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Home , Hurwitz zeta function, Psi (Greek)

, Psi , Bernoulli Functions via Hurwitz Function .

VIVEK V . RANE

A-3/203 , Anand Nagar ,

Dahisar (East)

Mumbai – 400 068

INDIA

e-mail address : v_v_rane@ yahoo.co.in

Abstract : Using three basic facts concerning Hurwitz zeta function , we give new natural proofs of the known results on , and also obtain Gauss’ expression for psi function at a rational point all in a unified fashion . We also give a new proof of the relation between log gamma and the derivative of the Hurwitz zeta function including that of Stirling’s expression for

log gamma .

Keywords : Hurwitz zeta function , Gamma function , , Dirichlet L-series , Bernoulli polynomials/Numbers , Gauss’ psi function .

Gamma , Psi , Bernoulli Functions via Hurwitz Zeta Function

VIVEK V . RANE

A-3/203 , Anand Nagar ,

Dahisar (East)

Mumbai – 400 068

INDIA

e-mail address : v_v_rane@ yahoo.co.in

The object of this paper is to indicate that the theory of Euler’s gamma function , that of Gauss ‘ psi function and that of Bernoulli functions can be developed naturally , with ease from the theory of

Hurwitz’s zeta function and its derivatives .

For a complex variable s and for a α≠0 , -1, -2 , ………. let ζ(s,α ) be the

∞ −푠 Hurwitz’s zeta function defined by ζ(s,α)= 푛=0(푛+∝) for Re s>1 ; and its . Let

ζ(s,1)=ζ(s) be the . If χ(mod q) is a Dirichlet character modulo an integer q≥1 and

푎 if L(s,χ) is the corresponding Dirichlet L-series , then in view of L(s,χ)=q-s 푞 휒 푎 휁(푠, ) , it is already 푎=1 푞 clear from author [2] and many other works of the author that the theory of Dirichlet L-series can be developed naturally from the theory of ζ(s,α) .

For integers n≥0 , let Bn(α) be the Bernoulli polynomials in α defined by

푧푒 ∝푧 푧푛 = ∞ 퐵 (훼) for |z|<2π ; and let B =B (0) be Bernoulli numbers . Note that B (α)=1 . Then it is 푒 푧−1 푛=0 푛 푛! n n 0 known that if n≥0 is an integer , Bn+1(α)=(-n-1)ζ(-n,α) and this fact has been proved in author [4] by comparing the Fourier series expressions of ζ(-n,α) and Bn+1(α) . As a function of s , ζ(s,α) is a with a simple pole at s=1 with residue 1 . As a function of α , ζ(s,α) is an analytic function of α except possibly for α=0,-1,-2 , ………. However , if m≥0 is an integer , then ζ(-m,α) is a polynomial in α and thus is an entire function of α (See author [2]) . For any integer r≥0 , we write : 2 :

휕푟 휕 ζ(r)(s,α)= 휁(푠, 훼) . In particular , ζ’(s,α)= ζ(s,α) . As a function of α , 휁(푟)(s,α) is an analytic function 휕푠푟 휕푠 of α except possibly for α=0 , -1 , -2 , ………. (See author [2] , author [3] .) If s≠1 and α≠0 , -1 , -2, ……….,

휕푟1 휕푟2 휕푟2 휕 푟1 then 휁 푠, 훼 = 휁 푠, 훼 , where 푟 푟 ≥ 0 are integers . See for example author [3] . 휕∝푟1 휕푠푟2 휕푠푟2 휕∝푟1 1 , 2

휕 휕 휕 휕 In particular , if s≠1 and if α≠0 ,-1 , -2 , ………. , then 휁 푠, 훼 = 휁 푠, 훼 . 휕훼 휕푠 휕푠 휕훼

Next , let Г(α) be the Euler’s gamma function defined by Weierstrass ‘ product namely

훼 1 훾훼 훼 − Г(훼) = 훼푒 Π (1+ )푒 푛 , where γ is Euler’s constant . Then it is known that ζ’(0,α)=log . In Г(훼) n≥`1 푛 2휋 what follows , this fact will be proved with ease afresh as Proposition 5 , wherein we shall also obtain

푑 푑 Г(훼) 휕 Stirling’s expression for log Г(α) . Next , let ψ(α)= log Г(α)= log = ζ’(0,α) . In Proposition 4 푑훼 푑훼 2휋 휕훼

푎 below , we shall obtain Gauss’ expression for ψ , where 1≤a

We shall show that by defining

1) 퐵푛+1 훼 = −푛 − 1 휁(−푛, 훼) for n≥ -1

휕 2) log Г(α)= ζ(s,α) at s=0 휕푠

휕 휕 3) ψ(α)= 휁 푠, 훼 at s=0 휕훼 휕푠 we can develop naturally the theory of Bernoulli polynomials , gamma function and Gauss’ ψ function in terms of ζ(s,α) and its derivatives as in the case of Dirichlet L-series .

Basically , there are three facts concerning ζ(s,α) .

휕 1) 휁 푠, 훼 = −푠휁(푠 + 1 , 훼) 휕훼 : 3 ∶

2) 휁 푠, 훼 − 휁 푠, 훼 + 1 = 훼−푠

휋푠 3) 퐹표푟 푅푒 s<1 and for 0<α<1 , ζ(s,α)=2푠휋푠−1Г(1-s) sin + 2휋푛훼 푛푠−1 푛≥1 2

2휋𝑖푛훼 s-1 =Г(1-s) |푛|≥1 푒 (2πin)

Fact I) resulted in the expansion of ζ(s,α+1) in author *1+ in the form

(−훼)푛 휁 푠, 훼 + 1 = 휁 푠 + s(s+1) ………. (s+n-1)ζ(s+n) and consequently in author [2] , we find 푛≥1 푛!

훼푚 +1 that for integral m≥0 , ζ(-m,α)= 푚 푚 ζ(-k)훼푚−푘 + 훼푚 − . In author [4] , it has been shown 푘=0 푘 푚 +1

푚′ 푚′ 푚 ′ −푘 that the above expression for ζ(-m,α) with m≥0 , is equivalent to 퐵푚′ (α)= 푘=0 푘 퐵푘 훼

푤𝑖푡ℎ 푚′ = 푚 + 1 . Thus fact II) and fact III) remain to be exploited . Using these facts , we prove

with ease the following four propositions .

Proposition 1 We have

푚−1 1) 퐵푚 (α+1) - 퐵푚 훼 = 푚훼 for m≥0 .

2) Г 훼 + 1 = 훼Г 훼

1 Corollary : ψ(α+1)=ψ (α) + 훼

Proposition 2 : We have

푚 1) 퐵푚 1 − 훼 = −1 퐵푚 (훼)

휋 2) Γ 훼 Γ 1 − 훼 = 푠𝑖푛휋훼

Corollary : ψ(1-α)=ψ (α) +π cot πα

Proposition 3 : We have for any integer m≥2 and for any integer n≥1 , : 4 :

1 2 푚 −1 1) 퐵 훼 + 퐵 훼 + + 퐵 (훼 + ) + ⋯ … … . +퐵 훼 + = 푚1−푛 퐵 푚훼 푛 푛 푚 푛 푚 푛 푚 푛

푛−1 1 1 푛−1 −푛훼 2) Γ 훼 Γ 훼 + … … … . Γ 훼 + = 2휋 2 푛2 Γ(푛훼) 푛 푛

푘 Corollary : We have 푚 −1 sinπ⁡(훼 + ) = 21−푚 sin 휋푚훼 . 푘=0 푚

Note that in the above three propositions , the results are in pairs . The results pertaining to Bernoulli polynomials are about sums and the results pertaining to gamma function are about products .

Next , we state remaining Propositions .

Proposition 4 : We have If 1≤a

푎 2휋푟푎 휋푟 휋 2휋푟푎 then 휓 = −(훾 + log 푞 ) + 푞−1 cos log 2 sin + 푞−1 푟 sin 푞 푟=1 푞 푞 푞 푟=1 푞

Next with Weierstrass product definition of Г(α)−1 and using Euler’s summation formula , we

1 state and prove Proposition 5 . In what follows , we shall write 휙 푢 = 푢 − 푢 − , where [u] denotes 2 integral part of the real variable u .

1 Proposition 5 : Let 휙 푢 = 푢 − 푢 − . 2

∞ 휙 (푢) ∞ 휙(푢) Γ(훼) Then log Г(α)=(α-1/2) log α – α +1 + 푑푢 − du and 휁′ 0, 훼 = log . 1 푢 0 푢+훼 2휋

Next , we prove our propositions . We shall prove Proposition 5 first .

1 Proof of Proposition 5 : From Weierstrass product expression for on taking logarithm , Γ(훼)

훼 훼 we have log Г 훼 = ( − log⁡(1 + )) − 훾훼 − log⁡ 훼 . 푛≥1 푛 푛 : 5 :

On using Euler’s summation formula

1 ∞ 훼 훼 log Γ( 훼) = 훼 − log 1 + 훼 − 훾훼 − log 훼 + ( − log⁡(1 + ))du + 2 1 푢 푢

∞ 푑 훼 훼 + ∅(푢) ( - log(1+ ))푑푢 , where 휙 푢 = 푢 − 푢 − 1/2 . 1 푑푢 푢 푢

∞ 훼 훼 푢 ∞ 푢 1 Next ( − log⁡(1 + ))du = 푢 + 훼 푙표𝑔 = lim푢 ∞ 푢 + 훼 푙표𝑔 − 1 + 훼 푙표𝑔 1 푢 푢 푢+훼 푢=1 푢+훼 1+훼

= -훼 + 1 + 훼 log 1 + 훼 .

1 1 ∞ 푑 훼 훼 Thus log Γ 훼 = (훼 + ) log (1+α) – (γ + ) α – log α + 휙 푢 − log 1 + 푑푢 . 2 2 1 푑푢 푢 푢

∞ 푑 훼 훼 ∞ ∅(푢) 1 1 푁푒푥푡 , 휙(푢) − log⁡(1 + ) du = 훼 − du 1 푑푢 푢 푢 1 푢 푢+훼 푢

∞ 1 1 ∞ ∅(푢) ∞ ∅(푢) ∞ ∅(푢) 1 = 휙(푢) − 푑푢 − 훼 푑푢 = du - du +α 훾 − 1 푢 푢+훼 1 푢2 1 푢 1 푢+훼 2

∞ ∅(푢) ∞ ∅(푢) 1 1 1 = du - du +α 훾 − + 1 +(훼 + ) log α – (α+ )log(1+α) . 1 푢 0 푢+훼 2 2 2

1 ∞ ∅(푢) ∞ ∅(푢) Thus log Γ 훼 = (훼 − ) log α – α+1 + du - du . 2 1 푢 0 푢+훼

∞ ∅(푢) This gives an expression for log Γ(훼) . Note that integrating by parts repeatedly , can be 0 푢+훼 developed into an asymptotic series in α .

Γ(훼) Next , we shall prove 휁′ 0, 훼 = log . For 0<α≤1 , Re s>1 and for arbitrary >0 , 2휋

−푠 −푠 −푠 we have 휁 푠, 훼 = 푛≥0(푛 + 훼) = 0≤푛≤푥−훼(푛 + 훼) + 푛>푥−훼(푛 + 훼) .

−푠 We use Euler’s summation for 푛>푥−훼(푛 + 훼) . : 6 :

This gives for x>0 , 0<α≤1 and for Re s>0 ,

푥 1−푠 1 ∞ 휙 푢−훼 휁 푠, 훼 = (푛 + 훼)−푠 + + (x-α-[x-α+- )푥−푠-s du . 0≤푛≤푥−훼 푠−1 2 푥 푢푠+1

훼1−푠 훼−푠 ∞ 휙(푢−훼) Choosing x=α , we get 휁 푠, 훼 = 훼−푠 + − −푠 du 푠−1 2 훼 푢푠+1

훼−푠 훼1−푠 ∞ 휙 푢−훼 so that 휁 푠, 훼 = + –s du . 2 푠−1 훼 푢푠+1

훼−푠 훼1−푠 휁 푠.훼 − − ∞ 휙(푢−훼) Thus lim 2 푠−1 = - lim 푑푢 . 푠 0 푠 푠 0 훼 푢푠+1

훼−푠 훼1−푠 1 1 1 As s 0 , 휁 푠, 훼 − − = 휁 0, 훼 − + α = ( -α) - + α 0 . 2 푠−1 2 2 2

0 Thus the left hand side is of the form as s 0 through real values . Hence by L’Hospital’s rule , 0

훼 −푠 훼1−푠 휁 푠,훼 − – 훼−푠 훼1−푠 log 훼 훼1−푠 We have lim 2 푠−1 = lim 휁′ 푠, 훼 + 푙표𝑔훼 + + 푠 0 푠 푠 0 2 푠−1 (푠−1)2

1 ∞ 휙 푢−훼 ∞ 휙 푢−훼 =휁′ 0, 훼 + log 훼 − 훼 log 훼 + 훼 and lim 푑푢 = 푑푢 . 2 푠 0 훼 푢푠+1 훼 푢

1 ∞ 휙(푢−훼) Thus we have 휁′ 0, 훼 + log 훼 − 훼 log 훼 + 훼 = − 푑푢 . 2 훼 푢

1 ∞ 휙(푢) That is 휁′ 0, 훼 + log 훼 − 훼 log 훼 + 훼 = − 푑푢 . 2 0 푢+훼

1 ∞ 휙(푢) That is 휁′ 0, 훼 = (훼 − )log 훼 − 훼 − 푑푢 . 2 0 푢+훼

This gives 휁′ 0, 훼 = 푙표𝑔 Г 훼 + 푐 with c a constant , on comparing with the above expression for

1 푙표𝑔 Г 훼 . Putting α=1 , we get c=ζ’(0)= - log 2π . 2 : 7 :

Γ(훼) Thus we have 휁′ 0, 훼 = 푙표𝑔 . 2휋

Proof of Proposition 1: I) We have 휁 푠, 훼 − 휁 푠, 훼 + 1 = 훼−푠 .

퐵 훼 퐵 훼+1 Taking s= - m , where m≥0 is an integer , we have – 푚 +1 + 푚 +1 =훼푚 . 푚+1 푚 +1

푚 That is , 퐵푚+1 훼 + 1 − 퐵푚 +1 훼 = 푚 + 1 훼

II) From 휁 푠, 훼 − 휁 푠, 훼 + 1 = 훼−푠 , on differentiation with respect to s , we have 휁′ 푠, 훼 − 휁′ 푠, 훼 + 1 = −훼−푠 log 훼 .

Putting s=0 , we have 휁′ 0, 훼 − 휁′ 0, 훼 + 1 = − log 훼 .

Γ(훼) Γ(훼+1) Thus 푙표𝑔 − 푙표𝑔 = - log α , in view of Proposition 5 . 2휋 2휋

That is log Г(α+1) – log Г(α) = log α or Г(α+1) = αГ(α) .

Proof of Corollary : Differentiating the expression log Г(α+1) – log Г(α) = log α ,

1 we get ψ(α+1)-ψ(α) = . 훼

Proof of Proposition 2 : I) We have for Re s<1 and for 0<α<1 ,

휋푠 휁 푠, 훼 = 2푠휋푠−1Γ(1 − 푠) sin + 2휋푛훼 푛푠−1 푛≥1 2

휋푠 So that 휁 푠, 1 − 훼 = 2푠휋푠−1Γ(1 − 푠) sin − 2휋푛훼 푛푠−1 . 푛≥1 2

Writing s= -m , where m≥0 is an integer ,

휋푚 we have 휁 −푚, 1 − 훼 = −2−푚 휋−푚−1Γ(1 + 푚) sin⁡ + 2휋푛훼 푛−푚 −1 푛≥1 2 : 8 :

−휋푚 whereas 휁 −푚, 훼 = 2−푚 휋−푚 −1Γ(1 + 푚) sin⁡ + 2휋푛훼 푛−푚 −1 푛≥1 2

This gives 휁 −푚, 1 − 훼 = (−1)푚 +1 휁 −푚, 훼

푚 +1 That is , 퐵푚+1 1 − 훼 = (−1) 퐵푚 +1 훼 .

휋푠 II) We have for Re s<1 and for 0<α<1 , 휁 푠, 훼 = 2(2휋)푠−1Γ(1 − 푠) sin⁡ + 2휋푛훼 푛푠−1 푛≥1 2

휋푠 Consider 휁 푠, 1 − 훼 = 2(2휋)푠−1Γ(1 − 푠) sin⁡ − 2휋푛훼 푛푠−1 . 푛≥1 2

Differentiating with respect to s,

휋푠 휁′ 푠, 1 − 훼 = 2(2휋)푠−1 log 2휋 ∙ Γ 1 − 푠 − Γ′ (1 − 푠)) sin⁡ − 2휋푛훼 ∙ 푛푠−1 푛≥1 2

휋 휋푠 휋푠 +Γ(1 − 푠) cos − 2휋푛훼 + sin − 2휋푛훼 ∙ log 푛 푛푠−1} 푛≥1 2 2 2

휋 1 (−푠𝑖푛 2휋푛훼 ) ( 푐표푠2휋푛훼 −log 푛∙sin 2휋푛훼 ) Thus 휁′ 0,1 − 훼 = { 푙표𝑔2휋 + γ + 2 } 휋 푛≥1 푛 푛≥1 푛

1 푠𝑖푛 2휋푛훼 Replacing α by 1-α , 휁′ 0, 훼 = {(푙표𝑔2휋+훾) + 휋 푛≥1 푛

1 휋 + cos 2휋푛훼 + log 푛 ∙ sin 2휋푛훼 } . 푛≥1 푛 2

푐표푠2휋푛훼 1 (푒 2휋𝑖푛훼 + 푒 −2휋𝑖푛훼 ) Thus ζ’(0,α) + 휁′ (0 ,1 − 훼) = = 푛 푛 2 푛≥1 푛

1 1 = - log 1 − 푒2휋𝑖훼 + log⁡(1 − 푒−2휋𝑖훼 ) = - log 푒휋𝑖훼 푒−휋𝑖훼 − 푒휋𝑖훼 + log 푒−휋𝑖훼 푒휋𝑖훼 − 푒−휋𝑖훼 2 2

1 = - 푙표𝑔 푒휋𝑖훼 − 푒−휋𝑖훼 + log −푒휋𝑖훼 + log 푒휋𝑖훼 − 푒−휋𝑖훼 + 푙표𝑔 푒−휋𝑖훼 2

1 (푒 휋𝑖훼 −푒 −휋𝑖훼 ) = - 2 푙표𝑔 + log −2𝑖푒휋𝑖훼 + 푙표𝑔 2𝑖푒−휋𝑖훼 2 2𝑖 : 9 :

1 = - {2푙표𝑔푠𝑖푛휋훼 + log −4𝑖2 } = - (푙표𝑔 푠𝑖푛 휋훼 + log 2) = −log 2 sin πα 2

Thus ζ’(0,α) + 휁′ 0,1 − 훼 + 푙표𝑔 2푠𝑖푛휋훼 = 0

Γ(훼) Γ(1−훼) That is 푙표𝑔 + 푙표𝑔 +log 2sinπα=0 . 2휋 2휋

Γ 훼 Γ 1−훼 푠𝑖푛휋훼 푠𝑖푛휋훼 That is , log = 0 . That is , Γ 훼 Γ 1 − 훼 = 1 휋 휋

휋 That is Γ 훼 Γ 1 − 훼 = . 푠𝑖푛휋훼

Proof of Corollary : From log Г(α)+log Г(1-α) = log π-log sin πα , on differentiation , we have ψ(α)=ψ(1-α) - π cot πα

−푠 Proof of Proposition 3 : I) For Re s>1 , 휁 푠, 훼 = 푛≥0(푛 + 훼) .

1 1 For an integer m≥2 , consider ζ(s,α+ ) = (푛 + 훼 + )−푠 = 푚푠 (푚푛 + 1 + 푚훼)−푠 푚 푛≥0 푚 푛≥0

푠 −푠 −푠 푚−1 −푠 = 푚 { 푛≥0(푛 + 푚훼) – 푛≡0 푚표푑 푚 (푛 + 푚훼) – 푟=2 푛≡푟 푚표푑 푚 (푛 + 푚훼) }

푠 −푠 −푠 푚 −1 −푠 = 푚 { 푛≥0(푛 + 푚훼) − 푘≥0(푚푘 + 푚훼) − 푟=2 푘≥0(푚푘 + 푟 + 푚훼) }

푟 = 푚푠{ 휁 푠, 푚훼 − 푚−푠 (푘 + 훼)−푠 − 푚−푠 푚 −1(푘 + 훼 + )−푠} 푘≥0 푟=2 푚

푟 = 푚푠 휁 푠, 푚훼 − 휁 푠, 훼 − 푚−1 휁(푠, 훼 + ) 푟=2 푚

푟 Thus 푚−1 휁(푠, 훼 + ) = 푚푠휁 푠, 푚훼 푟=0 푚

푟 Writing s= - (n-1) , where n≥1 is an integer , we have 푚−1 퐵 (훼 + ) =푚1−푛 퐵 (mα). 푟=0 푛 푚 푛 : 10 :

푟 Next differentiating 푚−1 휁(푠, 훼 + ) = 푚푠휁 푠, 푚훼 with respect to s , 푟=0 푚

푟 we have 푚−1 휁′(푠, 훼 + ) = 푚푠휁′ (푠, 푚훼) + 푚푠푙표𝑔 푚 ∙ 휁(푠, 푚훼) . 푟=0 푚

푟 Putting s=0 , we have 푚−1 휁′(0, 훼 + ) = 휁′ 0, 푚훼 + 푙표𝑔푚 ⋅ 휁(0 , 푚훼) . 푟=0 푚

r Γ α+ Γ(푚훼 ) 1 That is , 푚−1 log m = log + 푙표𝑔푚 ∙ ( –mα) . 푟=0 2휋 2휋 2

푟 Thus 푚−1 푙표𝑔Γ(훼 + ) = 푙표𝑔Γ 푚훼 + 푚 − 1 푙표𝑔 2휋+log 푚 –mα log m . 푟=0 푚

푚 −1 1 푟 (2휋) 2 푚 2Γ(푚훼 ) Thus log 푚 −1 Γ(훼 + ) = log . 푟=0 푚 푚 푚훼

푚 −1 1 푚 −1 푟 −푚훼 Thus Γ(훼 + ) = (2휋) 2 푚2 Г(푚훼) . 푟=0 푚

푘 푘 Proof of Corollary : Consider for 0<α<1 , 푛−1 Γ(훼 + ) ∙ 푛−1 Γ(−훼 + ) . 푘=0 푛 푘=0 푛

1 1 푛−1 −푛훼 +푛훼 푛−1 = (2휋) ∙ 푛2 ∙ Γ(푛훼) ∙ 푛2 ∙ Γ −푛훼 = (2휋) ∙ 푛 Γ(푛훼)Γ −푛훼

(2휋)푛−1 (2휋)푛−1 (2휋)푛−1 π = - Γ 푛훼 ∙ −푛훼 ∙ Γ −푛훼 = - Γ 푛훼 Γ 1 − 푛훼 = - 훼 훼 훼 푠𝑖푛휋푛훼

푘 푘 푘 푛−푘 On the other hand 푛−1 Γ(훼 + ) ∙ 푛−1 Γ(−훼 + ) = 푛−1 Γ(훼 + ) 푛 Γ(−훼 + ) 푘=0 푛 푘=0 푛 푘=0 푛 푘=1 푛

푛−1 푛−1 푘 푘 푛−1 π 휋 Γ(α)Γ(1−훼) = Г(α)Г(-α)∙ 푘=1 Γ(훼 + )Γ(1 − 훼 − ) = Г(α)Г(-α) 푘=1 푘 = - 푘 푛 푛 푠𝑖푛휋 (훼+ ) 훼 푛−1 푠𝑖푛휋 (훼+ ) 푛 푘=1 푛

푛 푛 휋 1 휋 푛−1 푘 −1 = - 푘 = − ( 푘=0 푠𝑖푛휋(훼 + )) . 훼 푠𝑖푛휋훼 ∙ 푛 −1 푠𝑖푛휋 (훼+ ) 훼 푛 푘=1 푛

−1 (2휋)푛−1 휋 휋푛 푘 Thus we have – ∙ = − 푛−1 푠𝑖푛휋(훼 + ) 훼 푠𝑖푛휋푛훼 훼 푘=0 푛 : 11 :

푛 −1 2 1 Thus = 푘 푠𝑖푛휋푛훼 푛−1 푠𝑖푛휋 (훼+ ) 푘=0 푛

푘 Thus 푛−1 푠𝑖푛휋(훼 + )=21−푛 sinπnα . 푘=0 푛

푑 휕 휕 Proof of Proposition 4 : We have ψ(α)= log Г(α) = 휁(푠, 훼) 푑훼 휕훼 휕푠 푠=0

휕 휕 휕 휕 휕 = 휁(푠, 훼) = −푠휁(푠 + 1, 훼) 푠=0 = (−(푠 − 1)휁(푠, 훼) = (1 − 푠)휁(푠, 훼) 휕푠 휕훼 푠=0 휕푠 휕푠 푠=1 휕푠 푠=1

Next , we have for Re s<1 and 0<훼 < 1

휋푠 휁(s,α)=2(2π)푠−1Г(1-s) 푠𝑖푛 + 2휋푛훼 ∙ 푛푠−1 푛≥1 2

휋푠 So that (1-s)ζ(s,α) = 2(2π)푠−1Γ 2 − 푠 sin⁡( + 2휋푛훼)푛푠−1 푛≥1 2

푎 휋푠 푎 This gives for Re s<0 , (1-s)ζ(s, )=2(2π)푠−1Γ 2 − 푠 sin⁡( + 2휋푛 )푛푠−1 푞 푛≥1 2 푞

휋푠 푎 =2(2π)푠−1Γ 2 − 푠 푞 sin⁡( + 2휋푟 ) ∙ 푛푠−1 푟=1 2 푞 푛≡푟(푚표푑 푞)

휋푠 푎 푟 =2(2π푞)푠−1Γ 2 − 푠 푞 sin⁡( + 2휋푟 ) ∙ 휁(1 − 푠, ) . 푟=1 2 푞 푞

휕 푎 휋푠 2휋푟푎 푟 Thus 1 − 푠 휁 푠, =2(2π푞)푠−1 log 2휋푞 − Γ′ (2 − 푠) ∙ 푞 sin + ∙ 휁 1 − 푠, 휕푠 푞 푟=1 2 푞 푞

π 휋푠 2휋푟푎 푟 휋푠 2휋푟푎 푟 + 2(2π푞)푠−1Γ 2 − 푠 푞 { cos( + ) ∙ 휁(1 − 푠, ) - sin( + )휁′(1-s, )} 푟=1 2 2 푞 푞 2 푞 푞

Γ′ 휋푠 2휋푟푎 푟 =2(2π푞)푠−1 log 2휋푞 − 2 − 푠 Γ(2 − 푠) ∙ 푞 sin( + )휁(1 − s, ) Γ 푟=1 2 푞 푞

π 휋푠 2휋푟푎 푟 휋푠 2휋푟푎 푟 +2(2π푞)푠−1Γ 2 − 푠 푞 { cos⁡( + )휁(1 − 푠, ) - sin( + )휁′(1-s, )} . 푟=1 2 2 푞 푞 2 푞 푞 : 12 :

Γ′ 휋푠 2휋푟푎 푟 Next 2(2휋푞)푠−1(log 2휋푞 − (2 − 푠)) ∙ Γ(2 − 푠) 푞 sin( + )휁(1 − s, ) Γ 푟=1 2 푞 푞

Γ′ 휋푠 2휋푟푎 푟 =(1-s)(log 2휋푞 − (2 − 푠)) ,2(2πq)푠−1Γ(1 − 푠) 푞 sin + 휁 1 − s, } Γ 푟=1 2 푞 푞

Γ′ 푎 Γ′ 푎 =(1-s) (log 2휋푞 − (2 − 푠))ζ(푠, ) = −(log 2휋푞 − (2 − 푠)) ((s-1)ζ(s, )) Γ 푞 Γ 푞

−(log 2휋푞 + 훾) ∙ 1 = −(푙표𝑔2휋푞 + 훾) as s 1 .

π 휋푠 2휋푟푎 푟 휋푠 2휋푟푎 푟 As s 1 , consider 2(2휋푞)푠−1 Γ(2 − 푠) 푞 { cos( + )휁(1 − s, ) - sin + 휁′ (1 − 푠, )} . 푟=1 2 2 푞 푞 2 푞 푞

π 휋 2휋푟푎 푟 휋 2휋푟푎 푟 This is equal to 2 푞 { cos( + )휁(o, ) - sin + 휁′ (표, )} 푟=1 2 2 푞 푞 2 푞 푞

r 휋 2휋푟푎 1 푟 2휋푟푎 Γ =2 푞 (− sin ∙ − − 푞−1 푐표푠 ∙ 푙표𝑔 q − 휁′ (표)) 푟=1 2 푞 2 푞 푟=1 푞 2휋

r 휋 2휋푟푎 1 푟 2휋푟푎 Γ = - 2{ 푞 sin ∙ ( - ) + 푞−1 푐표푠 푙표𝑔 q + 휁′ (표)} 푟=1 2 푞 2 푞 푟=1 푞 2휋

휋 2휋푟푎 푟 2휋푟푎 푟 = - 2{− 푞 (푠𝑖푛 ) ∙ + 푞−1 푐표푠 ∙ 푙표𝑔Γ( )} 푟=1 2 푞 푞 푟=1 푞 푞

휋 2휋푟푎 2휋푟푎 푟 = 푞 푟 sin - 2 푞−1 푐표푠 ∙ 푙표𝑔Γ( ) = S , say . 푞 푟=1 푞 푟=1 푞 푞

Case 1 : Let q ≥2 be odd . Then

휋 푞 2휋푟푎 2휋푟푎 푟 푟 푞−1 S= 푟=1 푟 sin – 2 푟≤ 푐표푠 푙표𝑔Γ + 푙표𝑔Γ(1 − ) 푞 푞 2 푞 푞 푞

휋 푞−1 2휋푟푎 2휋푟푎 푟 푟 푞−1 = 푟=1 푟 sin - 2 푟≤ 푐표푠 ∙ 푙표𝑔Γ( )Γ(1 − ) 푞 푞 2 푞 푞 푞

휋 푞−1 2휋푟푎 2휋푟푎 휋 = 푟 sin - 2 푞−1 푐표푠 ∙ 푙표𝑔 휋푟 푞 푟=1 푞 푟≤ 푞 푠𝑖푛 2 푞

: 13 :

휋 푞−1 2휋푟푎 2휋푟푎 휋푟 2휋푟푎 푞−1 푞−1 = 푟=1 푟 sin + 2 푟≤ 푐표푠 푙표𝑔 푠𝑖푛 − 2 log 휋 푟≤ 푐표푠 푞 푞 2 푞 푞 2 푞

휋 2휋푟푎 2휋푟푎 휋푟 2휋푟푎 = 푞−1 푟 sin + 푐표푠 log sin - log 휋 ∙ 푐표푠 푞 푟=1 푞 푟≤푞−1 푞 푞 푟≤푞−1 푞

휋 2휋푟푎 2휋푟푎 휋푟 = 푞−1 푟 sin + 푞−1 푐표푠 푙표𝑔 푠𝑖푛 + log 휋 푞 푟=1 푞 푟=1 푞 푞

푞 Case 2 : If q≥2 is even , we have to deal with the additional term corresponding to r= . However , the 2

푞 휋 2휋푟푎 term corresponding to r= does not contribute to S= 푞−1 푟 푠𝑖푛 + 푙표𝑔휋 + 2 푞 푟=1 푞

2휋푟푎 휋푟 + 푞−1 cos ∙ log sin and thus S remains unchanged . 푟=1 푞 푞

푎 휋 2휋푟푎 2휋푟푎 휋푟 Thus 휓 = −(훾 + log 2휋푞) + 푞−1 푟 sin + log 휋 + 푞−1 푐표푠 푙표𝑔 푠𝑖푛 푞 푞 푟=1 푞 푟=1 푞 푞

휋 2휋푟푎 2휋푟푎 휋푟 = - (훾 + 푙표𝑔2푞) + 푞−1 푟 sin + 푞−1 푐표푠 푙표𝑔 푠𝑖푛 푞 푟=1 푞 푟=1 푞 푞

: 14 :

References

[1] V.V. Rane , Dirichlet L-function and power series for Hurwitz zeta function , Proc. Indian acad.Sci.

(Math Sci.) , vol. 103 , No.1, April 1993, pp 27-39 .

*2+ V.V. Rane , Instant Evaluation and Demystification of ζ(n),L(n,χ) that Euler, Ramanujan Missed-I

(arXiv.org website)

*3+ V.V. Rane , Instant Evaluation and Demystification of ζ(n),L(n,χ) that Euler, Ramanujan Missed-II

(arXiv.org website)

[4] V.V. Rane , Instant multiple zeta values at non-positive integers (arXiv.org website)