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 Abstract : Using three basic facts concerning Hurwitz zeta function , we give new natural proofs of the known results on Bernoulli polynomials , gamma function 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 character , 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 complex number 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 analytic continuation . Let ζ(s,1)=ζ(s) be the Riemann zeta function . 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 ∝푧 푛 푧푒 ∞ 푧 = 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 meromorphic function 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<q are integers . 푞 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 2휋푛훼 s-1 =Г(1-s) |푛|≥1 푒 (2πin) Fact I) resulted in the Taylor series 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 푤푡ℎ 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 = 푠푛휋훼 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<q are integers , 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 −1 Next with Weierstrass product definition of Г(α) 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 푙표 = lim푢 ∞ 푢 + 훼 푙표 − 1 + 훼 푙표 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 x>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 , 1−푠 −푠 푥 1 −푠 ∞ 휙 푢−훼 휁 푠, 훼 = (푛 + 훼) + + (x-α-[x-α+- )푥 -s du .