑32 23 1− xjjjjjj===111 and more specifically

1 ttN log2 dt=−22(3) H (2) + ς ∫ N 0 (1− t )

(2) (this is also valid for N = 0, where we take H0 = 0 ).

Hence we get

117

∞ 1(1)n ⎛⎞n − k (4.4.24o) =−2(3)⎡ς H (2) ⎤ ∑∑⎜⎟ 2 ⎣ N ⎦ nk==00nkN+++1(⎝⎠k 1)

Differentiating (4.4.24n) p times we have

1 ttβ −+11logp log[ 1−− (1 tx ) ] ∞ xnk+1 n ⎛⎞n (1)− dt=−(1)(p p + 1)! ∫ ∑∑⎜⎟ p+2 0 (1−+tn ) nk==001⎝⎠k (k+β ) and with x =1 we get

1 ttβ −+12log pk∞ 1n ⎛⎞n (− 1) dt=−(1)(p p + 1)! ∫ ∑∑⎜⎟ p+2 0 (1−+tn ) nk==001⎝⎠k (k+β )

Differentiating (4.3.68) we see that

1 ttβ −+12log p ∫ dt =−ψ (2)p+ ()β 0 (1− t ) and hence we have

∞ 1(1)n ⎛⎞n − k (4.4.24p) (2)pp++() (1)(1p 1)! ψ β =− + ∑∑⎜⎟ p+2 nk==00nk++1()⎝⎠k β

We know from [126, p.22] that

∞ 1 ()nn()zn (1)++1 ! (1)n1 nnz !( 1,) ψ =− ∑ n+1 =−ς + k =0 ()kz+ and therefore we see that

∞ 1(1)n ⎛⎞n − k (2)pp++( ) ( 1)1 (ppp 1)! ( 1) p +1 ( 2)! ( 3, ) ψ βς=− +∑∑⎜⎟ p+2 =− + + β nk==00nk++1()⎝⎠k β

Hence we get

11∞ n ⎛⎞n (1)− k (4.4.24pi) (2,)p ςβ+=∑∑⎜⎟ p+2 pn++11nk==00⎝⎠k () k +β which is the formula (3.12a) originally derived by Hasse.

118

As will be noted in (4.4.135a), Larcombe et al. [95] showed that

⎛⎞⎛⎞mn+ nmn n (1)− k + 1 m ⎜⎟⎜⎟ 2 = nk∑∑ ⎝⎠⎝⎠kkm==0 ()mk+ k and hence we have

−1 ∞∞1(1)11n ⎛⎞nN− k ⎛++1 n ⎞ Nn+1+ 1 ==2(3)⎡ς −H (2) ⎤ ∑∑⎜⎟ 2 ∑⎜ ⎟ ∑ ⎣ N ⎦ nk==00nkNNn+++++1(⎝⎠kn 1)11 n= 0 ⎝ ⎠ kN=+ 1 k

With N = 1 we obtain

−1 ∞∞1(1)111nn⎛⎞nn− k ⎛+ 2 ⎞ +2

∑∑⎜⎟ 2 = ∑⎜ ⎟ ∑ nk==00nk++1(2)21⎝⎠kn n= 0 n +⎝ ⎠ k= 2 k and the right-hand side is equivalent to

−1 11∞+⎛⎞n + 2 n 2 1∞H −1 n+2 ∑∑∑⎜⎟ = 2 21nkn==02nknn+++⎝⎠n =0 (1)(2)

This then gives us

∞ H −1 (4.4.24q) n+2 2(3)1 ∑ 2 = []ς − n=0 (1)(2)nn++ which, from a structural point of view, initially appears to be unexpected. However, structurally similar series have been found by Ogreid and Osland (see [105(ii)] and [105(iii)]): for example

∞ Hn ∑ 2 =−2(3)ς ς (2) n=1 nn(1)+

∞ H ∑ n = ς (2) n=1 nn(1)+

Similar series are also given in the 2005 paper “Six Gluon Open Superstring Disk Amplitude, Multiple Hypergeometric Series and Euler-Zagier Sums” by Oprisa and Stieberger [105(iv), p.54].

119 Using the Larcombe identity (4.4.135) we get another rather unexpected result after making the substitution in (4.4.24k)

∞∞1(1)11n ⎛⎞n − k (4.4.24r) (2) 1 ς −=∑∑⎜⎟ = ∑ 2 nk==00nk++122(1)(2)⎝⎠k n= 0 nn ++

Integrating (4.4.24n) with respect to x we get

u 1 ttβ −1 log log[ 1−− (1 tx ) ] ∞ unk+2 n ⎛⎞n (1)− (4.4.24s) dx dt = ∫∫ ∑∑⎜⎟ 2 00 1(1)(2)()−+tnnk==00nk++⎝⎠k β and therefore we obtain

11ttββ−−11log log[ 1−− (1 tutt ) ] log log[ 1−− (1 tu ) ] 1ttβ −1 log (4.4.24t) udt−− dtudt ∫∫2 ∫ 001(−−−ttt1)1 0

∞ unk+2 n ⎛⎞n (1)−

= ∑∑⎜⎟ 2 nk==00(1)(2)nn++⎝⎠k ( k +β )

Dividing (4.4.24n) by x and then integrating with respect to x we get

1 β −1 ∞ nkn ttLitulog2 [ (1− ) ] u ⎛⎞n (1)− (4.4.24u) dt =− ∫ ∑∑22⎜⎟ 0 (1−++tnk )nk==00 ( 1)⎝⎠k (β )

With β =1 and u =1 we get

1 logtLi (1−− t ) ∞ 1n ⎛⎞n ( 1)k 2 dt =− ∫ ∑∑22⎜⎟ 0 1(−++tnknk==001)(⎝⎠k 1)

Since

1 1 logtLi (1− t ) 1 2 1 2 dt=−=− Li(1 t )ς 2 (2) ∫ []2 0 (1− t ) 2 0 2 we see that

∞ 1(1)1n ⎛⎞n − k (4.4.24v) 2 (2) ∑∑22⎜⎟ = ς nk==00(1)nk++⎝⎠k (1)2

120

From Larcombe’s formula we see that

n ⎛⎞n (1)− k H n+1 ∑⎜⎟ 2 = k=0 ⎝⎠k (1)kn++ 1 and hence we have

∞∞1(1)n ⎛⎞n − k H n+1 ∑∑223⎜⎟ = ∑ nk==00(1)nkn+++⎝⎠k (1)(1) n = 0 which gives us

∞ H n 1 2 (4.4.24vi) ∑ 3 = ς (2) n=1 n 2

Similarly, integrating (4.4.24u) with respect to u we get

u 1 β −1 ∞ nkn du ttLitulog2 [ (1− ) ] u ⎛⎞n (1)− (4.4.24w) dt =− ∫∫ ∑∑32⎜⎟ 00ut(1−++ )nk==00 ( nk 1)⎝⎠k (β ) and therefore we have

1 β −1 ∞ nkn ttLitulog3 [ (1− ) ] u ⎛⎞n (1)− dt =− ∫ ∑∑32⎜⎟ 0 (1−++tnk )nk==00 ( 1)⎝⎠k (β )

With β =1 and u =1 we get

1 logtLi (1− t ) ∞∞1(1)n ⎛⎞n − k H (4.4.24x) 3 dt =− =− n ∫ ∑ 324∑∑⎜⎟ 0 1(1)(1)−++tnknnk==00⎝⎠k n= 1

It is clear that this may be generalised to

1 β −1 ∞ nkn ttLitulogp [ (1− ) ] u ⎛⎞n (1)− (4.4.24y) dt =− ∫ ∑∑p ⎜⎟ 2 0 1(1)()−++tnknk==00⎝⎠k β and differentiation of (4.4.24y) with respect to β results in

1 β −+11q ∞ nkn ttLitulogp [ (1− ) ] u ⎛⎞n (1)− (4.4.24z) dt=+(1)!(1) q −q+1 ∫ ∑∑pq⎜⎟ +2 0 (1−++tn ) nk==00( 1)⎝⎠k (kβ )

121 With β =1 , q = 1, u =1 and p = 2 we get (see (4.4.64ci) et seq. in Volume III)

1 log2 tLi (1− t ) ∞ 1(1)n ⎛⎞n − k 2 dt = 2 ∫ ∑∑23⎜⎟ 0 (1−++tnk ) nk==00( 1)⎝⎠k ( 1) and integration by parts gives us

log2 tLi (1−− t ) log tLi (1 t ) 22dt= log t dt ∫∫(1−−tt ) (1 )

2 112 [Li2 (1− t )] =−−logtLi[]2 (1 t ) dt 22∫ t

Therefore we have

112212 12 logtLi (1− t ) 12 1ςς (2) 1[Li2 (1− t )] 1 (2) 2 dt=−−−log a Li (1 a ) dt dt + dt ∫∫[]2 ∫∫ aa(1− tttt ) 2 2 2 a2 a

1 2 2 1112 [Li2 (1−− t )] ς (2) =−Li(1 a ) log a −ς 2 (2)log a − dt []2 ∫ 222a t and, in the limit as a → 0 , we get

112 2 2 logtLi (1− t ) 1 ς (2)−−[Li2 (1 t )] ∫∫2 dt = dt 00(1− tt ) 2

In passing we note that if limf ′ (a ) is finite then ao→ fa() limf (aa )log= lim aa log ao→→ao a

f ()afa′ () lim= lim ao→→a ao 1 and therefore limfa ( )log a= 0 if f ′(0) is finite. ao→

Hence we see that (see (4.4.64ci) et seq. in Volume III)

2 1 2 ∞ n k ς (2)−−[Li2 (1 t )] 1(1)⎛⎞n − (4.4.24zi) dt = 4 ∫ ∑∑23⎜⎟ 0 tnknk==00(1)++⎝⎠k (1)

122

We note from the Larcombe identity (4.4.135b) that

2 ⎛⎞⎛⎞mn+ nmnmn n (1)− k ⎛⎞++ 1 1 2m ⎜⎟⎜⎟∑∑∑32=+⎜⎟ ⎝⎠⎝⎠nkkkmkm===0 ()mk+ ⎝⎠ k k and thus we may write (4.4.24zi) in terms involving the generalised harmonic numbers.

We have

112 2 ς (2)−−[Li2 (1 t )] (ςς(2)+−[Li22 (1 t )])( (2) −−[ Li (1 t )]) ∫∫dt= dt 00tt

()ς (2)+−[Li2 (1 t )] dt=+−+−+ς (2)log t Li (1 t )log t log(1 t )log2 t 2 Li ( t )log t − 2 Li ( t ) ∫ t 223 and with integration by parts we obtain

1 2 2 ς (2)−−[Li2 (1 t )] ∫ dt = 0 t

1 2 ςς(2)−−[]Li22 (1 t )⎡⎤ (2)log t +−+− Li (1 t )log t log(1 t )log t + 2 Li 23 ( t )log t − 2 Li ( t ) ()⎣⎦0

1 log t −+−+−+−⎡⎤ς (2)logtLit (1 )log t log(1 t )log2 t 2 Lit ( )log t 2 Lit ( ) dt ∫ ⎣⎦223 0 1− t

1 log t =−2ςς (2) (3) −⎡⎤ ς (2) logtLit + (1 − ) log t + log(1 − t ) log2 t + 2 Lit ( ) log t − 2 Lit ( ) dt ∫ ⎣⎦223 0 1− t

We have

log2 t dt=−log(1 − t ) log2 t − 2 Li ( t ) log t + 2 Li ( t ) ∫ 1− t 23 but the other integrals are more intractable. I considered trying Euler’s identity (1.6c)

ς (2)=−++− logx log(1xLixLix )22 ( ) (1 ) which gives us

123 2 2 []Li2 (1−= x )ςς (2) − 2 (2)log x log(1 −− x ) 2ς (2) Li2 ( x )

22 2 +−−−+ logx log (1xxxLixLix ) 2log log(1 )22 ( )[] ( )

However, the Wolfram Integrator was unable to evaluate the resulting integrals.

From (4.4.38b) we see that

(1−−tu ) ( 1)pp−−111 log y Li(1− t ) u = dy p [ ] ∫ Γ−−(pt )0 (1 (1 )uy ) and substitution in (4.4.24z) gives us

111β −+11q pq−−11β + 1 p − 1 ttLitulogp [ (1− ) ] uttty(−− 1) log (1 )log ∫∫dt = dt∫dy 000(1)−Γ−−tptt () (1) 1(1)−uy

uy(1)− pp−−1111 log = ∫∫tβ −+11logq t dtdy Γ−()pt00 1(1)−uy

For example, we now let β = 1 , u =1 and p = 3 to get

1 log2 y 1 1 dy=−⎡⎤log(1)log2()log2() Ax2 x + Li Ax x − Li Ax ∫ ⎣⎦23 0 1− Ay A 0

2 =− Li() A A 3

However, I believe that the analysis is becoming rather circular at this stage! Having reached the end of the alphabet, I will leave further consideration of this to another day!

The following result is well-known.

(ii) Theorem 4.2:

xu∞ s−1 (4.4.25) (xLi ) = du s ∫ u Γ−()se0 x

Proof:

The gamma function is defined in (4.3.1) as

124 ∞ (4.4.26) Γ=()stedt∫ st−−1 , Re s)( > 0 0 and, using the substitution = kut , we obtain

∞ (4.4.27) )( =Γ ∫ 1 −− kuss dueuks 0

Hence we have

11 ∞ (4.4.28) = 1 −− kus dueu s ∫ Γ sk )( 0

We now consider the finite sum set out below

n ⎛⎞n xk (4.4.29) Sx() n = ∑⎜⎟s k=1 ⎝⎠k k and combine (4.4.28) and (4.4.29) to obtain

n ⎛⎞n xk n ⎛n⎞ 1 ∞ (4.4.30) Sx()= = ⎜ ⎟ xk . 1 −− kus dueu n ∑⎜⎟s ∑⎜ ⎟ ∫ k=1 ⎝⎠k k k =1 ⎝k ⎠ Γ s)( 0

∞ n 1 s−1 ⎧⎛n⎞ − kku ⎫ (4.4.31) = ∫u ∑⎨⎜ ⎟ ⎬ duxe Γ s)( 0 k=1 ⎩⎝k ⎠ ⎭

∞ n 1 su−−1 ⎧⎫⎛⎞n k (4.4.32) = ∫uexd∑ ⎨⎬⎜⎟⎣⎦⎡⎤ u Γ()s 0 k =1 ⎩⎭⎝⎠k

From the binomial theorem we have

n ⎛n⎞ k n ∑⎜ ⎟ AA )1( −+= 1 k =1 ⎝k ⎠ and hence (4.4.32) becomes

k ∞ n ⎛⎞n x 1 n (4.4.33) Sx()= =+−uxedusu−−1 11 n ∑⎜⎟s ∫ {()} k=1 ⎝⎠k k Γ()s 0

We now revisit the expression for the polylogarithm derived in (3.55)

125

∞ 1 n ⎛n⎞ xk (4.4.34) xLi )(2 = ⎜ ⎟ s n ∑∑ ⎜ ⎟ s = 2 kn =11 ⎝k ⎠ k

and substitute the relation obtained for Sxn ( ) to deduce

∞ ∞∞11 n (4.4.35) 2()Li x =+usu−−1 1 xe− 1 du s ∑∑n ∫ {()} nk==112()Γ s 0

n −u ∞ 11+−xe 1 ∞ {()} (4.4.36) = udus−1 ∫ ∑ n Γ()s 0 n=1 2

(where we have assumed that interchanging the order of summation and integration is permissible). Use of the geometric series simplifies (4.4.36) to

x ∞ 1eu −− us xu∞ s−1 (4.4.37) (xLi ) = du = du s ∫ −u ∫ u s 0 1)( −Γ xe Γ−()se0 x

After deriving the above formula, I subsequently found out that it had been discovered by Appell many years ago: equation (4.4.37) was in fact posed as an exercise in Whittaker and Watson’s book [135, p.280] and, inter alia, is also reported in [126, p.121].

By letting x = 1 in (4.4.37) we obtain the familiar formula [25, p.222]

1 ∞ us−1 (4.4.38) Li )1( = ς ()sd= u s ∫ u Γ−()se0 1

This simpler expression can be derived directly in the following manner. Using (4.4.28) and making the summation we have

∞ 11 ∞ ∞ = 1 −− kus dueu ∑ s ∑ ∫ k =1 Γ sk )( k =1 0

1 ∞ ∞ = ∫ s 1∑ −− kudueu Γ s)( 0 k =1

126 After noting that e−u < 1 ∀∈ u (0, ∞ ) , we sum the geometric series and obtain (4.4.38). This identity is used in the proof of the functional equation for the Riemann zeta function in Appendix F of Volume VI.

When I started work on this in 2003 I was concentrating on the definition of the polylogarithm in (4.4.34) and I completely missed the rather more obvious approach outlined below.

ks∞∞−1 ∞∞xx1 k u ==uxedusu−−1 du ∑∑su∫∫() kk==11ksΓΓ()00 ()sex−

Making the substitution ut= log in (4.4.37) we have

x ∞ log s−1 t (4.4.38a) (xLi ) = dt s ∫ Γ−()stxt1 ( ) and the substitution ty=1/ gives us

x(1)− ss−−111 log y (4.4.38b) (xLi ) = dy s ∫ Γ−()sxy0 (1 )

We also note the double integrals

u Li() x (1)− ss−−11u 1 log y (4.4.38c) Li() u= s dx = dydx s+1 ∫ ∫∫ 0 x Γ−()sxy00 1

1 Li() x (1)− ss−−1111 log y (4.4.38d) ς (1)sdx+=∫ s = ∫∫ dydx 0 x Γ−()sxy00 1 and after carrying out the integration on x we get

(−− 1)ss1 log−1 yuy log(1 ) (4.4.38e) Li() u = dy s+1 ∫ Γ()sy0

(−− 1)ss1 log−1 yy log(1 ) (4.4.38f) ς (1)sd+= ∫ y Γ()sy0

(and in this regard, see (4.4.91ga)).

Letting x =1 in (4.4.38b) we get

127 (1)− ss−−111 log y (4.4.39) Li (1) = dy s ∫ Γ−()sy0 1 and hence

(1)− ss−−111 log y (4.4.40) ς ()sd= ∫ y Γ−()sy0 1

Letting x =−1 in (4.4.38b) we have

(1)− ss−−211 log y (4.4.41) Li (− 1) = dy s ∫ Γ+()sy0 1 and hence

(1)− ss−−111 log y 1 ∞ us−1 (4.4.42) ς ()sd= y= du a ∫ ∫ u Γ+()sy0 1 Γ+()se0 1

(equation (4.4.42) was employed in (3.83) as the starting point of Amore’s variational method).See also (3.86j).

From (4.4.38e) we obtain

ttLi() u (−− 1)ssdu1 log−1 y log(1 uy ) ∫∫∫s+1 du = dy 000usuyΓ() and hence we get

(1)− s+1 1 logs−1 yLi ( ty ) (4.4.42i) Li() t = 2 dy s+2 ∫ Γ()sy0

Integration by parts gives the following for s = 2

logyLi ( ty ) logy Li ( ty )−= Li ( ty ) 2 dy 34∫ y which concurs with the above. Equation (4.4.42i) is employed in Appendix E of Volume VI.

We also note that

128

ttLi() x 1 ∞ u s−1 s dx= dx du ∫∫∫u 000x Γ−()sex

1 ∞ =−−∫uusu−1[ log( etdu )] Γ()s 0 and we then obtain

1 ∞ Li() t=− usu−−1 log(1− te ) du s+1 ∫ Γ()s 0

Differentiating (4.4.37) with respect to s we obtain

∂ ψ ()su∞∞ss−−11 xulog u Li() x=− x du + du s ∫∫uu ∂Γ−Γ−ssexsex()00() and therefore

∞ xnslog nx∞ u−1 log u −=−+Li() xψ () s du ∑ sus ∫ n=1 nsΓ−()0 ex

Letting x =1 we obtain

∞ uus−1 log du=Γ() sςς′ () s + () s Γ () s ψ () s ∫ u 0 e −1

d (4.4.42a) =Γ()ssςς′′ () + () ss Γ () =[] Γ ()() ssς ds

With the substitution ut= log we have

∞∞uuss−−11log log t log log t du ==dtΓ+() sςς′ () s () sΓ′ () s ∫∫u 01et−−1(1t) and with ty=1/ this becomes

1 logs−1 yy log log(1/ ) (1)−=s ∫ dyΓ () sςς′ () s+ () sΓ′ () s 0 1− y

129 In [6a] Adamchik reports the following integral

∞ uulog 1 (4.4.42b) du =−ς ′(1) ∫ 2πu 0 e −12 and making the substitution x = 2πu this becomes

∞∞uulog 1 xx log log 2π ∞ x du =−dx dx ∫∫22πux 2 ∫ x 00eee−−14ππ 1 4 0 − 1

Reference to (4.4.37) shows that

∞ x dx = ς (2) ∫ x 0 e −1

From (4.4.42a) above we have

∞ xxlog (4.4.42bi) dx =Γ(2)ςς′ (2) + (2) Γ′ (2) ∫ x 0 e −1

Therefore, since Γ= (2) 1, we get

∞ uulog 1 log 2π du =+Γ−ς ′′(2)ςς (2) (2) (2) ∫ 22πu []2 0 e −14ππ4

In (E.20a) of Volume VI it is shown that Γ′(2)=− 1 γ and hence we get

∞ uulog 1 log 2π du =+−−ςγς′(2) (1 ) (2) ∫ 22πu [] 0 e −14π 24

Using the functional equation for ς (s ) in (F.7) of Volume VI it is shown that

11 ςγπς′′(1)−= (1 −− log2) + (2) 12 2π 2 and combining these equations we obtain the same result as Adamchik

∞ uulog 1 du =−ς ′(1) ∫ 2πu 0 e −12

130 We also have from (4.4.25)

xu∞ s−1 Li()−=− x du s ∫ u Γ+()se0 x and therefore we get

1 ∞ us−1 Li (1)−=− du s ∫ u Γ+()se0 1

and, since ς as (sLi )=− ( − 1) , this is equivalent to

1 ∞ us−1 ς ()sd= u a ∫ u Γ+()se0 1

Differentiating with respect to s we obtain

Γ′()su∞∞ss−−111 u log u (4.4.42c) ς ′ ()sdudu=− + a 2 ∫∫uu []Γ()s 00ese+Γ1()1 +

Therefore we have

1log∞ uus−1 duss=Γ()ςψς′ () − () ss () ∫ u aa 0 e +1

See also the analysis following (C.61) in Volume VI.

Let us now consider a generalised version of Theorem 4.2.

(ii) Theorem 4.2(a):

∞ n ks∞ −−+1(1)yu n ⎛⎞n xxtue (4.4.43) td⎜⎟ s = u ∑∑k ()(1)()ky+−Γ t s∫ ⎡ −u ⎤ nk==11⎝⎠ 0 ⎣11−+()xe t⎦

11⎛⎞xt =Φ⎜⎟,,sy − s 1(1)(1)−−tt⎝⎠ − ty

xt⎛⎞ xt = 2 Φ+⎜⎟,,sy 1 (1−−tt )⎝⎠ (1 )

131 ∞ zn where Φ=(,,zsa ) ∑ s is the Hurwitz-Lerch zeta function defined in (4.4.44fi). n=0 ()na+

We shall see in (4.4.44) that

∞ n ks∞ −−1 yu n ⎛⎞n xu1 e td⎜⎟ s = u ∑∑k ()ky+Γ () s∫ ⎡ −u ⎤ nk==00⎝⎠ 0 ⎣11−+()xe t⎦

Proof:

We again start with the gamma function

∞ Γ=()stedt∫ st−−1 , Re s)( > 0 0 and, using the substitution tuky=+ ( ) , we obtain

11∞ (4.4.43a) = uesuky−−1() + du s ∫ ()ky+Γ () s0

We now consider the finite sum set out below

n ⎛⎞n xk (4.4.43b) Sxy(, ) n = ∑⎜⎟ s k =1 ⎝⎠k ()ky+ and combine (4.4.43a) and (4.4.43b) to obtain

n ⎛⎞n xk n ⎛n⎞ 1 ∞ Sxy(, )= = ⎜ ⎟ xk . uesuky−−1() + du n ∑⎜⎟ s ∑⎜ ⎟ ∫ k =1 ⎝⎠k ()ky+ k =1 ⎝k ⎠ Γ()s 0

∞ n 1 su−−1(⎧⎛⎞n k+y)k⎫ = ∫uexdu∑ ⎨⎜⎟ ⎬ Γ()s 0 k =1 ⎩⎭⎝⎠k

∞ n 1 suy−−−1 ⎧⎫⎛⎞n k u (4.4.43c) = ∫uexdu∑ ⎨⎬⎜⎟⎣⎦⎡⎤e Γ()s 0 k=1 ⎩⎭⎝⎠k

Applying the binomial theorem we have

132 ∞ 1 n (4.4.43d) Sxy ( , ) = uxeedusuyu−−−1 11+− n ∫ {()} Γ()s 0

We now consider the following function Zs (xyt , , ) which, for t < 1, is defined by

1 ∞ n ⎛⎞n xk (4.4.43e) Z (xyt , , ) t n s = ∑∑⎜⎟ s 2()nk==11⎝⎠k ky+

and substitute the relation obtained for Sxyn ( , ) to deduce

∞ 11∞ n Z (xyt , , ) = tuxeeduns−−+−1()11+− uyyu s ∑ ∫ {()} 2()n=1 Γ s 0

∞ 11 ∞ n (4.4.43f) =+−∫uxetedusunyu−−−1)∑{()11} 2()Γ s 0 n=1

(where we have again assumed that the conditions required for interchanging the order of summation and integration are satisfied). Use of the geometric series simplifies (4.4.43f) to

∞ n ks∞ −−+1(1)yu 11n ⎛⎞n xxtue (4.4.43g) Zs (,xyt ,)== t ⎜⎟ s du 2()2(∑∑k ky+−Γ1)() t s∫ ⎡ −u ⎤ nk==11⎝⎠ 0 ⎣11−+()xe t⎦

We now recall the identity in (3.67d)

∞ n ⎛⎞n xk 1 ⎛⎞xt tLin ∑∑⎜⎟s = s ⎜⎟ nk==11⎝⎠k kt11− ⎝⎠− t and with y = 0 this shows that

xtue∞ su−−1 ⎛⎞ xt du= Lis ⎜⎟ Γ−()st∫ ⎡⎤−u ⎝⎠1 0 ⎣⎦11−+()xe t

With t = 1/2 we recover (4.4.25). An alternative proof is shown below. Letting y = 0 in (4.4.43g) we have

∞ n ks∞ −−1 u 11n ⎛⎞n xxtue (4.4.43h) Zs (,0,)xt== t ⎜⎟s du 22∑∑k kts(1−Γ)()∫ ⎡ −u ⎤ nk==11⎝⎠ 0 ⎣11−+()xe t⎦

133 and therefore

∞ n ⎛⎞n xxtuks∞ −1 tdn = u ∑∑⎜⎟su2 ∫ nk==11⎝⎠k ktsextt(1−Γ ) ( )0 −[] /(1 − )

With t =1/2 we recover (4.4.37).

From (4.4.25) we see that

[xt/(1− t )] ∞ u s−1 Li xt/(1−= t ) du s [] ∫ u Γ−−()sextt0 [] /(1 ) and we accordingly obtain (3.67d) again

∞ n ⎛⎞n xk 1 ⎛⎞xt (4.4.43i) tLin ∑∑⎜⎟s = s ⎜⎟ nk==11⎝⎠k kt11− ⎝⎠− t

With x =−1 we have

∞ n ⎛⎞n (1)−−k 1 ⎛⎞t (4.4.43ia) tLin ∑∑⎜⎟ s = s ⎜⎟ nk==11⎝⎠k ktt11− ⎝⎠−

and employing Landen’s identity (3.111) in the case where s = 2

⎛⎞−t 1 2 Li2 ⎜⎟=−log (1 −tLit ) − 2 ( ) ⎝⎠1− t 2 we see that

∞ n ⎛⎞n (1)−−k 1log(1)2 ttLi() t 1 ⎛⎞ − (4.4.43ib) tLn 2 i ∑∑⎜⎟ 2 =− − = 2 ⎜⎟ nk==11⎝⎠k ktttt21− 1−− 1⎝⎠ 1 −

Dividing (4.4.43ib) by t and then integrating gives us

x Li() t 2 dt=−log(1 − x ) Li ( x ) − log x log2 (1 − x ) ∫ 2 0 tt(1− )

−− 2Li233 (1 x )log(1 −+−+− x ) 2 Li (1 x ) Li ( x ) 2ς (3)

134 x log2 (1− t ) dt=−2log(1 x ) Li (1 −+ x ) log x log2 (1 − x ) ∫ 2 0 tt(1− )

1 −−−−++ log3 (1xLixLix ) 2 (1 ) ( ) 2ς (3) 3 33 and we obtain

∞ xnkn ⎛⎞n (1)− (4.4.43ic) ∑∑⎜⎟ 2 = nk==11nk⎝⎠k

11 log(1−+−−+−+−x )Li ( x ) Li (1 x )log(1 x ) log x log23 (1 x ) log (1x ) 22 26

3 −−−Li(1 x ) Li ( x ) +ς (3) 332

Using Landen’s identity (3.115) in the case where s = 3

⎛⎞−t 1123 Li3 ⎜⎟= ςς(2)log(1−−t ) log t log (1 −− tLit )33 (1 −+ ) (3) + log (1 −− tLit ) ( ) ⎝⎠1− t 26 gives us

(4.4.43id)

∞ n ⎛⎞n (1)− k 1 1 tn (2)log(1 t ) log t log23 (1 t ) Li (1 t ) (3) log (1 t ) Li ( t ) ∑∑⎜⎟ 3 =−−−−−++−−ςς33 nk==11⎝⎠k k 26

Letting tx= in (4.4.43i) we obtain

∞ n ⎛⎞n xk 1 ⎛⎞x2 (4.4.43ie) n ∑∑xL⎜⎟s = is ⎜⎟ nk==11⎝⎠k kx11− ⎝⎠− x and x =1/2 gives us

∞ 111n ⎛⎞n ⎛⎞ (4.4.43if) Li ∑∑nks+1 ⎜⎟ = s ⎜⎟ nk==1122⎝⎠k k ⎝⎠ 2

We have with t =1/2 in (4.4.43i)

135 ∞ 1 n ⎛⎞n xk Li() x ∑∑ns+1 ⎜⎟ = s nk==112 ⎝⎠k k which we derived by alternative means in (3.64) in Volume I.

Substituting the Flajolet and Sedgewick identity (3.16c)

k n ⎛⎞n (1)− 113 1 − =+HHHH(1) (1) (2) + (3) ∑⎜⎟ 3 ()nnnn k =1 ⎝⎠k k 62 3 in (4.4.43id) gives us for t < 1

(4.4.43ig)

∞ 3 ⎡⎤11(1) (1) (2) 1 (3) n −++=∑ ⎢⎥()HHHHtnnnn n=1 ⎣⎦62 3

11 ςς(2)log(1−−t ) log t log23 (1 −− tLit ) (1 −+ ) (3) + log (1 −− tLit ) ( ) 2633

Dividing this by t and integrating will result in more expressions for Euler sums.

We have from (4.4.43g) with t = 1/2

∞ 1 n ⎛⎞n xxueks∞ −−1 yu (4.4.43j) Z (,xy ,1/2)==du s ∑∑nsu+1 ⎜⎟ ∫ nk==112()()⎝⎠k ky+Γ s0 e − x and, with x =1 and y →−y 1, this becomes

∞ 111n ⎛⎞n ∞ uesyu−−−1(1) (4.4.43k) Z (1,yd−= 1,1 / 2) = u s ∑∑nsu+1 ⎜⎟ ∫ nk==112(1)()1⎝⎠k ky+− Γ s0 e −

From [126, p.92] we have the following formula for the Hurwitz zeta function

1 ∞ uesyu−−−1(1) (4.4.43l) ς (,sy )= du ∫ u Γ−()se0 1 and hence we obtain

∞ 11n ⎛⎞n (4.4.43m) (,sy ) ς = ∑∑ns+1 ⎜⎟ nk==112(1)⎝⎠k ky+−

136

∞ 11n ⎛⎞n (,1)ss () ςς==∑∑ns+1 ⎜⎟ nk==112 ⎝⎠k k

Letting y =1 in (4.4.43j) we get

∞ 1 n ⎛⎞n xxueks∞ −−1 u Z (,1,1/2)xd==u s ∑∑nsu+1 ⎜⎟ ∫ nk==112(1)()⎝⎠k ksex+Γ0 − and using partial fractions this becomes

11∞∞u s−1 =−du usu−−1 e du ∫∫u Γ−Γ()sex00 () s

Reference to (4.4.25) and (4.4.26) then shows

Li() x =−s 1 x and therefore we get

∞ 1 n ⎛⎞n xk Li() x (4.4.43ma) Zx(,1,1/2)s 1 s ==−∑∑ns+1 ⎜⎟ nk==112(1)⎝⎠k kx+

∞ 11n ⎛⎞n (4.4.43mb) Zs(1,1,1/ 2) ( ) 1 s ==−∑∑ns+1 ⎜⎟ ς nk==112(1)⎝⎠k k +

Hence we see that

∞ 111n ⎛⎞⎡n ⎤ (4.4.43mc) 1 ∑∑nss+1 ⎜⎟⎢ − ⎥ = nk==112(1)⎝⎠⎣k kk+ ⎦ and with s = 2 we have

∞ 121n ⎛⎞n k + (4.4.43md) 1 ∑∑n+122⎜⎟ = nk==112(1)⎝⎠k kk+

Differentiation of (4.4.43mc) results in

∞∞1nn⎛⎞nn logkk 1⎛⎞ log(+ 1) (4.4.43me) ∑∑nsn++11⎜⎟= ∑∑ ⎜⎟ s nk==1122(1)⎝⎠kkkk nk == 11 ⎝⎠ +

137

As before, we get with y = 0 in (4.4.43j)

∞ 1 n ⎛⎞n xxuks∞ −1 (4.4.43n) Z (,0,1/2)xLix== () = du ss∑∑ns+1 ⎜⎟ ∫ u nk==112(⎝⎠k ksexΓ−)0 where we have recalled the series expansion for the polylogarithm in (3.55) and made reference to (4.4.25).

Letting y =1 in (4.4.43g) we get

∞ n ks∞ −−12u 11n ⎛⎞n xxtue (4.4.43o) Zs (,1,)xt== t ⎜⎟ s du 2(1∑∑k kts+−Γ)2(1)()∫ ⎡ −u ⎤ nk==11⎝⎠ 0 ⎣11−+()xe t⎦

We now divide (4.4.43g) by t and integrate to obtain

∞ qxnkn ⎛⎞n q xue∞ syu−−+1(1) (4.4.43p) ⎜⎟ s = dudt ∑∑nkytsk ()(1)()+−Γ∫∫⎡⎤−u nk==11⎝⎠ 00⎣⎦11−+()xe t

Reversing the order of integration we get

q xue∞∞syu−−+1(1) x q 1 du dt= usyu−−+1(1) e du dt ∫∫(1−Γts ) ( )⎡⎤−−uuΓ (s ) ∫ ∫⎡⎤ 00⎣⎦11−+()xet 0 0⎣⎦ 11−+()xett (1) −

Straightforward integration gives us

qq 111⎧ −A ⎫ −u dt=+⎨ ⎬ dt where Axe=+(1 ) ∫∫⎡⎤−u 111−−−AAtt⎩⎭ 00⎣⎦11−+()xe t (1) − t

1 =−−−[]log(1Aq ) log(1 q ) 1− A and hence

1 ∞ ∞ qxnkn ⎛⎞n −−+−−=uesyu−−1 log⎡⎤ 1 1xeq − u log(1 qdu ) ∫ {}⎣⎦() ∑∑⎜⎟ s Γ+()sn0 nk==11⎝⎠k ()ky

∞ Γ()y From (4.4.28) we have uesyu−−1 du= and therefore we get for y > 0 ∫ s 0 y

138 (4.4.43q)

Γ()yqx∞ ∞ nkn ⎛⎞n log(1−−que )syu−−1 log⎡⎤ 1 −+ 1 xeqdus − u =Γ ( ) s ∫ ⎣⎦() ∑∑⎜⎟ s y 0 nk==11nky⎝⎠k ()+

Letting q =1/2 we get

∞ ∞ 1 n ⎛⎞n xk (4.4.43r) −−=uesyu−−1 log⎡⎤ 1 xedu − u Γ ( s ) ∫ ⎣⎦ ∑∑ns⎜⎟ 0 nk==11nky2()⎝⎠k +

With the substitution p = e−u this becomes

1 ∞ 1 n ⎛⎞n xk (4.4.43s) ppxpdpsys−−11log log(1−=−Γ ) ( 1) s ( ) ∫ ∑∑ns⎜⎟ 0 nk==11nky2()⎝⎠k +

and with x =1 we have

1 ∞ 11n ⎛⎞n (4.4.43sa) pppdpsys−−11log log(1−=−Γ ) ( 1) s ( ) ∫ ∑∑ns⎜⎟ 0 nk==11nky2()⎝⎠k +

We immediately see that

d s−1 11 p yys−−−111log(1−=pdp ) p log p log(1− pdp ) s−1 ∫∫ dy 00 and this then directs our attention to the Beta function.

From (4.4.7) we have

1 B(,)yz=−∫ pyz−−11 (1 p ) dp , (Re (y ) > 0, Re (z ) > 0) 0

ΓΓ()()y z = Γ+()y z

Therefore we have

∂∂s−1 1 B(yz , )=− pys−−11 log p log(1 pdp ) s−1 ∫ ∂∂zy z=1 0

139 and accordingly we get

s−1 ∞ n ∂∂ Γ()()yz Γ s 1⎛⎞n 1 s−1 =−(1)() Γs ns⎜⎟ ∑∑k ∂∂zy Γ() y + z z=1 nk==11n2⎝⎠ () k+ y

Differentiation results in

∂Γ()()y Γzyzyzyzyz Γ ( + ) Γ′ ()() Γ −Γ ()() Γ Γ′ ( + ) = ∂Γyyz() + Γ2 ()yz +

(4.4.43t) =−+B(,)yz[ψψ () y ( y z )]

The second derivative results in

∂Γ2 ()()yz Γ (4.4.43u) =−++−+B(,)yz[][]ψψ′′ () y ( y z ) B ′ (,) yzψψ () y ( y z ) ∂Γ+yyz2 ()

=−++−+Byz(,){[]ψψ′′ () y ( y z )[] ψψ () y ( y z )2}

Similarly we see that

∂Γ3 ()()y Γz (4.4.43v) = ∂Γ+yyz3 ()

B(,)yz{[][]ψψ () y−++ ( yz )3 3 ψψ () y −+ ( yz )[] ψ′ () y − ψ ′ ( yz ++ )[] ψ ′′ () y − ψ ′′ ( yz + )}

Differentiation of (4.4.43t) with respect to z then gives us for s = 2

∂∂Γ()()yz Γ ∂ =−B(,)()(,)()()()()yzψψψψψ y ++ z Byz[] z −+ y z[] y −+ y z ∂∂zy Γ() y + z∂ z

ΓΓ()(1)1y and hence since By(,1)== we get Γ+(1)y y

∂∂Γ()()yz Γ 1 =−++−+{}ψψψψψ′(1)(1)(1)()(1)yyyy[][] −+ ∂∂zy Γ() y + zz=1 y

We have

140 1 ψ ()yy−+=−ψ ( 1) y

ψ (1)−+=−−+ψγψ (yy 1) ( 1)

∞ 1 From (E.16) we see that ′(1)y and we therefore obtain ψ +=∑ 2 k =0 (1)y ++k

∂∂Γ()()yz Γ 1⎧ ∞ 1γψ+ (y + 1)⎫

=−⎨ ∑ 2 + ⎬ ∂∂zy Γ() y + zz=1 y⎩⎭k =0 (1) y ++ k y and hence

11⎧⎫∞∞γψ++ (1)11y n ⎛n⎞ (4.4.43w) ⎨⎬−+=∑∑22n ∑⎜⎟ y ⎩⎭knk===011(1)yk++ y n 2⎝k ⎠ ( ky+ )

1 =−∫ p y−1 logppdp log(1 ) 0

With y =1, and using ψ (2)=−γ + 1, we obtain

∞∞11n ⎛⎞n 1 1 −=∑∑∑22n ⎜⎟ knk===011(2++kn ) 2⎝⎠k ( k 1) and this may be written as

∞ 11n ⎛⎞n 1 (4.4.43x) 2−=ς (2) =logx log(1 −xdx ) ∑∑n ⎜⎟ 2 ∫ nk==11nk2(1)⎝⎠k + 0

This is readily verified as follows: employing integration by parts we see that

logx log(1−=+−−+−−−−xdx ) 2 x (1 x )log(1 x ) x (1 x )log(1 x ) log x − Lix ( ) ∫ []2 and hence we get

1 ∫ logxxdx log(1−=− ) 2ς (2) 0

141 It should be noted that, having evaluated ∫ logx log(1− xdx ) , it is easy to determine

∫∫log2 x log(1−=xdx ) log x [log x log(1 − x )] dx with integration by parts.

In addition we may consider the integral

11∞∞11 ∫∫xxdxxdxyn−+−11log(1−=− ) ∑∑y =− 00nn==11nn()n+ y and differentiating s −1 times with respect to y we obtain

1 ∞ 1 xxxdxsys−−11log log(1−=−− ) ( 1)s ( 1)! ∫ ∑ s 0 n=1 nn()+ y

By the geometric series we have

s−1 1(−+x 1)s ∑(1)x +=r r=0 x and dividing by (x + 1)s we easily see that

ss−−11 11r 11 ssrs∑∑(1)x +=− = − (1)x +++rr==00(1)(1)xxxx and accordingly we obtain the partial fraction expansion

11ss−−12 111 1 s =−∑∑sr− =− − sr− xx(++++ 1) xrr==00 ( x 1) x x 1 ( x 1)

We therefore have

∞∞1111⎡⎤⎧ ∞∞ 1 ∞ 1⎫ =−− + ++... ∑∑ss⎢⎥⎨ ∑∑s−12 ∑⎬ nn==11nn(1)+++++⎣⎦ n n 1⎩⎭ nn == 11 (1)(1) n n n = 1 (1)n

ss =−1()1()∑∑[]ς rsr − =− ς rr==22 and hence we obtain

142 1 s ss−1 ⎡ ⎤ ∫ logx log(1−=−−−xdx ) ( 1) ( s 1)!⎢ s∑ς ( r )⎥ 0 ⎣ r=2 ⎦

Incidentally, as s →∞ we get Goldbach’s theorem

∞ ∑[]ς ()r −= 1 1 r=2

as noted in [126, p.142].

The above partial fraction expansion may be generalised as follows (see [30a] for more details)

1(ts−−11⎛⎞sr++11−−1)(rs ⎛⎞tr++ 1)

ts=+∑∑⎜⎟ trsr− ++ ⎜⎟ trsr− xx()++αααrr==00⎝⎠st−−11x ⎝⎠()xα and in particular we get

1111⎡⎤⎡ 1 1 1⎤ ss=−⎢⎥⎢ −s +21s−− ++... s 12⎥ nny()+++++ y⎣⎦⎣ nny yny() yny () y () ny⎦

Therefore we have

∞∞1 111111⎡⎤⎧ ∞∞ 1 1 ∞ 1⎫ ... ∑∑ss=−−⎢⎥⎨ ∑∑s +2112s−− ++ s ∑⎬ nn==11nny()+++++ y⎣⎦⎩ nny y nn == 11 () ny y () ny y n = 1 () ny ⎭

We have from [135, p.14]

∞∞⎡⎤11 1 1 ∑∑⎢⎥−=yy =++ψγ() nn==11⎣⎦nny++ nny() y

∞∞1111 and (,ry ) ∑∑rr=−=−rς r nn==10()ny++ () ny y y

Therefore we obtain

∞ 11⎡⎤⎡⎤ 1s 11 ()yr (,)y ∑∑ss=++−−⎢⎥⎢⎥ψγ ς rsr− +2 nr==12nn()+ y y ⎣⎦⎣⎦y y y

We also have

143 2 ∂∂ Γ()()yz Γ ∂ 2 =−++−+Byz(,)[]ψψ′′ () y ( y z )[] ψψ () y ( y z ) ∂∂zy2 Γ() y + z ∂ z {}

=−++−+−+Byz(,){ ψψψψψ′′ ( y z ) 2[ () y ( y z )][ ′′ () y ( y z )]}

+−+−++−+Byz(,)[]ψψ () z ( yz ){}[] ψ′′ () y ψ ( yz )[] ψψ () y ( yz )2

Therefore we see that

∂∂2 Γ()()yz Γ 2 = ∂∂zy Γ() y + z z=1

1 {}−+−−+ψψψψ′′(1)2()(1)(1)yyyy[]′ + y

1 2 +−+−++−+[]ψψ(1)(1)()(1)()(1)yyyyy[] ψ′′ ψ[] ψ ψ y { }

21111⎧⎫∞∞ (1)y =+−++⎨⎬∑∑322[]γψ yykyyky⎩⎭kk==00(1)++ (1) ++

Therefore we obtain

(4.4.43y)

11111⎧⎫∞∞ ∞1n ⎛n⎞ 1 (1)y ⎨⎬∑∑32+−++=2[]γψ −∑∑n ⎜⎟ 3 y ⎩⎭kk==00(1)ykyyky++ (1) ++ nk=1n2=1⎝k ⎠ ( ky+ )

With y =1, and using ψ (2)=−γ + 1, we obtain

⎧⎫⎛∞∞1111 ∞n n⎞ 1 −+=⎨⎬∑∑33 ∑∑n ⎜⎟ 3 ⎩⎭⎝kk==00(2)(2)kk++ nk == 11 nk 2k ⎠ (1)+ or alternatively ∞ 11n ⎛⎞n 3(2)(3) −−=ςς∑∑n ⎜⎟ 3 nk==11nk2(1)⎝⎠k +

We may also obtain (4.4.43y) by differentiating (4.4.43w) whence we get:

144

11⎧⎫⎧⎫∞∞γψ++ (1)11(1)(1)yyψ′ y+−−+ γψy 2 −−22⎨⎬∑∑ + +⎨32+ ⎬= yyky⎩⎭kk==00(1)++ yyk⎩ (1)++ y ⎭

∞ 11n ⎛⎞n 2 − ∑∑n ⎜⎟ 3 nk==11nky2()⎝⎠k +

In (4.4.43i) let tt→− and x =−1 divide by t and integrate to obtain

∞ utnkn ⎛⎞n (1)− u 1 ⎛⎞ (4.4.43z) (1)−=n Li dt ∑∑⎜⎟ s ∫ s ⎜⎟ nk==11nkttt⎝⎠k 0 (1++ )⎝⎠ 1

Reference to Volume I shows that

u 1 ⎛⎞tu ⎛ ⎞ Li dt= Li ∫ ss⎜⎟+1 ⎜ ⎟ 0 tt(1++ )⎝⎠ 1 t ⎝ 1 + u ⎠ and we accordingly obtain

⎛⎞uu∞ nkn ⎛⎞n (1)− (4.4.43za) Li (1)n s+1 ⎜⎟=−∑∑⎜⎟ s ⎝⎠1+ unknk==11⎝⎠k

With, for example, u =1/2 (4.4.43za) becomes

∞ (1)−−nkn ⎛⎞n (1) (4.4.43zb) Li 1/3 s+1 ()= ∑∑ns⎜⎟ nk==11nk2 ⎝⎠k and compare this with (4.4.43if).

u Letting t = , we have 1+ u

n ∞ 1(⎛⎞t n ⎛⎞n −1)k (4.4.43zc) Li() t ( 1)n s+1 =−∑∑⎜⎟⎜⎟ s nk==11nt⎝⎠1− ⎝⎠k k

Note that in (4.4.45) we have shown that

∞ 1 n ⎛⎞n t k Li t s+1 ()= ∑∑ns⎜⎟ nk==11nk2 ⎝⎠k

145 and we therefore have

n ∞∞1(⎛⎞ttnn⎛⎞nn−1)kk1⎛⎞ (4.4.43zd) Li() t ( 1)n s+1 =−∑∑∑∑⎜⎟⎜⎟ s = ns⎜⎟ nknk====1111nt⎝⎠12− ⎝⎠kk knk⎝⎠

n k +1 ⎛⎞n (1)− (1) Letting s =1 in (4.4.43za) and noting that ∑⎜⎟ = H n we get k =1 ⎝⎠k k

∞ (1) ⎛⎞u nnHn (4.4.43ze) Li2 ⎜⎟=−∑(1) u ⎝⎠1+ unn=1

We now recall (3.31) and (3.111a)

∞ (1) 1 2 Hn n log (1−+uLiu )2 ( ) =∑ u , u <1 2 n=1 n

⎛⎞u 1 2 Li22⎜⎟=−log (1 + u ) − Li ( − u ) ⎝⎠12+ u and note the equivalence via (3.111b).

Alternatively, letting s = 2 in (4.4.43za) and noting that

k n ⎛⎞n (1)− 1 2 =+⎡ HH(1) (2) ⎤ ∑⎜⎟ 2 ⎢()nn⎥ k =1 ⎝⎠k k 2 ⎣ ⎦ we get

n ⎛⎞u 1(1)∞ − 2 (4.4.43zf) Li =+⎡ H(1) H (2) ⎤ un 3 ⎜⎟∑ ⎢()nn⎥ ⎝⎠12+ unn=1 ⎣ ⎦

With u =1 we obtain

n 1(1)∞ − 2 (4.4.43zg) Li 1/2 =+⎡ HH(1) (2) ⎤ 3 () ∑ ⎢()nn⎥ 2 n=1 n ⎣ ⎦

We also note from (3.46a) that

n ∞ 2 (1)− (1)n 1 3 ∑ ()Hun = −++−−−+log(1)() uLiuLiu32 ()log(1) u n=1 n 3

146 and (3.34) gives us

u Li() x ∞ H (2) ∫ 2 dx= ∑ n un 0 xx(1− ) n=1 n

We have uuuLi() x Li () x Li () x ∫∫∫222dx=+ dx dx 000xx(1−− ) x1 x

uu2 Li() x u log (1− x ) 2 dx=−log(1 − x ) Li ( x ) − dx ∫∫2 o 011− xx

The final integral is evaluated in (4.4.100gii) in Volume III and we see that u Li() x 2 dx=−− Li( u ) log(1 u ) Li ( u ) − log u log2 (1 −−− u ) 2log(1 u ) Li (1 −+−− u ) 2 Li (1 u ) 2ς (3) ∫ 32 23 0 xx(1− ) and therefore we have

(4.4.43zh)

∞ H (2) n n 2 ∑ uLiu=−−32( ) log(1 uLiu ) ( ) − log u log (1 −−− u ) 2log(1 uLiu ) 23 (1 −+−− ) 2 Liu (1 ) 2ς (3) n=1 n

Letting uu→− we have an expression involving complex numbers

∞ (2) nnHn 2 ∑(1)−=−−+−−−+uLiu32 ( )log(1) uLiu ( )log()log(1) u u n=1 n

− 2log(1 +uLi ) (1 ++ u ) 2 Li (1 +− u ) 2ς (3) 23 From (4.4.44ic) we have

n 11∞ u 2 ⎡⎤H (1)+=−−Hu (2) log2 (1 ) loguu −− log(1 )Li (1 −+−−−u )Li (1uL )ς (3)i (u ) ∑ ⎢⎥()nn 23 3 22n=1 n ⎣⎦

The second leg of (4.4.43) was demonstrated in (3.67e) in Volume I where we showed that n ∞∞n ⎛⎞n xxk 11⎡t ⎤ t n ∑∑⎜⎟ ss= ∑ ⎢ ⎥ nk==11⎝⎠k ()1()(1)ky+−+− t n =1 ny⎣ t ⎦

147

n 11∞ ⎡⎤xt 1

=−∑ s ⎢⎥ s 1()(1)(1)−+−tnyn=0 ⎣⎦ t − ty and therefore

∞ n ⎛⎞n xxk 11⎛t ⎞ tsn ,,y ∑∑⎜⎟ s =Φ⎜ ⎟ − s nk==11⎝⎠k ()1ky+−− t⎝ (1) t ⎠ (1) − ty

The following identity [75aa] is easily derived

1 zzsyΦ+=Φ−()(),, 1 zsy ,, ys and therefore we have

11⎡⎤⎛⎞xt xt ⎛ xt ⎞ ⎢⎥Φ−=Φ+⎜⎟,,sy s 2 ⎜ ,,sy 1 ⎟ 1−−tt⎣⎦⎝⎠ (1) yt (1)(1) − ⎝ − t ⎠

Hence we obtain

∞ n ⎛⎞n xxtxtk ⎛ ⎞ tsyn ,, 1 ∑∑⎜⎟s = 2 Φ+ ⎜ ⎟ nk==11⎝⎠k ()(1)(1)ky+− t⎝ − t ⎠

Letting t =1/2 we get

∞ 1 n ⎛⎞n xk (4.4.43zi) xxsy,, 1 ∑∑ns+1 ⎜⎟ = Φ+() nk==112()⎝⎠k ky+

With xy=−1 and =0 we get

∞ n ⎛⎞n (1)−−−k tt⎛⎞ tsn ,,1 ∑∑⎜⎟ s =Φ2 ⎜⎟ nk==11⎝⎠k ktt(1−− )⎝⎠ 1

and since Lis () z=Φ z() z ,,1 s we have another proof of (4.4.43ia)

∞ n ⎛⎞n (1)−−k 1 ⎛⎞t tLn i ∑∑⎜⎟ s = s ⎜⎟ nk==11⎝⎠k ktt11− ⎝⎠−

148 From (4.4.43sa) we see that with y = 0 we have

1 logs−1 p ∞ 1n ⎛⎞n 1 log(1−=−Γpdp ) ( 1)s ( s ) ∫ ∑∑ns⎜⎟ 0 pnnk==112 ⎝⎠k k

We note from (4.4.45) that

∞ 1 n ⎛⎞n xk Li() x s+1 = ∑∑ns⎜⎟ nk==11nk2 ⎝⎠k and hence we get

1 logs−1 p ∫ log(1−=−Γ+pdp ) ( 1)s ( s )ς ( s 1) 0 p

which concurs with (4.4.91ga). Similarly, using (4.4.43s) we note that

1 logs−1 p (4.4.43zj) log(1−=−Γxpdp ) ( 1)s ( sLi ) ( x ) ∫ s+1 0 p

in accordance with (4.4.38e).

Differentiating the above with respect to x results in

1 logs−1 p Li() x (4.4.43zk) ∫ dp=−(1)s+1 Γ () s s 0 1− xpx

which agrees with (4.4.38b). A further differentiation results in

1 pplog s−1 Li() x− Li () x (4.4.43zl) dp=−(1)s+1 Γ () s ss−1 ∫ 22 0 (1− xp ) x

Integration of (4.4.43zj) gives us

u dx1 log s−1 p log(1−=−Γxpdp ) ( 1)s ( sLi ) ( u ) ∫∫ s+2 00xp

and we therefore have

149 1 logs−1 pLi ( pu ) (4.4.43zm) 2 dp=−(1)s+1 Γ () s Li () u ∫ s+2 0 p in accordance with (4.4.42i). 

We easily determine that

2 ∂Γ()()yz Γ 2 =−++−+Byz(,)[]ψψ′′ () y ( y z )[] ψψ () y ( y z ) ∂Γ+yyz2 () {}

2 ∂Γ()()yz Γ 1 2 =−++−+ψψ′′(1) (1zz ) ψψ (1) (1 ) 2 {}[][] ∂Γ+yyzz()y=1

[ψψ′(1)−+′ (1zz )] [ ψψ (1) −+ (1 )] =+[]ψψ(1) −+ (1z ) zz

As noted in [30a, p.34] we may consider this in the case where z → 0 : we have B(,1)1/yy= .

We therefore have the limit

∂Γ2 ()()yz Γ lim=−ψ ′′ (1) = 2ς (3) z→0 2 ∂Γ+yyz()y=1 and we also see that

∂Γ2 ()()yz Γ 1 =−pppdpyz−−12log (1 ) 1 2 ∫ ∂Γ+yyz()0

∂Γ22()()y Γzp1 log lim = dp z→0 2 ∫ ∂Γ+yyz()y=1 0 (1) − p

Therefore we see that

1 log2 p ∫ dp = 2(3)ς 0 (1− p )

Let us now consider another generalisation of Theorem 4.2.

150 (ii) Theorem 4.2(b):

We have for s > 0 and t < 1

∞ n ⎛⎞n xueks1 ∞ −−1 yu (4.4.44) tdn = u ∑∑⎜⎟ s ∫ −u nk==00k ()ky+Γ () s ⎡ xe t⎤ ⎝⎠ 0 ⎣11−+()⎦ where the summation starts at nk= 0 and = 0 in this case.

Proof:

As before we have 11∞ (4.4.44a) = uesuky−−1() + du s ∫ ()ky+Γ () s0

We now consider the finite sum set out below (where the summation now starts at k = 0 and we specify that y is neither zero nor a negative integer)

n ⎛⎞n xk (4.4.44b) Sxy0 (, ) n = ∑⎜⎟ s k =0 ⎝⎠k ()ky+

(where the superscript highlights the fact that the summation starts this time at k = 0 )

Now combine (4.4.44a) and (4.4.44b) to obtain

n ⎛⎞n xk n ⎛⎞n 1 ∞ Sxy0 (, )= = xk . uesuky−−1() + du n ∑⎜⎟ s ∑⎜⎟ ∫ k =0 ⎝⎠k ()ky+ k =0 ⎝⎠k Γ()s 0

∞ n 1 sukyk−−+1(⎧⎛⎞n )⎫ = ∫ uexdu∑ ⎨⎜⎟ ⎬ Γ()s 0 k =0 ⎩⎭⎝⎠k

∞ n 1 suyu−−−1 ⎧⎫⎛⎞n k = ∫ uexed∑ ⎨⎬⎜⎟ ⎣⎦⎡⎤ u Γ()s 0 k =0 ⎩⎭⎝⎠k

Applying the binomial theorem we have

∞ 1 n (4.4.44c) Sxy0 (, )= uxeedusuy−−−1 1+ u n ∫ () Γ()s 0

151 0 We now consider the following function Zs (,xyt ,)which, for t < 1, is defined by the series

1 ∞ n ⎛⎞n xk (4.4.44d) 0 t n Zs (,xyt ,)= ∑∑⎜⎟ s 2()nk==00⎝⎠k ky+

(where the superscript again highlights the fact that the summation starts this time at 0 n = 0 ). We then substitute the relation obtained for Sxyn (, ) to deduce

∞ 11∞ n Z 0 (,xyt ,)= tuxeeduns−−−1 1+ uyu s ∑ ∫ () 2()n=0 Γ s 0

∞ 11 ∞ n (4.4.44e) =+∫ uxetedusu−−−1 ∑()1 nyu 2()Γ s 0 n=0

(where we have again assumed that interchanging the order of summation and integration is permitted). Use of the geometric series simplifies (4.4.44e) to

11∞ n ⎛⎞n xueks1∞ −−1 yu (4.4.44f) Z 0 (,xyt ,)== tn du s ∑∑⎜⎟ s ∫ −u 2()2()nk==00k ky+Γ s ⎡ xe t⎤ ⎝⎠ 0 ⎣11−+()⎦

We have for t =1/2

∞ 11n ⎛⎞n xueks∞ −−−1(1)yu Z 0 (,xy ,1/2)==du s ∑∑nsu+1 ⎜⎟ ∫ nk==002()()⎝⎠k ky+Γ s0 e − x

We then note from [126, p.121] that the Hurwitz-Lerch zeta function Φ (x ,sy , ) is defined by ∞ xn 1 ∞ uesuy−−1(1) − (4.4.44fi) Φ=(,,xsy ) = du ∑ s ∫ u n=0 ()ny+ Γ−()sex0 and hence we see that

∞∞1 n ⎛⎞n xxkn (4.4.44fii) Zxy0 (, ,1/2) (,, xsy ) s ==Φ=∑∑ns+1 ⎜⎟ ∑ s nk==002()⎝⎠k ky++ n = 0 () ny

With y =1 we get

152 ∞ 11n ⎛⎞n xuks∞ −1 Z 0 (,1,1/2)xd===uLi ()x s ∑∑nsu+1 ⎜⎟ ∫ s nk==002(1)()⎝⎠k ksex+Γ0 −

With partial fractions we obtain

∞∞∞∞uesu−−12 11 u s −1 us −1 1 us −1 du=+ du du + du ∫∫∫∫uuuu22 2 0000ex−− xe x e xex

111 1 =Γ()ssLixs + Γ () + () Γ () x 2s xx23s

Therefore we get

11∞ uesu−−12 111 du=++ Li() x ∫ us23s Γ−()sex0 x2 xx

We have ∞ xuens1 ∞ −−−1(1)au Φ=(,,x sa ) = du ∑ su∫ n=0 ()na+Γ () s0 e − x and with a = 3 we get

1 ∞ uesu−−12 ∞ xn du=Φ(,,3) x s = ∫ us∑ Γ−()sex0 n=0 (n + 3)

∞ xn 11 1 1 Li() x ∑ ss=++23s n=0 (3)2nxxx+

∞ xxxxxxxxn 1123 ⎧ 3456⎫

∑ sssss=++++=...3 ⎨ ssss ++++ ...⎬ n=0 (3)3456nx+ ⎩⎭ 3456

11⎧ x x2 ⎫ =−+33Li3 () x ⎨ ss⎬ xx⎩⎭12

1111 =−−Li() x x323 xx2s

We note from (4.4.44fi) (and we will also see this in (4.4.79)) that

153 ∞ 1(1)n ⎛⎞n − k ∞ (1)− n (1,,)sy ∑∑ns+1 ⎜⎟ = ∑ s = Φ− nk==002()⎝⎠k ky+ n=0 ()ny+ where Φ(zsy , , ) is the Hurwitz-Lerch zeta function. With y = 1 this becomes the Hasse/Sondow identity (3.11)

∞ 1(1)1n ⎛⎞n − ks∞ u −1 ς ()sd==u a ∑∑nsu+1 ⎜⎟ ∫ nk==002(1)()1⎝⎠k kse+Γ0 +

Differentiating (4.4.44f) with respect to s we get

(4.4.44g)

∞ n ⎛⎞n xkyskslog(+Γ )′ ( ) ∞∞ ue−−11yu1 ueus −−yulog tdn =−udu ∑∑⎜⎟ s 2 ∫∫−−uu nk==00k ()ky+ΓΓ()s ⎡ xe t⎤⎡()s xe t ⎤ ⎝⎠ []00⎣11−+()⎦⎣11 −+() ⎦

Letting ts==1/ 2, 1 and x =−1 we have

∞ 1n ⎛⎞n (−+ 1)ku log(ky ) ∞∞ e−−(1)y e −−u(1)ylog u =−γ du − du ∑∑nu+1 ⎜⎟ ∫∫u nk==002()1⎝⎠k ky+++00 e e 1

Letting y =1 we get from the Hasse/Sondow identity

∞ 1(1)log(1)1n ⎛⎞n −+k ku∞∞ log ςγ′ (1) ==−−du du a ∑∑nuu+1 ⎜⎟ ∫∫ nk==002(1)11⎝⎠k kee+++00

We easily see that

∞∞−u 1 −e ∞ du=− du =−log(1 + e−u ) = log 2 ∫∫uu− 0 00ee++11

1 ∞ u s−1 Since ς ()sdu= we also see that a ∫ u Γ+()se0 1

∞ 1 ς (1)==du log 2 a ∫ u 0 e +1

154 We note that differentiating (3.11) results in (see also (4.4.112x))

∞ 1n ⎛⎞n (−+ 1)k log(k 1) ′ ()s ∑∑ns+1 ⎜⎟ =−ς a nk==002(1)⎝⎠k k + and hence

∞ 1n ⎛⎞n (−+ 1)k log(k 1) ′ (1) ∑∑n+1 ⎜⎟ =−ς a nk==0021⎝⎠k k +

We therefore see that

∞ log u (4.4.44h) du =−γςlog 2 + ′ (1) ∫ u a 0 e +1

We know from (C.61) of Volume VI that

⎡⎤log 2 ςγa′ (1)=− log 2 ⎣⎦⎢⎥2 and hence we have

∞ logu log2 2 (4.4.44i) du =− ∫ u 0 e +12

It should be noted that the above result could have been obtained directly from (4.4.42c) by letting s =1.

Making the substitution x = e−u we get

∞ logux1 log log(1/ ) du= dx ∫∫u 00ex++11

As shown in Appendix C of Volume VI, Adamchik [2a] has considered logarithmic integrals of this type and has shown that

1 x p−1 ⎛⎞1γ ++ log(2np ) ⎡ ⎛ ⎞ ⎛ np ⎞⎤⎡1 ⎛ p ⎞ ⎛ np + ⎞⎤ log log dx =−+−ψψ ς′′1, ς 1, ∫ n ⎜⎟ ⎢ ⎜ ⎟ ⎜ ⎟⎥⎢ ⎜ ⎟ ⎜ ⎟⎥ 0 12+ xx⎝⎠ n⎣ ⎝2 nnnnn ⎠ ⎝2 ⎠⎦⎣2 ⎝2 ⎠ ⎝2 ⎠⎦ and in particular we have another proof of the above integral

155 1 11log2⎛⎞ 2 ∫ log log ⎜⎟dx =− 0 12+ xx⎝⎠

A further differentiation of (4.4.44g) with respect to s gives us

∞ n ⎛⎞n xk log2 (ky+ ) t n −=∑∑⎜⎟ s nk==00⎝⎠k ()ky+

22 [ΓΓ−ΓΓ()ss] ′′ () 2 () ss[ ′ ()] ∞∞uesyu−−11Γ′() s uesyu−− log u du+ 2 du ΓΓ()ss42∫∫⎡ −−uu⎤⎡⎤ () [] 00⎣11−+()xe t⎦⎣⎦[] 11 −+() xe t

1l∞ uesyu−−12og u − du Γ()s ∫ ⎡⎤−u 0 ⎣⎦11−+()xe t

Letting tsy===1/ 2, 1, 1 and x = −1 we have

∞ n k 2 ∞∞∞2 1⎛⎞n (−+ 1) log (kd 1) 2 ulogu log u −=Γ−Γ+Γ−′′(1) 2 ′ (1) 2 ′ (1) du du ∑∑n+1 ⎜⎟ {}[]∫∫∫uuu nk==002(1⎝⎠k ke++++) 000111ee

From (E.16d) we have Γ=+′′(1)γ 2 ς (2) and hence we obtain

∞ 1n ⎛⎞n (−+ 1)k log22 (ku 1) ∞ log =−⎡⎤γς22(2) log 2 − γ log 2 + du ∑∑n+1 ⎜⎟ ⎣⎦ ∫ u nk==002(1⎝⎠k ke++) 0 1

Therefore we get ∞ log2 u (4.4.44j) du =−−++ςγςγ′′ (1)⎡⎤2 (2) log 2 log 2 ∫ u a ⎣⎦ 0 e +1

Making the substitution x = e−u we get

2 ∞ log2 u 1 1⎡⎤⎛⎞ 1 du= log log dx ∫∫u ⎢⎥⎜⎟ 00exx++11⎣⎦⎝⎠

This type of integral is examined in Appendix C of Volume VI where we obtained the same result in a different way.

We note from [126, p.121] that the Hurwitz-Lerch zeta function Φ (x ,sy , ) is defined by

156 ∞ xn 1 ∞ uesuy−−1(1) − Φ(,,x sy )= = du ∑ s ∫ u n=0 ()ny+ Γ−()sex0 and hence ∞ (1)− nuy∞ e−−(1) Φ− ( 1,1,y ) ==du ∑ ∫ u n=0 ()ny++0 e 1

Differentiating the Hurwitz-Lerch zeta function with respect to s we get

∂ ∞ xn log(ny+ ) Γ′()sue∞∞suy−−1(1) − 1 uesuy−−1(1) −log u Φ(,,x sy )=− =− du + du ∑ s 2 ∫∫uu ∂s n=0 ()ny+ []Γ()s 00ex−Γ() s ex − and with s =1 and x =−1 we have

∂−+∞ ( 1)nu log(ny ) ∞∞ e−−(1)yu e −−(1)ylog u Φ=−=(,,x s y ) γ du − du ∑ ∫∫uu ∂+++sns=1 n=0 ()y00ee 1 1

∞ 1(1)n ⎛⎞n − k ∞ (1)− n Since (1,,)sy ∑∑ns+1 ⎜⎟ = ∑ s = Φ− nk==002()⎝⎠k ky+ n=0 ()ny+ we deduce that

∞∞1n ⎛⎞n (−+ 1)kn log(ky ) ( −+ 1) log(ny ) (4.4.44k) ∑∑n+1 ⎜⎟ = ∑ nk==002 ⎝⎠k ky++ n = 0ny

We now integrate (4.4.44f) to obtain

∞ qxnk+−−11n ⎛⎞n 1 q ∞ ue syu = dudt ∑∑⎜⎟ s ∫∫ −u nk==00nkys++Γ1()()k ⎡⎤xe t ⎝⎠ 00⎣⎦11−+()

Reversing the order of integration we get

111qq∞∞uesyu−−1 dudt= usyu−−1 e du dt ΓΓ()ss∫∫⎡⎤−−uu ()∫ ∫ ⎡⎤ 00⎣⎦11−+()xe t 0 0⎣⎦11 −+() xe t

syu−−1 −u 1 ∞ ue log⎡ 1−+( 1 xeq) ⎤ =− ⎣ ⎦ du ∫ −u Γ+()sx0 1 e Therefore we have

157 syu−−1 −u 1 ∞ ue log⎡⎤ 1−+( 1 xeq) ∞ qxnk+1 n ⎛⎞n (4.4.44l) −=⎣⎦du ∫ −us∑∑⎜⎟ Γ+()sx0 1enk==00 nk ++ 1⎝⎠k (y )

With the substitution zxe=+ 1 −u we obtain

y ⎛⎞11−−zzs−1 ⎛⎞ ∞+syu−−1 ⎡⎤−u 1 x ⎜⎟log ⎜⎟ log[] 1− zq uelog 1−+() 1 xeq xx ⎣ ⎦du=−(1)s−1 ⎝⎠ ⎝⎠ dz ∫∫−u 011(1)+−xe z z

∞ qxnk+1 n ⎛⎞n

= ∑∑⎜⎟ s nk==00nky++1()⎝⎠k

With q =1, x =−1 we get

∞ ues − yu ∞ 1(1)n ⎛⎞n − k ς (1,)su+=−us du = ⎜⎟ ∫ 11(1)−++en∑∑k k 0 nk==00⎝⎠

Differentiating (4.4.44l) with respect to q results in (4.4.44f)

∞ syu−−1 ∞ n k 1 ue n ⎛⎞n x du= q ⎜⎟ s ∫ −u ∑∑k Γ+()sk0 11−+()xe q nk==00⎝⎠ (y )

We note from (4.4.44) that

∞ n ⎛⎞n xueks1 ∞ −−1 yu tdn = u ∑∑⎜⎟ s ∫ −u nk==00k ()ky+Γ () s ⎡ xe t⎤ ⎝⎠ 0 ⎣11−+()⎦ and integration gives us

∞ vxnkn ⎛⎞n 1 v ∞ ue syu−−1 = dt du ∑∑⎜⎟ s ∫∫ −u nk==00nkys++Γ1()()k ⎡ xe t⎤ ⎝⎠ 00⎣11−+()⎦

1 ∞ v dt = uesyu−−1 du Γ()s ∫∫⎡ −u ⎤ 00⎣11−+()xet⎦

158 −u 1 ∞ uesyu−−1 v −+(1 xedt) =− du Γ()s ∫∫1+ xe−u ⎡ −u ⎤ 00()⎣11−+()xet⎦

−u 1 ∞ uesyu−−1 v −+(1 xedt) =− du Γ()s ∫∫1+ xe−u ⎡ −u ⎤ 00()⎣11−+()xet⎦

We have

v −+()1 xe−u dt =−+log⎡⎤ 1() 1 xev−u ∫ ⎡⎤−u ⎣⎦ 0 ⎣⎦11−+()xe t and therefore

syu−−1 −u ∞ vxnkn ⎛⎞n 1 ∞ ue log⎡ 1−+( 1 xev) ⎤ =− ⎣ ⎦ du ∑∑⎜⎟ su∫ − nk==00nkys++Γ1()()1⎝⎠k 0 + xe

Letting x →−x gives us

∞ 1 n ⎛⎞n xuks1∞ −−1exyu log[e −u ] (1)−=−k du ∑∑⎜⎟ su∫ − nk==00nk++Γ−1()()1⎝⎠k ysx0 e

1l∞∞uesyu−−−og x us1 e yu =−du du ∫∫−−uu Γ−()sxesxe00 1 Γ−() 1

1l∞∞uesyu−−(1) ogx us −1 e −− (1) yu =−du du ∫∫uu Γ−Γ()sex00() s ex − having regard to (4.4.44fi)

1 ∞ uesyu−−−1(1) Φ=(,,x sy ) du ∫ u Γ−()sex0 we then obtain

∞ 1 n ⎛⎞n xk (1)k sxsy (, 1,) log xxsy (,, ) ∑∑⎜⎟−=Φ+−Φs nk==00nky++1()⎝⎠k and with x =1 we have

159

∞ 1(1)n ⎛⎞n − k ssy(1, 1, ) ∑∑⎜⎟ s =Φ + nk==00nky++1()⎝⎠k

Letting ss→−1 gives us

∞ 1(1)n ⎛⎞n − k (1)(1,,)ssy ∑∑⎜⎟ s−1 =−Φ nk==00nky++1()⎝⎠k and since

∞ xn (,, ) Φ=xsy ∑ s n=0 ()ny+ we obtain the Hasse identity

11∞∞n ⎛⎞n (1)1− k (,sy ) ∑∑⎜⎟ ss−1 == ∑ ς sn−+11nk==00⎝⎠k ()() kyny + n = 0 +

Differentiation results in

∞ vxkynkn ⎛⎞n log(+ )

∑∑⎜⎟ s = nk==00nky++1()⎝⎠k

syu−−11−u syu−− −u 1(∞∞uelog u log⎡⎤ 1−+() 1 xevψ s) ue log ⎡⎤ 1 −+( 1 xev) −+⎣⎦du ⎣⎦du ∫∫−−uu Γ+()sx001 e Γ+() sx 1 e

∞ vknkn ⎛⎞n (−+ 1) log(yu ) 1∞∞sy logues−−(1)uψ () u se−−(1)yu =−du du ∑∑⎜⎟ su∫∫ u nk==00nk++Γ−Γ−1()()1()1⎝⎠k ysese00

We note from (4.4.13b) that

1 ttys−1 log ∞ n ⎛⎞n xk dt=−(1)!sn s v (1)− k ∫ ∑∑⎜⎟ s+1 0 1(1)−−vxt nk==00⎝⎠k (ky+ )

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