Some integrals involving the Stieltjes constants: Part II
Donal F. Connon
11 April 2011
Abstract
Some new integrals involving the Stieltjes constants are developed in this paper.
CONTENTS Page
1. Introduction 1
2. A useful formula for the Stieltjes constants 2
3. An application of the Abel-Plana summation formula 6
4. A family of integral representations of the Stieltjes constants 16
5. An application of the alternating Hurwitz zeta function 35
1. Introduction
The Stieltjes constants γ p ()x are the coefficients of the Laurent expansion of the Hurwitz zeta function ς (,sx ) about s =1
∞∞11(1)− p (1.1) (,sx ) ()(x s 1)p ςγ==+∑∑s p − np==00()nx+− s 1 p !
where γ p ()x are known as the generalised Stieltjes constants and we have [41]
(1.2) γ 0 ()x =−ψ ()x
where ψ ()x is the digamma function.
With x =1 equation (1.1) reduces to the Riemann zeta function
∞ p 1(1)− p ςγ()ss=+∑ p ( − 1) sp−1!p=0
As previously noted in [20], using (1.1) it is easily seen that the difference of two Stieltjes constants may be represented by p p ∂ (1.3) γγpp()x −=− ()ys (1)lim [(,)ςx − ς (,sy )] s→1 ∂s p
2. A useful formula for the Stieltjes constants
We recall Hasse’s formula [31] for the Hurwitz zeta function which is valid for all s ∈C except s =1 (in this form, it is valid in the limit as s →1)
∞ 1(j ⎛⎞j −1)k (2.1) (1)(,)ssx −=ς ∑∑⎜⎟ s−1 jk==00jx++1()⎝⎠k k and differentiation with respect to x gives us
∂−∞ 1(j ⎛⎞j 1)k (,sx ) ς =−∑∑⎜⎟ s ∂+x jk==00jx1(⎝⎠k +k) and we then have
∂∂nk∞ 1(1)log(j ⎛⎞j −nx +k) (2.2) (,sx ) (1)n+1 nsς =− ∑∑⎜⎟ ∂∂sx jk==00j+1()⎝⎠k xk+
We note that the partial derivatives commute in the region where ς (,)sx is analytic and hence we have
∂∂nn ∂∂ ςς(,sx )= (, sx ) ∂∂sxnn ∂∂ xs
Evaluation of (2.2) at s = 0 results in
∞ j ∂ ()nn+1 1 ⎛⎞j kn (2.3) ς (0,x )=− ( 1) ∑∑⎜⎟(− 1) log (xk + ) ∂+xjjk==001 ⎝⎠k
We may write (1.1) as
∞ n 1(1)− n (2.4) ςγ(1,)sx+=+∑ n ()xs snn=0 ! and we have the Maclaurin expansion
2 1 ∞ (1)− n+1 R ()x (2.5) ς (,sx )=+∑ n sn sn−1!n=0 where
n n+1 ∂ Rxn ()=− (1)n ς (,)sx ∂s s=0
R()xRx= 2 () is referred to as the Deninger R − function, after Deninger [28] who introduced it in 1984.
Differentiating (2.5) with respect to x gives us
∂ ∞ (1)− n+1 R′ ()x (2.6) ς (,sx )= ∑ n sn ∂xnn=0 ! We note that
∂ ςς(,sx )=− s ( s + 1,) x ∂x and, using (2.4), this is equal to
∞ n+1 (1)− n+1 (2.7) =−1( +∑ γ n x)s n=0 n! and comparing the coefficients of (2.6) and (2.7) Chakraborty, Kanemitsu and Kuzumaki [12] deduced the important identity
∂∂n ′ n+1 (2.8) Rnn()xs=− (1)n ςγ (,)x=−n−1 ()x ∂∂sx s=0
This may be compared with the more familiar formula for the Stieltjes constants where the limit is evaluated at s =1
∂n−1 ⎡⎤1 (2.9) (1)−−=n+1 ςγ (,)sx −()x n−1 ⎢⎥n−1 ∂−ss⎣⎦1 s=1
We see from (2.3) that (2.8) is equal to
∞ j 1 ⎛⎞j kn =−∑∑⎜⎟(1)log(x +k ) jk==00j +1 ⎝⎠k
3 and hence we easily deduce that
∞ j 11⎛⎞j kn+1 (2.10) γ n ()x =− ∑∑⎜⎟(1)log(−xk + ) nj++11jk==00⎝⎠k which was previously obtained in 2007 in [20].
In [20, Eq. (4.3.231)] we also noted that Rnn′()xnx= − γ −1 () in the equivalent form for x > 0
x (1)− n+1 γςς()tdt=−⎡ (1)nn++(0,x )(1) (0)⎤ ∫ n ⎣ ⎦ 1 n +1 but the usefulness of this simple identity was not then fully appreciated by the author.
The formula (2.8) features throughout the rest of this paper where it is used to simplify the derivation of some known identities and also to produce some new ones.
Remark (i)
The equivalent formula to (2.5) as reported by Chakraborty et al. [12] did not include the 1 term ; it seems to me that it should be so included if only to concur with the analysis s −1 previously carried out by Sitaramachandrarao [37] in 1986 where he considered the Maclaurin series for the Riemann zeta function
1 ∞ (1)− n δ (2.11) ς ()ss+=∑ n n 1!− snn=0 where
m m ⎡⎤nn1 n (2.12) δn =−−lim logkxdx log log m m→∞ ⎢⎥∑ ∫ ⎣⎦k =1 1 2
nn() =−(1)⎣⎦⎡⎤ς (0) +n !
1 This additional term of course vanishes when equation (2.5) is differentiated with s −1 respect to x and thus (2.8) continues to remain valid.
4 Remark (ii)
We note from (2.8) that with n =1
∂∂ R10′()xsxx==ςγ (,)− ()=ψ ()x ∂∂sx s=0 or equivalently
∂ ςψ′(0,x )= (x ) ∂x and integration then results in
(2.13) ς ′′(0,tt )−=Γς (0) log ( )
We have Legendre’s duplication formula for the gamma function [38, p.7]
⎛⎞11 logΓ=Γ+Γ++− (2tt ) log ( ) log⎜⎟ tt (2 1)log 2 − logπ ⎝⎠22 and substituting (2.13) gives us
⎛⎞11 ς ′′′(0, 2tt )=++−+−−ςς (0, )⎜⎟ 0, t ς ′ (0) (2 t 1) log 2 logπ ⎝⎠22 and with t =1/2 we obtain
⎛⎞11 ς ′′⎜⎟0,=+ςπ (0) log ⎝⎠22
We recall the identity [30]
⎛⎞1 s ς ⎜⎟ss,[21](=−ς ) ⎝⎠2 and differentiation results in
⎛⎞1 ss ςςς′′⎜⎟sss,[21]()()2log=− + 2 ⎝⎠2 so that ⎛⎞11 ςς′⎜⎟0,==− (0)log 2 log 2 ⎝⎠22
5 This then gives us the well-known result 1 (2.14) ς ′(0)=− log(2π ) 2 and hence we have obtained Lerch’s identity [7] in a very direct manner without the need to resort to the functional equation for the Riemann zeta function
1 (2.15) ς ′(0,tt )=Γ− log ( ) log(2π ) 2
Remark (iii)
It should be noted that the Stieltjes constant γ1 referred to in Eq. (2.15) of Deninger’s paper [28] should be increased by a factor of 2 (this difference arises because Deninger [28, p.174] employed a different definition in the Laurent expansion of the Hurwitz zeta function; this value was also inconsistently employed in Eq. (1.26) of the paper by Chakraborty et al. [12]).
3. An application of the Abel-Plana summation formula
Adamchik [2] has recently reported that the Hermite integral for the Hurwitz zeta function may be derived from the Abel-Plana summation formula [38, p.90]
∞ 1(∞∞fix)−− f( ix) (3.1) f ()kf=+ (0) fxdxi () + dx ∑ ∫∫2π x k=0 2100e − which applies to functions which are analytic in the right-hand plane and satisfy the convergence condition limefxiy−2π y (+ )= 0 uniformly on any finite interval of x . y→∞ Derivations of the Abel-Plana summation formula are given in [39, p.145] and [40, p.108].
The Hermite integral for the Hurwitz zeta function may be derived as follows. Letting f ()kku=+ ( )−s we obtain
∞ 1(uu−−ss1 ∞ uixuix+−−)−s() − s (3.2) ς (,su )==++i dx ∑ sx∫ 2π k =0 ()2ku+− s 10 e − 1
Then, noting that
(u+−−= ix)()()()−−−−s u ixsisi reθθ − re −s
6 =−re−−sisis[]θ eθ
2 (3.2.1) ()()uix+−−=−−ss uix sin(tan(/s−1 xu)) iu()22/2+ x s we may write (3.2) as Hermite’s integral for the Hurwitz zeta function ς (,)su
uu−−ss11∞ sin( s tan− ( xu / )) (3.3) ς (,su )=+ +2 dx ∫ 22/22sxπ 21()(1)suxe−+0 −
Differentiating (3.2) with respect to u
∂+su−−ss11∞ ()()u ix−− − u− ix −−s1 (3.4) ς (,su )=− −u−s − is dx ∫ 2π x ∂−ue210
We note that if f ()()ssgs= then the Leibniz differentiation formula results in
f (1)nn++()ssgs=++ (1) () ( n 1) gs()n () so that
(3.5) fng(1)nn+ (0)=+ ( 1)() (0) and hence we obtain
n+1 ∂∂n +1 nn n++11 n n+1 ς (,su )=− (1)log −u − (1) − log u ∂∂sus=0 2 u
∞ ()log()()log()uixuixuixuix−+−+−nn −−in(1)(n + 1) dx ∫ 222π x 0 ()(1)uxe+− and comparing this with (2.8) we readily obtain the integral formula originally obtained by Coffey [17] in 2007 for the Stieltjes constants
11∞ ()log()()log(uixuixuixuix−+−+−nn) (3.6) γ ()uu=− lognn log+1 ui + dx n ∫ 222π x 21un++0 ()uxe(1−)
This derivation is slightly more direct than the one originally provided by Coffey [17] (and is also simpler than my previous proof in [22]). □
Chen [13] has recently shown that for u > 0 and s > 0
7 ∞ −1 2s2 in[sxutan(/)] (3.7) ey−−uy21 s sin( xydy 2 ) = ∫ 22/2s Γ+()su0 ()x and hence we have using Hermite’s integral (3.3)
−−ss1 ∞∞ uu 41 2 ς (,s u )=+ + dx e−−uy y21 s sin( xy 2 ) dy ∫∫2π x 21()1sse−Γ00 −
−−ss12∞∞ uu 4s2 in(xy) =+ + e−−uy y21 s dy dx ∫∫2π x 21()ss−Γ00 e−1
Using Legendre’s relation [39, p.122]
∞ sin(xt ) 1 1 1 1t 1 (3.7.1) 2cdx =−+=−oth ∫ 2π xt 0 eet−−11222t this results in −−ss1 ∞ uu 2112 ⎡ 1⎤ ς (,su )=+ + e−−uy y21 s −+dy ∫ ⎢ y2 2 ⎥ 21()ss−Γ0 ⎣e −1 y2⎦
With the substitution vy= 2 this becomes
uu−−ss1 111∞ ⎡ 1⎤ (3.8) ς (,su )=+ + e−−uv v s 1 −+dv ∫ ⎢ v ⎥ 21()ss−Γ0 ⎣ ev − 12⎦ which is reported in [38, p.92] as being valid for Re ()s > −1.
With u =1 we have
11 1∞ ⎡ 1 11⎤ (3.9) ς ()se=+ + −−vsv1 −+ dv ∫ ⎢ v ⎥ 21()ss−Γ0 ⎣ ev − 1 2⎦ which is also reported in [38, p.100] as being valid for Re ()s > −1. □
We may write (3.7) as
∞∞ 2222s ∫∫ey−−uy21 s sin( xydy 2 ) = ey−−uy21 s sin( xyd 2 ) y ΓΓ()ss00(1+ )
8 and we have the limit
∞∞ 222 s 2 lim ey−−uy21 s sin( xydy 2 )= lim ey−−uy21 s sin( xyd 2 ) y ss→→00∫∫ ΓΓ()ss00(1+ )
2sin(se∞ −uy xy) = lim dy s→0 ∫ Γ+(1sy ) 0
We have the well-known integral (rigorous derivations of which are contained in [3, p.285] and [6, p.272])
∞ exy−uy sin( ) (3.10) tan−1 ()x / ud= ∫ y 0 y and hence we see that ∞ 2 2 lim ey−−uy21 s sin( xydy 2 )= 0 s→0 ∫ Γ()s 0 showing that Chen’s result is valid for s ≥ 0 .
□
For completeness we present another proof of (3.7). Using the definition of the gamma function we have
∞ Γ()s eydy−−zy s 1 = ∫ s 0 z and with zuix=± we obtain
∞ ∫ ey−−uy s 1[cos( xyi )− sin( xydyuixs )]=+ ( )−s Γ ( ) 0
∞ ∫ ey−−uy s 1[cos( xyi )+ sin( xydyuixs )]=− ( )−s Γ ( ) 0
This gives us
∞ −−uy s 1 −s −s ie∫ ysin( xydy )= −+⎣⎦⎡⎤ ( u ix ) −− ( u ix ) Γ ( s ) 0
9 ∞ −−uy s 1 −s −s ∫ eycos( xydyuix )= ⎣⎦⎡⎤ (++−Γ ) ( uix ) ( s ) 0
Then, noting (3.2.1) we obtain
1s∞ in[sxutan−1(/)] ey−−uy s 1 sin( xydy ) = ∫ 22/2s Γ+()su0 ()x and 1c∞ os[sxutan−1(/)] ey−−uy s 1 cos( xydy ) = ∫ 22/2s Γ+()su0 ()x □
We have from (3.8)
⎡⎤uu−−ss1 ∞ ⎡ 111⎤ (3.11) Γ−−=()ssuς (, ) ev−−uv s 1 −+dv ⎢⎥∫ ⎢ v ⎥ ⎣⎦21sev−−0 ⎣ 12⎦ and we consider the limit as s → 0 . With simple algebra we may write
uu−−s 11ss⎛⎞1 uu−−s ςς(,)su −− =(,)su −ς (0,) u +−−−⎜⎟ u 21ss−−⎝⎠2 21
⎛⎞11−−ss1 ⎛ ⎞ =−+−−−+ςς(,su ) (0,) u⎜⎟ u() 1 u u ⎜ 1 ⎟ ⎝⎠21 ⎝s − ⎠ and we have −s ⎡⎤uu−−ss1 ⎡ςς(,suu )− (0,)⎛⎞ 1 ()1− u u1−s ⎤ Γ−−=Γ++−−()ssu⎢⎥ς (, ) (1s )⎢ ⎜u⎟⎥ 21sss−−2 s1 ⎣⎦⎣⎢ ⎝⎠ ⎦⎥
Taking the limit as s → 0 we see that
⎡⎤uu−−ss1 ⎛⎞1 lim()Γ−−=+−ssu⎢⎥⎜ςς (,) ′(0,)u u⎟ log u+u s→0 ⎣⎦21s − ⎝⎠ 2 and hence we obtain
⎛⎞11∞ e−uv ⎡11 ⎤ (3.12) ς ′(0,uuuu )+− log += −+dv ⎜⎟ ∫ ⎢ v ⎥ ⎝⎠210 ve ⎣− v2 ⎦
10 Using Lerch’s identity (2.15), it may be seen that this is equivalent to Binet’s first formula for logΓ (u ) [39, p.249]
⎛⎞11∞ e−uy ⎡ 111⎤ (3.13) logΓ=− (uu ) log uu −+ log(2π ) + −+dy ⎜⎟ ∫ ⎢ y ⎥ ⎝⎠220 ye⎣ − 1 y2⎦
We note Alexeiewsky’s theorem [38, p.32]
t 11 (3.14) ∫ logΓ=−+ (udu ) t (1 t ) t log(2π ) − logG (1 ++Γ t ) t log ( t ) 0 22
where Gx() is the Barnes double gamma function Γ2 ()x = 1/()Gx defined, inter alia, by the Weierstrass canonical product [38, p.25]
k 2 x ∞ ⎧ ⎫ 2 ⎡⎤⎛⎞1 22 ⎪ xx⎛⎞⎪ (3.15) G(1+= x ) (2πγ ) exp⎢⎥ − (xxx ++ )∏⎨⎜⎟ 1 + exp⎜⎟ −x⎬ ⎣⎦⎝⎠22k =1 ⎩⎭⎪ kk⎝⎠⎪ and we note that GG(1)== (2) 1.
Hence integrating (3.13) results in
11∞ 1− e−tv ⎡ 111⎤ (3.16) logGtt (1+= ) log Γ ( t ) + t2 − tt −1 log t − −+ dv () ∫ 2 ⎢ v ⎥ 42 0 ve⎣ −1 v 2⎦ and with t =1 we immediately obtain
∞ 1111− e−v ⎡⎤1 −+dv = ∫ 2 ⎢⎥v 0 ve⎣⎦−12 v 4
Subtracting this from (3.16) results in
11 ∞ ee−−tv− v ⎡ 111⎤ (3.17) logGtt (1+= ) log Γ ( t ) + ( t2 −− 1) tt − 1 log t + −+ dv () ∫ 2 ⎢ v ⎥ 42 0 ve⎣ −12 v⎦
It may be possible employ Pringsheim’s artifice with this integral (as was employed in [39, p.249]).
□
We obtain by differentiating (3.2)
11 ⎛⎞1t∞ an−1(xu/) ς ′(0,uu )=− log uu −+ 2 dx ⎜⎟ ∫ 2π x ⎝⎠210 e −
and differentiating this with respect to u gives us
∂ 1 ∞ x ς ′(0,uu )=−− log 2 dx ∫ 222π x ∂uu2 0 ()()uxe+−1
∂ Since from (2.8) ςγ′(0,u )=− (u ) we obtain ∂u 0
1 ∞ x −=−−γ ()uu log 2 dx 0 ∫ 222π x 2u 0 ()()uxe+−1
and since γ 0 ()u=−ψ ()u this is equivalent to
1 ∞ x ψ ()uu=−− log 2 dx ∫ 222π x 2u 0 ()()uxe+−1
as reported in [39, p.251].
Similarly, we have
22− 1 ⎛⎞1 ∞ log(ux+ ) tan( xu / ) ς ′′(0,uuuuuu )=− log2 + 2 log −− 2 2 dx ⎜⎟ ∫ 2π x ⎝⎠210 e −
so that
∂ 1 ς ′′(0,uuu )=− log log2 ∂uu
22 ∞∞x xuxlog ()+ ∞tan−1 ()ux / −++224π dx dx u dx ∫∫222ππxx222 ∫222πx 00()()uxe+−11()() uxe +− 0()()uxe +−1
∂∂2 Then using 2 ς(,su )= 2γ1 () u we obtain ∂∂su s=0
12 22 11 ∞∞xuxlog ( + ) tan−1 ()xu / (3.18) γ ()uuu=−+ log log2 dxu−2 dx 1 ∫∫222ππxx222 22uu00 ()(1)()(1+−xeu +−xe)
It may be noted that Shail [36, p.799] made reference to these “seemingly intractable” integrals in 2000.
□
We have the well-known Hurwitz’s formula for the Fourier expansion of the Hurwitz zeta function ς (,)st as reported in Titchmarsh’s treatise
⎡⎤⎛⎞π sntsn∞∞cos 2 ππ⎛⎞sin 2 πt (,)st 2 (1 s )sin cos ς =Γ− ⎢⎥⎜⎟∑∑11−−s + ⎜⎟ s ⎣⎦⎝⎠2nn==11 (2)ππnn⎝⎠ 2 (2) where Re ()s < 0 and 0 < t ≤1. In 2000, Boudjelkha showed that this formula also applies in the region Re ()s < 1. It may be noted that when t =1 this reduces to Riemann’s functional equation for ς ()s .
Unfortunately, it appears that (2.8) cannot be applied to Hurwitz’s formula for the Fourier expansion of the Hurwitz zeta function because this would result in divergent series. This however may give an indication of Ramanujan’s erroneous thinking in this area as exemplified in Berndt’s book, Ramanujan’s Notebooks, Part I [8, p.200]. □
It is instructive to consider the original derivation of the formula for the Stieltjes constants which was derived by Stieltjes himself. This is recorded in a letter dated June 1885 from Stieltjes to Hermite [32] and the proof proceeds as follows:
From (3.9) we have 11 1∞ ⎡⎤ 1 11 ς (1+−=+se ) − xs−+ xdx ∫ ⎢⎥x ssex2(1)Γ+0 ⎣⎦ − 1 2
11∞ ⎡⎤e− x 11 =+ ex−xs−+ dx ∫ ⎢⎥−x 2(1)Γ+sex0 ⎣⎦ 1 − 2
11∞ ⎡⎤ 111 =+ ex−xs−− dx ∫ ⎢⎥−x 2(1)Γ+sex0 ⎣⎦ 1 − 2 so that 11∞ ⎡⎤ 11 (3.19) ς (1+−=se ) −xs− xdx ∫ ⎢⎥−x ssΓ+(1 )0 ⎣⎦ 1 − ex
13 Using the Maclaurin expansion
∞ log j x xs = ∑ s j j=0 j! we see that 11∞ a ς (1+−=ss ) ∑ j j ssjΓ+(1 )j=0 !
∞ ⎡⎤11 where ae=−− xjlog xdx. j ∫ ⎢⎥− x 0 ⎣⎦1− ex
∞ ∞ Let Ft()= e−tx log n xdx. We have Γ=()nxn(1)ex− log dx and with x → tx this n ∫ ∫ 0 0 becomes ∞ Γ=()nt(1)te∫ − x ( log x + log t ) n dx 0
We write Ftn () as ∞ Ft( )=+− e−tx (log x log t log t ) n dx n ∫ 0
n 1 jnj⎛⎞n ()− j =−∑(1)⎜⎟ Γ (1)logt t j=0 ⎝⎠j and we then have
n j ∞ jnj⎛⎞n ()−−log t txn (3.20) ∑(1)−Γ⎜⎟ (1) =∫ ex log dx j=0 ⎝⎠j t 0
We have the summation
nrj r∞ jnj⎛⎞n ()−−log t txn ∑∑∑(1)−Γ⎜⎟ (1) =∫ exlog dx jt==01⎝⎠j t t=10
∞ ⎡⎤ee−−+xrx− (1) = logn x dx ∫ ⎢⎥−x 0 ⎣⎦1− e
and the right-hand side may be written as afrgrnn+ ()()− n where
14 ∞ ⎡⎤11 (3.21) gr()=−e−+(1)rxlog n xdx n ∫ ⎢⎥− x 0 ⎣⎦1− ex
∞ ⎡⎤ee−−+xrx(1) (3.22) f ()rx=− logn dx n ∫ ⎢⎥ 0 ⎣⎦xx
where we note that limgrn ( )= 0 . Now treating r as a continuous variable we obtain r→∞
n ⎛⎞n logj (r + 1) ′ jnj()− frnn()=+=− Fr ( 1)∑ (1)⎜⎟ Γ (1) j=0 ⎝⎠j r +1
and since fn (0)= 0 we have by integration
n j+1 jnj⎛⎞n ()− log (r + 1) (3.23) frn ()=−∑ (1)⎜⎟ Γ (1) j=0 ⎝⎠j j +1 so
∞ ⎡⎤11 nr⎛⎞n ⎛⎞loglog(1)jjmr+1 + aex−−−+(1)rxlog ndx =−Γ ( 1)j() nj− (1) − n ∫ ⎢⎥−x ∑∑⎜⎟ ⎜⎟ 0 ⎣⎦11−+ex jm==01⎝⎠j ⎝⎠m j
nr⎛⎞j logjj++11rr++−⎡ log ( 1) log j + 1r⎤ jnj⎛⎞n ()− log m ⎣ ⎦ =−(1)⎜⎟ Γ (1)⎜⎟− ∑∑j ⎜⎟mj+1 jm==01⎝⎠ ⎝⎠
Therefore letting r →∞ we obtain
nrjj+1 jnj⎛⎞n ()− ⎛⎞logmr log an =−(1) Γ (1)lim − ∑∑⎜⎟ r→∞ ⎜⎟ jm==01⎝⎠j ⎝⎠mj+1 since lim⎡ logjj++11 (rr+− 1) log⎤ = 0. This may be written as r→∞ ⎣⎦ n ⎛⎞n jnj()− (3.24) aCnj=−Γ∑⎜⎟(1) (1) j=0 ⎝⎠j where ⎛⎞r logjjmr log +1 C j =−lim r→∞ ⎜⎟∑ ⎝⎠m=1 mj+1
Eq. (3.24) reminds us of the Cauchy product of two series
15 ∞∞ ∞()k a j j ⎛⎞mjCm ⎛⎞Γ (1) j ∑∑ss=−⎜⎟(1) ⎜⎟ ∑s jm==00jmk!!⎝⎠⎝⎠ k = 0!
∞ Γ()k (1) where Γ+(1ss ) =∑ j . k =0 k!
We have from (3.19)
⎡⎤11∞ ⎡⎤1 (3.25) ς (1+−sse ) Γ+= (1 ) −xs − xdx ⎢⎥∫ ⎢⎥− x ⎣⎦se0 ⎣⎦1− x
and we note from (2.4) that
∞ m 1(1)− m ςγ(1)ss+−=∑ m smm=0 ! and we conclude that
⎛⎞r logjjmr log +1 (3.26) γ jj==C lim − r→∞ ⎜⎟∑ ⎝⎠m=1 mj+1
A more direct derivation of (3.26) is given, inter alia, by Bohman and Fröberg [10].
4. A family of integral representations of the Stieltjes constants
With a view to utilising (2.8), we first of all differentiate (3.8) with respect to u to give us ∂ su−−s 1 111∞ ⎡ 1⎤ ς (,su )=− − u−−su − ev vs− + dv ∫ ⎢ v ⎥ ∂Γ−us2()10 ⎣ev2⎦ and differentiating this n times with respect to s (with the assistance of (3.5)) gives us
(4.1) ∂∂n n ∞ ⎡ 111⎤ ς (su , )=− ( − 1)nn−−11 logu − ( − 1) nn log u − F() n (0, ve ) − uv − + dv n ∫ ⎢ v ⎥ ∂∂su s=0 21u 0 ⎣e− v2⎦ where, for convenience, we have designated Fsv(,) as
vsvss (4.2) Fsv(,)== ΓΓ+()ss (1 )
16 ∂n and we note that Fv(0, )= 0 . We employ the notation Fsv(, )= F()n (, sv ) and ∂sn n ∂ ()n n Fsv(, ) = F(0,) v. ∂s s=0
Then using (2.8) n ∂∂ n n ςγ(,su )=− (1) nn−1 () u ∂∂su s=0 we have thus determined that for n ≥1
11(1)111− n ∞ ⎡ ⎤ (4.3) γ ()uuuFve=−− lognn−−1( log nu)(0,) v−+dv n−1 ∫ ⎢ v ⎥ 21un n0 ⎣ ev− 2⎦ and, in particular, for n ≥ 2 we have a family of integral representations of the Stieltjes constants (1)− n+1 ∞ ⎡ 1 1 1⎤ (4.4) γγ==(1) Fve()nu(0, ) − v−+dv nn−−11 ∫ ⎢ v ⎥ ne0 ⎣ −12v⎦ and for n =1 we obtain
11∞ ⎡ 11⎤ γ =+Fve(1) (0, ) −v −+ dv ∫ ⎢ v ⎥ 210 ⎣ev− 2⎦
We shall initially consider the first two derivatives of Fsv(,) . Differentiating (4.2) with respect to s results in
svs ⎡⎤1 Fsv′(, )=+− logvψ (1+ s ) Γ+(1ss ) ⎣⎦⎢⎥
svs =−[]logvsψ ( ) Γ+(1s ) and so we have
(4.5) Fsv′(, )=− Fsv (,)log[ vψ () s]
Since ssψ ( )=+− sψ (1 s ) 1 we see that limssψ ( )= − 1 and hence we deduce that s→0
(4.6) Fv′(0, )= 1
17 With n =1in (4.3) we have
11∞ ⎡ 11⎤ −=−+−γ ()uue log −uv −+dv 0 ∫ ⎢ v ⎥ 21uev0 ⎣ − 2⎦
or equivalently, since ψ ()u=−γ 0 ()u, we obtain the well-known integral
11∞ ⎡ 11⎤ (4.7) ψ ()uue=− + log −−uv − + dv ∫ ⎢ v ⎥ 21uev0 ⎣ − 2⎦
This integral, which appears in [38, p.16], may be easily verified by differentiating Binet’s first formula for logΓ (u ) [39, p.249], which we saw above in (3.13).
We now consider the second derivative
F′′(, sv )=− Fsv (, )ψψ′ () s + Fsv (, )log[ v− () s]2
22 =−Fsv( , )ψψ′ ( s ) + Fsv ( , )⎣⎡ log v− 2 ( s ) log v +ψ ( s )⎦⎤ and we have the limit
F′′(0,) v=−− lim(,) Fsv⎡⎤ψψ2 () s′ () s 2loglim(,) v Fsvψ () s ss→→00⎣⎦
⎡⎤ψ (1+ s ) 1 1 ⎧⎫2 ⎧′ ⎫ =+limFsv ( , )⎢⎥⎨⎬ψψ (1 s )−+ 2 22−+⎨(1s )+⎬+ 2 log v s→0 ⎣⎦⎩⎭ss⎩ s⎭
⎡⎤2 ψ (1+ s ) =+limFsv ( , )⎢⎥ψψ (1 s )−−+ 2 ′(1s )+ 2 log v s→0 ⎣⎦s
Since Fv(0, )= 0 this gives us ⎡⎤ψ (1+ s ) =−2limFsv ( , )⎢⎥+2logv s→0 ⎣⎦s and we therefore obtain
(4.8) Fv′′(0, )=+ 2(γ log v )
Referring back to (4.3) we have with n = 2
11 1∞ ⎡ 111⎤ γ ()uuuFve=−− log log2(2) (0,)−uv −+dv 1 ∫ ⎢ v ⎥ 222ue0 ⎣ − 12v⎦
18 and substituting (4.8) this becomes
11 ∞ ⎡ 111⎤ γγ(uuuve )=−−+ log log2 ( log ) −uv −+dv 1 ∫ ⎢ v ⎥ 22ue0 ⎣ −12v⎦
Using (4.7) we obtain
11⎡⎤⎡⎤ 1 ∞ 111 (4.9) γψγ()uuuuue=−++−− log log2 () log −uv −+logvdv 1 ⎢⎥⎢⎥∫ v 22uue⎣⎦⎣⎦ 2 0 −12v
and for u =1 we have
⎡⎤1111∞ ⎡ ⎤ (4.10) γγγ=−+ −ev−v − + log dv 1 ⎢⎥∫ ⎢v ⎥ ⎣⎦2120 ⎣ev− ⎦
∞ ∞ Since Γ=()muvm()uvevd∫ −−1 log v we have Γ=()mvm(1)∫ ev− log dv and, in particular, we 0 0 ∞ see that Γ=−=′(1) γ ∫ evd−v log v. Hence we may write (4.10) as 0 ∞ ⎡⎤11 (4.11) γγγ=−2 + −ev−v − log dv 1 ∫ ⎢⎥v 0 ⎣⎦ev−1
We note that 11ev ==+1 11−−ee−vv e v−1
with the result that
∞∞⎡⎤11 ⎡⎤11 ∞ e−−vv−=log v dv e −+logv dv e−v log v dv ∫∫⎢⎥−vv ⎢⎥∫ 00⎣⎦11−−ev ⎣⎦e v 0
∞ ⎡⎤11 = ev−v −−log dvγ ∫ ⎢⎥v 0 ⎣⎦ev−1
We then see that
∞ ⎡⎤11 (4.11.1) γγ=−2 − ev−v − log dv 1 ∫ ⎢⎥−v 0 ⎣⎦1− ev
19 as previously determined by Coppo [27] in 1999.
Having obtained integrals for γ 0 and γ1 , we now consider the general case for γ n . We will approach this by reference to the (exponential) complete Bell polynomials, the salient features of which are summarised in the attached Appendix.
It is well known that [34]
m d fx() fx () (1) (2) ()m (4.12) m eeYfxfxfx= m ()( ), ( ),..., ( ) dx
where the (exponential) complete Bell polynomials Yxn(1 ,..., xn ) are defined by Y0 =1 and for n ≥ 1
kk12 kn n! ⎛⎞⎛⎞xx12 ⎛⎞xn (4.13) Yxnn(1 ,..., x ) = ∑ ⎜⎟⎜⎟... ⎜⎟ π ()n kk12! !... kn !⎝⎠⎝⎠ 1! 2!⎝⎠n !
where the sum is taken over all partitions π ()n of n , i.e. over all sets of integers k j such that
kkk123++++=2 3 ... nkn n
For example, with n =1 we see that the only possibility is k1 =1and kjj =∀≥02 which results in
Yx11()= x 1
With n = 2 , we see that the possible outcomes are ( k1 = 2 and k2 = 0 ) and ( k1 = 0 and k2 =1) which results in
2 Yxx212(, )=+ x 1 x 2
Suppose that hx′()()()= hxgx and let f (xh )= log (x ) . We see that
hx′() f ′()xg== ()x hx() and then using (4.12) above we have
mm ddloghx ( ) (1) (m− 1) (4.14) mmhx() == e hxY( )m () gx ( ), g ( x ),..., g ( x ) dx dx
20 As a variation of (4.14) above, suppose that jx′( )= jxgx ( )[ ( )+α ] where α is independent of x and let f (xj )= log (x ). We see that
jx′() fx′()==+ gx () α jx() and f (1)kk++()xg= (1) () x for k ≥1 and therefore we obtain
mm ddlogjx ( ) (1) (m− 1) (4.15) mmj()xejxYgxgxg==( )m () ( ) +α , ( ),..., (x ) dx dx
We note from (4.8) above that Fsv′(, )=− Fsv (,)log[ vψ () s] and therefore we obtain
()nn(1) (− 1) (4.16) FsvFsvYv( , )=−−− ( , )n ( logψψ ( s ), ( s ),..., ψ ( s ))
We saw in (4.3) above that
11(1)111− n ∞ ⎡ ⎤ (4.17) γ ()uuuFve=−− lognn−−1( log nu)(0,) v−+dv n−1 ∫ ⎢ v ⎥ 21un n0 ⎣ ev− 2⎦ and hence we deduce that
(4.18) ∞ n+1(1)(nv−−1)⎡⎤111 (−= 1) nFsvYvssγψnn−1 ( , )() log −−− ( ),ψ ( ),...,ψ ( se ) v −+ dv ∫ ⎢⎥ev−12 0 ⎣⎦s=0
Prima facie, it is quite remarkable that this limit actually exists considering that the digamma function and all of its derivatives diverge at s = 0 .
It is shown in Appendix A that
n ⎛⎞n nk− Yxnn(12+=αα , x ,..., x ) ∑⎜⎟ Yx k(1 ,..., xk ) k =0 ⎝⎠k and we then determine that
d m m ⎛⎞m (4.19) j()()xjx= α mk−− Ygxgxg( ),(1) ( ),...,(k 1) ( x ) m ∑⎜⎟ k () dx k=0 ⎝⎠k
Now, referring back to (4.5), we see that with gs( )= log v−ψ ( s )
21 n ()nn⎛⎞n −−k(1) (k 1) (4.19.1) FsvFsv( , )=⋅ ( , )∑⎜⎟ log vYk ()−ψψ ( s ),− ( s ),...,− ψ ( s ) k =0 ⎝⎠k and thus we have
(4.20) n n ∞ 11(1)nn−−1 − ⎛⎞n nku−v⎡⎤111 γ nk−1(uuu )=−− log log ⎜⎟Y ( sFsvve ) ( , )log ⋅−+v dv 21un n∑ k ∫ ⎢⎥ev− 2 k=0 ⎝⎠ 0 ⎣⎦s=0
where, for convenience, we denote Yk ()s as
(1) (k− 1) Ykk(sY )=−()ψψ ( s ), − ( s ),..., − ψ ( s )
and we note that Yk ()s is independent of the integration variable v . □
We now adopt a slightly different approach so as to eliminate all of the apparently troublesome factors ψ ()j ()s in the limit as s → 0 . To this end we write (4.2) in the equivalent form vsvss Fsv(,)== ΓΓ+()ss (1 ) and first of all we employ the Leibniz differentiation formula to obtain
∂∂nnn ⎛⎞n −kd k1 Fsv(, )= []svs nn∑⎜⎟ −kk ∂∂ssdk =0 ⎝⎠k sΓ(1+s )
We see that
d 11 =−ψ (1 +s ) dsΓ+(1 s ) Γ+ (1 s ) and applying (4.14) we determine that
k d 11 (1) (k − 1) (4.21) k =−+−+−Ysk ()ψψ(1 ), (1 s ),..., ψ (1+s ) dsΓ+(1 s ) Γ+ (1 s )
We also have
∂ j []svssjsj=+ sv log v jv log−1 v ∂s j
22 so that
∂nsv n ⎛⎞n Fsv(, ) slognk−− v ( n k )lognk−1 vY (1 s ) n =+∑⎜⎟()−k + ∂Γ+ss(1 ) k=0 ⎝⎠k where, as before, we denote
(1) (k − 1) Ykk(1+=sY )() −ψψ (1 + s ), − (1 + s ),..., − ψ (1 + s )
When s = 0 this becomes
∂n n ⎛⎞n Fv(0, ) ( nk )lognk−−1( vY (1),1)( (1),...,k−1) (1) n =−∑⎜⎟ k () −−−ψψ ψ ∂s k=0 ⎝⎠k
n−1 ⎛⎞n nk−−1(1)(k−1) =−∑⎜⎟(nk )log vYk () −−−ψψ(1), (1),..., ψ (1) k =0 ⎝⎠k
Using the elementary binomial identity [29, p.157]
⎛⎞nn ⎛−1 ⎞ ()nk−=⎜⎟ n ⎜ ⎟ ⎝⎠kk ⎝ ⎠ this becomes n−1 ⎛⎞n −1 nk−−1(1)(k−1) =−nv∑⎜⎟log Yk ()ψψ(1),− (1),...,− ψ (1) k =0 ⎝⎠k
We therefore conclude that
11 (4.22) γ ()uu=− lognn−1 logu n−1 2un
n−1 ⎛⎞n −1 ∞ ⎡ 111⎤ +−( 1)nk−−1(Yv−ψψ(1), −1) (1),..., − ψ(1) (1) logn−k−1⋅e−uv − + dv ∑⎜⎟k ()∫ ⎢ v ⎥ k=0 ⎝⎠k 0 ⎣ev−12⎦ and
(4.23) n−1 ⎛⎞n −1 ∞ ⎡ 111⎤ γψ=−( 1)nk−−1(Yv−(1), −ψ1) (1),..., −ψ(1) (1) logn−k−1⋅e−v − + dv nk−1 ∑⎜⎟()∫ ⎢ v ⎥ k=0 ⎝⎠k 0 ⎣ev−12⎦
We note that
23 ∞ ∫ lognk−−1(ve⋅=Γ − v dv nk−−1)(1) 0
and from (A.6) we have
()kk(1) (− 1) Γ=(1)Yk ()ψψ (1), (1),..., ψ (1)
We will see in (A.9) that for n ≥1 ( n = 0 implies a value of 1)
n ⎛⎞n (4.24) ∑⎜⎟Yxj (11 ,..., xYjnj )−− (−−= x ,..., xnj ) δ n,0 j=0 ⎝⎠j and hence we may eliminate the factor of ½ in the integrand of (4.22) to obtain
11 (4.25) γ ()uu=− lognn−1 logu n−1 2un
n−1 ⎛⎞n −1 ∞ ⎡⎤11 +−( 1)nk−−1(Yv−ψψ(1), −1) (1),..., − ψ(1) (1) logn−k−1⋅e−uv − dv ∑⎜⎟k ()∫ ⎢⎥v k=0 ⎝⎠k 0 ⎣⎦ev−1 (4.26) n−1 ⎛⎞n −1 ∞ ⎡⎤11 γψ=−( 1)nk−−1(Yv−(1), −ψ1) (1),..., −ψ(1) (1) logn−k−1⋅e−v − dv nk−1 ∑⎜⎟()∫ ⎢⎥v k=0 ⎝⎠k 0 ⎣⎦ev−1 or equivalently for n ≥1
n ⎛⎞n ∞ ⎡⎤11 (4.27) γψ=−( 1)nkYv −(1), −ψψ(1) (1),..., − (−−− 1) (1) lognk⋅ev − dv nk∑⎜⎟ ()∫ ⎢⎥v k=0 ⎝⎠k 0 ⎣⎦ev−1
We could also represent this in terms of the partial exponential Bell polynomials but this only seems to add an extra layer of complexity.
With the substitution te= −v in (4.27) we obtain
(4.28) n ⎛⎞n 1 ⎡t 1 ⎤ γψ=−( 1)nkYt −(1), −ψψ(1) (1),..., − (−− 1) (1) lognk [log(1/ )] + dt nk∑⎜⎟ ()∫ ⎢ ⎥ k=0 ⎝⎠k 0 ⎣1log− tt ⎦
n ⎛⎞n 1 ⎡ 11⎤ =−( 1)nkYt −ψψ(1), −(1) (1),..., − ψ(−− 1) (1) lognk [log(1/ )]−+ 1 + dt ∑⎜⎟k ()∫ ⎢ ⎥ k=0 ⎝⎠k 0 ⎣ 1log− tt⎦
24 1 Since ∫ lognk− (−=Γ log(1/tdt ))()nk− (1) we see, in the same manner as before, that for 0 n ≥1 the terms involving −1 in the integrand cancel out and we conclude that
n ⎛⎞n 1 ⎡11 ⎤ (4.29) γψ=−( 1)nkYt −(1), −ψψ(1) (1),..., − (−− 1) (1) lognk [log(1/ )] + dt nk∑⎜⎟ ()∫ ⎢⎥ k=0 ⎝⎠k 0 ⎣logtt 1− ⎦
□
In his 2001 dissertation, Brede [11] showed that there exists a polynomial pn ()z of degree n such that 1 ⎡ 11⎤ (4.30) γ =−p [ log log(1/td )] + t nn∫ ⎢ ⎥ 0 ⎣logtt 1− ⎦
Brede [11] stated, for example, that
pz0 ()= 1
pz1()=− z γ
22 pz2 ()=− z 2γ z +−γς (2)
but he did not specify the precise form of the general polynomial pn ()z .
Using these polynomials for n =1,2,3 we have
1 ⎡⎤11 γ =+∫ ⎢⎥dt 0 ⎣⎦logtt 1−
11⎡⎤⎡⎤11 11 γγ=− log[log(1/td )] +t − + dt 1 ∫∫⎢⎥⎢⎥ 00⎣⎦⎣⎦log1tt−− log1 tt so that 1 ⎡⎤11 log[log(1/td )] + t=−γ −γ 2 ∫ ⎢⎥1 0 ⎣⎦logtt 1−
11⎡⎤11 ⎡⎤11 γγ=+log2 [log(1/td )] t+2 log[log(1/ t )] +dt 2 ∫∫⎢⎥ ⎢⎥ 00⎣⎦logtt 1−− ⎣⎦logtt 1
25 1 2 ⎡⎤11 +−[(2)]γς ∫ ⎢⎥ + dt 0 ⎣⎦logtt 1− so that
1 ⎡⎤11 log22 [log(1/td )] + t=+−−−2γ ⎡⎤γγ[ γς2 (2)] γγ ∫ ⎢⎥⎣⎦12 0 ⎣⎦logtt 1− □
Reindexing to knm=− we may write (4.29) as
n ⎛⎞n 1 ⎡11 ⎤ γψ=−( 1)nnYt−(1), −ψ(1) (1),..., −ψ(−−m 1) (1) logm [log(1/ )] + dt nn∑ ⎜⎟−m()∫ ⎢⎥ mn= ⎝⎠nm− 0 ⎣logtt 1− ⎦
⎛⎞⎛nn⎞ and since ⎜⎟⎜= ⎟this becomes after reversing the order of summation ⎝⎠⎝nm− m⎠ (4.31) n 1 ⎛⎞n k ⎡11 ⎤ γψ=−( 1)nk−−Yt −−− (1),ψ(1) (1),...,ψ(nk− 1) (1) − log[log(1/ )] +dt nn∑⎜⎟ −k()∫ ()⎢⎥ k =0 ⎝⎠k 0 ⎣logtt 1− ⎦
Comparing this with Brede’s representation (4.30) we are therefore able to specify the precise form of Brede’s polynomial
n ⎛⎞n nk−−(1) (nk− 1) k (4.32) pnn(zY )=−∑⎜⎟ ( 1)−k() −−−ψψ (1), (1),..., ψ (1) z k=0 ⎝⎠k or equivalently
n ⎛⎞n kk(1) (− 1) n−k (4.33) pnk(zY )=−−−−∑⎜⎟ ( 1)()ψψ (1), (1),..., ψ (1) z k=0 ⎝⎠k
With regard to the above, we may note that
k d 1 (1) (k− 1) k =−Ysk ()ψψ( ), − ( s ),..., − ψ ( s ) dsΓ() s
Differentiating (4.33) gives us
n ⎛⎞n ′ kk(1) (− 1) n−−k1 pznk( )=−∑ ( nk )⎜⎟ ( − 1) Y() −−ψψ (1), (1),..., − ψ (1) z k=0 ⎝⎠k
26 n−1 ⎛⎞n kk(1) (− 1) n−−k1 =−∑(nk )⎜⎟ ( − 1) Yk () −−ψψ (1), (1),..., − ψ (1) z k =0 ⎝⎠k
n−1 ⎛⎞n −1 kk(1) (− 1) n−−k1 =−−−−nY∑⎜⎟( 1)k ()ψψ (1), (1),..., ψ (1) z k =0 ⎝⎠k and we therefore see that
(4.34) pnn′ ()znpz= −1 ()
Since pnn′ ()znpz= −1 () we see that pn ()z is an Appell polynomial and therefore we have the relations
n ⎛⎞n k pnn()zp= ∑⎜⎟ −k(0)z k=0 ⎝⎠k
n ⎛⎞n nk− pnk()xy+=∑⎜⎟ pxy () k=0 ⎝⎠k
This concurs with Brede’s result [11]
∞ xnz=−px(log) zedz− ∫ n 0
We also note that
n ⎛⎞n ′ nk−−(1) (nk− 1) k− 1 pnn(zk )=−∑⎜⎟ ( 1)Y−k() −−−ψψ (1), (1),..., ψ (1) z k=0 ⎝⎠k
□
Having expended some energy getting to (4.22), it was somewhat disappointing to subsequently discover that this result could have been derived in a more succinct manner 1 using (3.8). Subtracting a factor of from both sides of (3.8), we may write that s −1 equation in the following form
11111uu−−ss1 − ∞ ⎡ 1⎤ (4.50) ς (,su )−=+ + e−−uv v s 1 −+dv ∫ ⎢ v ⎥ sssev−−Γ−12 1 ()0 ⎣ 1 2⎦
27 which we then differentiate and evaluate at s =1 this time to obtain
∂−∂nnn⎡⎤1(1) 1∞ ⎡⎤ 111 ς (,su )−=lognnu ++ f() (1) e−−uvs v 1 −+dv ∂−ssn ⎢⎥12 u∂Γ− ssevnv()12∫ ⎢⎥ ⎣⎦s=1 0 ⎣⎦s=1 where, as a useful artifice, we have denoted f ()s as
u1−s −1 fs()= s −1
We can represent f ()s by the following integral
u1−s −1 u f ()sx==−∫ −sdx s −1 1 so that u f ()nn()sx=− (1) − ∫ −s lognxdx 1 and thus we see that u logn x f ()nn(1)=− ( − 1) ∫ dx 1 x
logn+1 u =−(1) − n n +1
Hence substituting (2.9)
∂n ⎡⎤1 ςγ(,su )−=− (1)n ()u n ⎢⎥n ∂−ss⎣⎦1 s=1 we obtain
(1)−∂nnlog+1 u n 1 ∞ ⎡⎤1 1 1 (1)−=nnγ ()uu log −−+ (1)nev−− uvs1 −+dv n 21(uns+∂Γnvsev)1∫ ⎢⎥ − 2 0 ⎣⎦s=1
Referring to the definition (4.2) of Fsv(, ) we see that
∂∂nn1111∞∞⎡⎤e−uv ⎡⎤111 ev−−uv s 1 −+dv = Fsv(, ) −+ dv ∂Γssnv()∫∫⎢⎥ ev− 12 ∂ s nv vev ⎢⎥−12 00⎣⎦s=11 ⎣⎦s=
28 and using (4.19.1) this becomes
n ⎛⎞n ∞ ⎡ 111⎤ =−−−Ysψψ( ),(1) ( s ),..., ψ(ku−−− 1) ( sev )vn log k−+dv ∑⎜⎟k ()∫ ⎢ v ⎥ k =0 ⎝⎠k 0 ⎣ev−12⎦
Hence we obtain
11 (4.51) γ ()uu=− lognn log+1 u n 21un+
n ⎛⎞n ∞ ⎡ 111⎤ +−( 1)nkYs−ψψ( ), −(1) ( s ),..., − ψ(−−− 1) ( sev )uv lognk− + dv ∑⎜⎟k ()∫ ⎢ v ⎥ k=0 ⎝⎠k 0 ⎣ev−12⎦ which corresponds with (4.22). Writing (4.50) as
⎡⎤11uu−−ss1 − ∞ ⎡ 111⎤ (4.51.1) ς (,su )−−− Γ=()s e−−uv v s 1 −+dv ⎢⎥∫ ⎢ v ⎥ ⎣⎦ss−−12 1 0 ⎣ ev −1 2⎦ and, using the Leibniz formula to differentiate this, we obtain an inversion formula
(4.52)
n ⎛⎞n ⎡⎤1logk+1 u ∞ ⎡ 111⎤ (−−+Γ= 1)kknγ (uu ) log ()−−ku(1) evv logn −+dv ∑⎜⎟ ⎢⎥k ∫ ⎢ v ⎥ k=0 ⎝⎠k ⎣⎦21uk+−0 ⎣ev12⎦ and with u =1 we have
n ⎛⎞n 11∞ ⎡ 11⎤ (4.53) (1)−Γknkγ ()−− (1) −Γ= () n (1)ev vn log −+dv ∑⎜⎟ k ∫ ⎢ v ⎥ k=0 ⎝⎠k 210 ⎣ev− 2⎦ or equivalently
n ⎛⎞n ∞ ⎡ 11⎤ (4.54) (1)−Γknkγ ()−− (1) =ev vn log −dv ∑⎜⎟ k ∫ ⎢ v ⎥ k=0 ⎝⎠k 0 ⎣ev−1 ⎦
Differentiating (4.51.1) with respect to u gives us
⎡⎤su−−s 1 ∞ ⎡ 111⎤ −++ssς (1,) u +Γ=− u−−su() s evvs −+dv ⎢⎥∫ ⎢ v ⎥ ⎣⎦210 ⎣ev− 2⎦ which may be written as
29 ⎡⎤11 ∞ ⎡1 11⎤ ς (1)ss+−− Γ+= (1) e−vsv −+ dv ⎢⎥∫ ⎢v ⎥ ⎣⎦se210 ⎣− v2⎦ and differentiating this will also result in (4.52)
n ⎛⎞n ⎡⎤1logk+1 u ∞ ⎡⎤111 (4.55) (−−+Γ= 1)kknγ (uu ) log ()−−ku(1) evv logn −+dv ∑⎜⎟ ⎢⎥k ∫ ⎢⎥v k=0 ⎝⎠k ⎣⎦21uk+−0 ⎣⎦ev12
□
Referring back to the method originally employed by Stieltjes, in particular (3.22), we have ∞ ⎡⎤ee−−+vuv(1) f ()uv=− logn dv n ∫ ⎢⎥ 0 ⎣⎦vv
n ⎛⎞n logk (u + 1) ′ knk()− funn()=+=− Fu ( 1)∑ (1)⎜⎟ Γ (1) k=0 ⎝⎠k u +1
and since fn (0)= 0 we have by integration
n k+1 knk⎛⎞n ()− log (u + 1) fun ()=−∑ (1)⎜⎟ Γ (1) k=0 ⎝⎠k k +1 so that n k+1 knk⎛⎞n ()− log u fun (1)(1)−=∑ −⎜⎟ Γ (1) k=0 ⎝⎠k k +1
∞ −−vuv n k+1 ⎡⎤ee nkn⎛⎞n ()−klog u ∫ ⎢⎥−=−Γlogvdv ∑ ( 1)⎜⎟ (1) 0 ⎣⎦vv k =0 ⎝⎠k k+1
Referring back to (4.52)
n ⎛⎞n ⎡⎤1logk+1 u ∞ ⎡ 111⎤ (−−+Γ= 1)kknγ (uu ) log ()−−ku(1) evv logn −+dv ∑⎜⎟ ⎢⎥k ∫ ⎢ v ⎥ k=0 ⎝⎠k ⎣⎦21uk+−0 ⎣ev12⎦ we see that
n ⎛⎞n kk⎡⎤1 ()n−k (4.56) ∑⎜⎟(1)−−⎢⎥γ k ()uu log Γ (1) k=0 ⎝⎠k ⎣⎦2u
30 ∞∞⎡⎤111 ⎡⎤ee−−vuv =−ev−uvlog n + dv−−logn vdv ∫∫⎢⎥v ⎢⎥ 00⎣⎦ev−12 ⎣⎦ vv
∞ ⎧⎫⎡⎤11e−v =+−ev−uv logn dv ∫ ⎨⎬⎢⎥v 0 ⎩⎭⎣⎦ev−12 and with u =1 we come back to (4.53) above.
n ⎛⎞n 11∞ ⎡ 11⎤ (1)−Γknkγ ()−− (1) −Γ= () n (1)ev vn log −+dv ∑⎜⎟ k ∫ ⎢ v ⎥ k=0 ⎝⎠k 210 ⎣ev− 2⎦
With n = 0,1,2 we obtain 1111∞ ⎡ ⎤ γ −=ed−v −+ v ∫ ⎢ v ⎥ 210 ⎣ev− 2⎦
11∞ ⎡ 11⎤ −−+γγ2 γ =ev−v log −+ dv 1 ∫ ⎢ v ⎥ 210 ⎣ev− 2⎦
⎛⎞11∞ ⎡11 ⎤ γγςγγγ−+++=[(2)]222ev−v log −+dv ⎜⎟ 12∫ ⎢ v ⎥ ⎝⎠210 ⎣ev− 2 ⎦
The approximate values of the first three Stieltjes constants are [5]
γ =⋅0.5772 ⋅⋅ γ1 =−0.0728 ⋅⋅⋅ γ 2 = −⋅0.0096 ⋅⋅ and inserting these values into the above three equations numerically demonstrates that the corresponding three integrals have positive values. □
We now wish to show that the integral In , which is defined below, is strictly positive for all positive integers and for n = 0
∞ ⎡ 111⎤ I =⋅lognvve− −+dv n ∫ ⎢ v ⎥ 0 ⎣ev−12⎦ We see that
1 ⎡⎤111 ∞ ⎡⎤111 I =lognvve ⋅−− −+dv +lognv ve ⋅ −+ dv n ∫∫⎢⎥vv ⎢⎥ 01⎣⎦ev−−12 ⎣⎦ev12
31 =+JKnn
−v With the substitution te= we obtain for the first component Jn
11⎡⎤111 /e ⎡ 1 11⎤ lognvve⋅−+=− dv logn (log(1/t ) ++dt ∫∫⎢⎥v ⎢ ⎥ 00⎣⎦ev−−12 ⎣1/1log2t t⎦ and with the substitution t=1/u this becomes
1/e ⎡⎤⎡111∞ 111⎤du (4.56) lognn (log(1/td ) ++tu =log (log ) − + ∫∫⎢⎥⎢⎥2 0 ⎣⎦⎣1/tt−− 1 log 2 e uu1 log 2⎦u
It is clear that logn (logu )≥ 0 for all ue≥ and we now consider the other part of the integrand in (4.56). Let
111 φ()u =− + uu−1log2
(1)log22uuu+−+ = 2(uu− 1)log
11 and we note that φ()e =− > 0. The denominator of φ()u is positive for all ue≥ e −12 and we now consider the numerator
hu()= ( u+− 1)log u 2 u+ 2
We note that he()=− 3 e > 0 and we have the derivative
1 hu′()= log u+− 1 u and he′()= e−1 > 0. We see that
u −1 hu′′()= u2 and therefore hu′′() > 0 for u > 1. This enables us to conclude that hu′() is monotonic increasing for u > e . Since he′() is positive we then deduce that hu() is also monotonic increasing for u > e . Accordingly, hu( )≥ 0 for all u≥ e. Finally, we have φ(u )≥ 0 for all u≥ e. Hence, since the integrand is positive, we conclude that Jn is positive.
32 We now consider Kn and we wish to determine the sign of f ()v where
111 fv()=−+ evv −12
2(2)(1vv+− ev −) = 2(vev − 1)
The denominator of f ()v is positive for all v ≥1 and we now consider the numerator
gv( )=+− 2 v ( v 2)( ev − 1)
We note that g(1)=− 3 e > 0 and we have the derivative
gv′( )=− ( v 1) ev + 1 so that gv′() > 0 for v ≥1. Accordingly, gv( )≥ 0 for all v ≥1. Finally, we have fv( )≥ 0 for all v ≥1. Hence, since the integrand is positive, we conclude that Kn is positive.
∞ ⎡ 111⎤ It has therefore been demonstrated that I =⋅lognvve− −+dv is positive for n ∫ ⎢ v ⎥ 0 ⎣ev−12⎦ all positive integers and for n = 0 .
⎡111⎤ Having regard to (3.7.1) we easily see that lim − +=0 ; alternatively, this ⎢v ⎥ v→0 ⎣⎦ev−12 may be easily determined using L’Hôpital’s rule.
It is possible that such representations for the Stieltjes constants may be useful in connection with the proof of the Riemann Hypothesis via the Li/Keiper constants (see for example [26]). This is because we note from (4.53) that
n ⎛⎞n 11∞ ⎡ 11⎤ (4.57) (1)−Γknkγ ()−− (1) −Γ= () n (1)ev vn log −+dv > 0 ∑⎜⎟ k ∫ ⎢ v ⎥ k=0 ⎝⎠k 210 ⎣ev− 2⎦ and reference to the Appendix then shows that if n is even then Γ()n (1) is positive and thus n ⎛⎞n knk()− (4.58) ∑⎜⎟(1)−Γγ k (1) > 0 k=0 ⎝⎠k □
33 Whilst not specifically employed in this paper, I accidentally came across the following lemma whilst I was trying to prove that In was positive
Suppose that ft( )≤ 0 and ft′( )≤ 0 for all t ∈[1,∞ ) . We see that
∞ I = ft()logn tdt n ∫ 0
1 ∞ =+∫∫f ()logttdtfttnn ()log dt 01 and with the substitution t=1/ y we have
1 ∞ ∫∫f (t ) lognn t dt=− ( 1) f (1/ y ) y−2 log n y dy 01
We therefore obtain
∞ I =+−⎡⎤ft() (1)nn t−2 f (1/)log t tdt n ∫ ⎣⎦ 1 and it is clear that the integrand is negative in the interval [1,∞ ) in the case where n is an even integer. We now consider the case where n is an odd integer and we want to prove that the following integral is also negative
∞ I =−⎡⎤ft() t−+22 f (1/) t logn 1 tdt 21n+ ∫ ⎣⎦ 1
As a consequence of the Mean Value Theorem of calculus we have
ftftttf()−=− (1/)()−1 ′ (α ) where α is such that t >α > t −1 . We see that
f (ttftftftfttft )−=−+−−−22 (1 / ) ( ) (1 / ) (1 / ) (1 / )
=−f (tftft ) (1 / ) + (1 / )( 1 − t−2 )
⎛⎞1 −2 =−⎜⎟tf′()α + ftt (1/)1() − ⎝⎠t
34 =+[tf′()α f (1/)1 t]( − t−2 )
and this is clearly negative in the interval (1,∞ ) . We therefore conclude that I21n+ is also negative.
We could also express In as
1 I =+−⎡⎤ft() (1)nn t−2 f (1/)log t tdt n ∫ ⎣⎦ 0
and hence In will also be negative if ft( )≤ 0 and ft′( )≤ 0 for all t ∈(0,1].
5. An application of the alternating Hurwitz zeta function
The alternating Hurwitz zeta function ς a (,)sx is defined by
∞ (1)− n (,sx ) ς a = ∑ s n=0 ()nx+
Upon a separation of terms according to the parity of n we see that for Re(s ) > 1
∞∞(1)− n 1 ∞1 ς a (,sx )==∑∑s ss − ∑ nn==00()nx+++ (2) nx n = 0 (21) n+ x
⎡⎤∞∞11 2−s =−⎢⎥∑∑ss ⎣⎦nn==00(/2)((1)/2)nx+++ n x
and we therefore see that ς a (,)st is related to the Hurwitz zeta function by the well- known formula [30]
−s ⎡⎤⎛⎞⎛x 1+ x ⎞ (5.1) ςςςa (,sx )=− 2⎢⎥⎜⎟⎜ s , s , ⎟ ⎣⎦⎝⎠⎝22 ⎠
Differentiation gives us
∂+−−s 1 ⎡⎤⎛⎞⎛x 1 x ⎞ ςςςa (,sx )=− 2 s⎢⎥⎜⎟⎜ s + 1, − s + 1, ⎟ ∂x ⎣⎦⎝⎠⎝22⎠
nn+1 ∂∂ 1 ∂ −s nn+1 ς a (,sx )=− ( n + 1) [ f (, sx )2] ∂∂sx s==002 ∂s s
35 and using the Leibniz formula we obtain
n 1 ⎛⎞n nk−− s nk − () k =−(1)(1)2log2nf +⎜⎟ − ⋅ (,)sx 2 ∑ k k =0 ⎝⎠ s=0
⎛⎞⎛x 1+ x ⎞ where fsx(, )=+ςς⎜⎟⎜ s 1, −+ s 1, ⎟ ⎝⎠⎝22⎠
We have
()kk()⎛⎞x () k ⎛1+ x ⎞ fx(0, )=−ςς⎜⎟ 1, ⎜ 1, ⎟ ⎝⎠22 ⎝ ⎠
k ⎡⎤⎛⎞x ⎛1+ x ⎞ =−(1)⎢⎥γγkk⎜⎟ − ⎜ ⎟ ⎣⎦⎝⎠22 ⎝ ⎠ and therefore
n+1 n ∂∂ 11nn+−1 ⎛⎞n k⎡ ⎛⎞x ⎛+x ⎞⎤ (5.2) n+1 ςγak(,sx )=− (1)(n + 1)⎜⎟ log 2⋅ −γk ∑ k ⎢ ⎜⎟ ⎜ ⎟⎥ ∂∂sx s=0 22k =0 ⎝⎠ ⎣ ⎝⎠ ⎝2 ⎠⎦
We now refer to the Hasse identity for the alternating Hurwitz zeta function (see equation (4.4.79) in [21])
∞ 1(i ⎛⎞i −1)j (5.3) ς (,sx )= a ∑∑is+1 ⎜⎟ ij==002(⎝⎠j x + j)
Differentiation gives us
∂−∞ 1(1i ⎛⎞i )j ς (,sx )=− s a ∑∑is+11⎜⎟ + ∂+x ij==002(⎝⎠j xj)
n+1 ∞ i jn ∂∂ n+1 1(1)log(⎛⎞i −x +j) ni++11ς a (,sx )=− (1)(n + 1) ⎜⎟ ∑∑j ∂∂sx s=0 ij==002 ⎝⎠ xj+ and equating this with (5.2) gives us
11ni⎛⎞ni⎡⎤⎛⎞x ⎛+ xx ⎞ ∞ 1⎛⎞(−+1)jnlog(j) (5.4) lognk− 2⋅−γγ = ∑∑⎜⎟ ⎢⎥kk⎜⎟ ⎜ ⎟ i+1 ∑⎜⎟ 2222ki==00⎝⎠kj⎣⎦⎝⎠ ⎝ ⎠ j=0⎝⎠ xj+
We have
36 ∞∞(1)log(−+inx ix ) 1i ⎛⎞i (1)log(−jn+j ) ς ()nn(,sx )=− (1) =−(1)n a ∑∑si+1 ∑⎜⎟ s ii==00()xi++ 2j=0⎝⎠j (x j )
()n and hence we may obtain expressions for ς a (1,x ) in terms of the Stieltjes constants
n ()nn11⎛⎞n −k⎡ ⎛⎞x ⎛+ x ⎞⎤ (5.5) ςγak(1,x )=⋅−∑⎜⎟ log 2 ⎢ ⎜⎟γk ⎜ ⎟⎥ 22k =0 ⎝⎠k ⎣ ⎝⎠ ⎝2 ⎠⎦
With x =1 and n = 0 in (5.4) we obtain
11⎡⎤⎛⎞ ∞ 1i ⎛⎞i (1− )j γγ−=(1) ⎢⎥00⎜⎟ ∑∑i+1 ⎜⎟ 22⎣⎦⎝⎠ ij==00 2⎝⎠j 1+ j which gives us
∞ 1(i ⎛⎞i −1)j log 2 = ∑∑i+1 ⎜⎟ ij==0021⎝⎠j + j
This is equivalent to limς a (s )= log 2 . s→1
With x =1 and n =1 in (5.4) we get
1⎡⎤⎡⎤⎛⎞ 1 1⎛⎞ 1 ∞ 1i ⎛⎞i (− 1)j log(1+ j ) γγ−+−=(1) log 2 γγ ⎢⎥⎢⎥00⎜⎟ 11⎜⎟ ∑∑i+1 ⎜⎟ 22⎣⎦⎣⎦⎝⎠ 22⎝⎠ ij==00 2⎝⎠j 1+ j or 11⎡⎤⎛⎞ ∞ 1i ⎛⎞i (1)log(1− j + j) log2 2 +−=γγ ⎢⎥11⎜⎟ ∑∑i+1 ⎜⎟ 22⎣⎦⎝⎠ ij==00 2⎝⎠j 1+ j
We have the relationship [23]
qp−1 p+1 ⎛⎞rqpjlog ⎛⎞p j (5.6) ∑∑γγpp⎜⎟=− +qq(1) − +⎜⎟(1) − γp− j logq rj==10⎝⎠qp+1 ⎝⎠j and with q = 2 this becomes
p+1 p ⎛⎞1lpjog2⎛⎞p j (5.7) γγpp⎜⎟=− +2( − 1) + 2∑⎜⎟ ( − 1)γp− j log 2 ⎝⎠21p + j=0 ⎝⎠j and in particular we have
37 ⎛⎞1 2 (5.8) γγ11⎜⎟=− −log 2 + 2 γγ 1 − 2 log 2 ⎝⎠2
We then obtain
1 1∞ 1i ⎛⎞i (− 1)j log(1+ j ) (5.9) γ =−log 2 ∑∑i+1 ⎜⎟ 2log22ij==00⎝⎠j 1+ j
This expression for Euler’s constant was originally derived by Coffey [17] in 2006 (a different derivation is contained in equation (4.4.116g) in [20])).
Similarly we may also obtain with n = 2
1 1∞∞ 1 ii⎛⎞iilog(1+ j ) 1 1 ⎛⎞ log2 (1+ j ) γ =−log2 2 + (− 1) jj− (− 1) 1 ∑∑ii++11⎜⎟ ∑∑⎜⎟ 12 2ij==00 2 ⎝⎠jj1++j 2log 2ij==00 2 ⎝⎠ 1 j which was also previously determined by Coffey [17] in a different manner.
Letting x =1and x = 2 in (5.4) results in
111ni⎛⎞ni⎡⎤⎛⎞ ∞ ⎛⎞(−1)jnlog(1+ j) lognk− 2⋅−=γγ ∑∑⎜⎟ ⎢⎥kk⎜⎟ i+1 ∑⎜⎟ 2221ki==00⎝⎠kj⎣⎦⎝⎠ j=0⎝⎠ + j
13ni⎛⎞ni⎡⎤⎛⎞ ∞ 1⎛⎞(−1)jnlog(2+ j) lognk− 2⋅−γγ = ∑∑⎜⎟ ⎢⎥kk⎜⎟ i+1 ∑⎜⎟ 22ki==00⎝⎠kj⎣⎦⎝⎠ 2j=0⎝⎠ 2+ j
We note from [38, p.89] that for m∈ N
m−1 1 (,sm x ) (, sx ) ςς+= −∑ s j=0 ()j + x and thus
∂+kkm−1 log (j x ) [(,sm x ) (,)](1) sx k+1 ksςς+− =− ∑ ∂+sjj=0 ()x
Therefore, referring to (1.3) we obtain
m−1 logk (j + x ) γγkk()()mx+− x =−∑ j=0 j + x
38 and in particular we have
logk x γγ(1+−xx ) ( ) =− kk x
Therefore we have
⎛⎞31 ⎛⎞ kk+1 γγkk⎜⎟−=− ⎜⎟(1)2log2 ⎝⎠22 ⎝⎠
APPENDIX
Some aspects of the (exponential) complete Bell polynomials
It is well known that [33]
m d fx() fx () (1) (2) ()m (A.1) m eeYfxfxfx= m ()( ), ( ),..., ( ) dx
where the (exponential) complete Bell polynomials Yxn(1 ,..., xn ) are defined by Y0 =1 and for n ≥ 1
kk12 kn n! ⎛⎞⎛⎞xx12 ⎛⎞xn (A.2) Yxnn(1 ,..., x ) = ∑ ⎜⎟⎜⎟... ⎜⎟ π ()n kk12! !... kn !⎝⎠⎝⎠ 1! 2!⎝⎠n !
where the sum is taken over all partitions π ()n of n , i.e. over all sets of integers k j such that
kkk123++++=2 3 ... nkn n
For example, with n =1 we see that the only possibility is k1 =1and kjj =∀≥0 2 which results in
Yx11()= x 1
With n = 2 , we see that the possible outcomes are ( k1 = 2 and k2 = 0 ) and ( k1 = 0 and k2 =1) which results in
2 Yxx212(, )=+ x 1 x 2
Suppose that hx′()()()= hxgx and let f (xh )= log (x ) . We see that
39 hx′() f ′()xg== ()x hx() and then using (A.1) above we have
mm ddloghx ( ) (1) (m− 1) (A.3) mmhx() == e hxY( )m () gx ( ), g ( x ),..., g ( x ) dx dx
As an example, letting hx()()=Γ xin (A.3) we obtain
m d logΓ− (xm ) ( ) (1) (m 1) (A.4) m exxYxx=Γ( ) =Γ ( )m ()ψψ ( ), ( ),..., ψ (x ) dx
∞ = ∫text−−1 log m tdt 0 and since [26, p.22]
(A.5) ψ ()pp()x =− (1)+1pp !(ς + 1,) x we may express Γ()m ()x in terms of ψ ()x and the Hurwitz zeta functions. In particular, Kölbig [33] noted that
()m (A.6) Γ=−(1)Yxxmm (γ ,11 ,...,− )
p+1 ()m where xppp =−(1) !(ς + 1). Values of Γ (1) for m ≤10 are reported in [38, p.265] and the first three are
Γ=−(1) (1) γ
Γ=+(2)(1)ς (2) γ 2
Γ=−+(3) (1) [2ς (3) 3γς (2) + γ 3 ]
As shown in [22], we note that Γ()n ()x has the same sign as (− 1)n for all x∈(0,α ] where α is the unique positive root of ψ ()x = 0 (Gauss determined that α ≈1.4616321...). This was also reported as an exercise in Apostol’s book [3, p.303] for the particular case of Γ()n (1) .
We have from (A.5) and (4.21)
40 k d 11 k (A.6.1) k =−+−+−−−Yxk ()ψς(1 ), 1! (2,1 x ),..., ( 1) ( kk 1)!ς ( ,1+x ) dxΓ+(1 x ) Γ+ (1 x ) so that with x = 0
k d 1 k (A.6.2) k =−Ykk ()γς, 1! (2),..., −−− ( 1) ( 1)!ς (k ) dxΓ+(1 x ) x=0
The complete Bell polynomials have integer coefficients and the first six are set out below [19, p.307]
(A.7) Yx11()= x 1
2 Yxx212(, )= x 1+ x 2
3 Yxxx3123(, , )= x 1++ 3 xx 12 x 3
42 2 Yxxxx41234(, , , )= x 1++++ 6 xx 12 4 xx 13 3 x 2 x 4
53 2 2 Yxxxxx512345( , , , , )= x 1++++++ 10 xx 12 10 xx 13 15 xx 12 5 xx 14 10 xx 23 x 5
6323 Yxxxxxx6123456( , , , , , )=+ x 1 6 xx 15 + 15 xx 24 + 10 x 2 + 15 xx 14 + 15 x 2 + 60 xxx 123
3224 +20x13xxxxx+++ 45 12 15 11x 6
The complete Bell polynomials are also given by the exponential generating function (Comtet [19, p.134])
⎛⎞∞∞ttj n (A.8) exp⎜⎟∑∑xYxxjn= (1 ,...,n ) ⎝⎠jn==10j!!n
We see that
dtnj⎛⎞∞ exp x = Yx( ,..., x ) n ⎜⎟∑ jn1 n dt j=1 j! ⎝⎠t=0
We note that a ∞∞ttnj⎛ ⎞⎛⎞⎛⎞∞ t j⎡ ∞tj⎤ ∑∑∑Yaxnn(1 ,..., ax )== exp⎜ax j ⎟⎜⎟⎜⎟exp a xj =⎢exp ∑xj⎥ nj==01nj!!⎝ ⎠⎝⎠⎝⎠j=1 j !⎣⎢ j=1j!⎦⎥
41 and thus we have
a ⎡⎤∞∞ttnn ⎢⎥∑∑Yxnn(11 ,..., x ) = Yax n( ,..., ax n ) ⎣⎦nn==00nn!!
We see with a =−1 that
∞∞ttnn 1=−−∑∑Yxnnn (11 ,..., x ) Yxx ( ,...,n ) nn==00nn!! and using the Cauchy product formula this becomes
∞ n ⎛⎞n t n =−∑∑⎜⎟Yxj(11 ,..., xY j ) nj−− ( x ,...,− xnj ) nj==00⎝⎠j n!
Hence we deduce that for n ≥1
n ⎛⎞n (A.9) ∑⎜⎟Yxj(11 ,..., xY j ) nj−− (− x ,...,−= xnj ) 0 j=0 ⎝⎠j
From the definition of the (exponential) complete Bell polynomials we have
2 mm Yaxaxmm(12 , ,..., ax )= aYxr (1 ,..., xm ) and thus with a =−1 we have
mm Yxxmm(−−=−12 , ,...,( 1) x ) ( 1) Yxxm (1 ,...,m )
We also note that
mm+1 Yxmm(12 ,−− x ,...,( 1) x ) =−−− ( 1) Ym ( x1 ,..., xm ) but no discernible sign pattern emerges here.
We have the recurrence relation [35]
nn⎛⎞nn⎛⎞ Yxxn++11( ,...,n 1 ) ==∑∑⎜⎟ Yxxxnk −(1 ,..., nk −+ ) k1 ⎜⎟ Yxxxk(1 ,..., k ) nk−+1 kk==00⎝⎠kk⎝⎠
n ⎛⎞n =+xYxxnkk+−11∑⎜⎟ ( ,..., )xnk+1 k =1 ⎝⎠k
42 We note that
⎛⎞⎛⎞⎛∞∞ttjj∞tj⎞ Yxnn(11++=+= y ,..., x yn ) exp⎜⎟⎜⎟⎜∑∑ ( x j yj ) expx j exp ∑ yj⎟ ⎝⎠⎝⎠⎝jj==11j!!jjj=1!⎠
∞∞ttnn =⋅∑∑Yxnn(11 ,..., x ) Yy n ( ,..., y n ) nn==00nn!! and, as before, we apply the Cauchy series product formula to obtain
n ⎛⎞n (A.10) Yxnn(11++= y ,..., x ynn ) ∑⎜⎟ Y−−kn( x1 ,..., xk ) Yyk (1 ,..., yk ) k =0 ⎝⎠k and we note that
k Yk (α ,0,...,0) = α
With yn=−xn we obtain from (A.10)
n ⎛⎞n (A.11) YYnn(0,...,0)== 0 ∑⎜⎟ −−k(x11 ,...,xnk )Yk (−x ,..., −xk ) k =0 ⎝⎠k as in (A.9) above.
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Donal F. Connon Elmhurst Dundle Road Matfield Kent TN12 7HD [email protected]
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