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Hypergeometric Summation Representations of the Stieltjes

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Hypergeometric Summation Representations of the Stieltjes Hypergeometric summation representations of the Stieltjes constants Mark W. Coffey Department of Physics Colorado School of Mines Golden, CO 80401 (Received 2011) March 9, 2011 Abstract The Stieltjes constants γk appear in the regular part of the Laurent expan- sion of the Riemman and Hurwitz zeta functions. We demonstrate that these coefficients may be written as certain summations over mathematical constants and specialized hypergeometric functions pFp+1. This family of results general- izes a representation of the Euler constant in terms of a summation over values of the trigonometric integrals Si or Ci. The series representations are suitable for acceleration. As byproducts, we evaluate certain sine-logarithm integrals and present the leading asymptotic form of the particular pFp+1 functions. Key words and phrases arXiv:1106.5148v1 [math-ph] 25 Jun 2011 Riemann zeta function, Stieltjes constants, generalized hypergeometric function, Gamma function, digamma function, Euler constant, series representation, integral representation, Hurwitz zeta function, cosine integral, sine integral 2010 AMS codes 11M06, 11M35, 11Y60, 33C20 1 Introduction and statement of results Recently we developed series representations of the Euler constant γ and values of the Riemann zeta function at integer argument, together with other mathemati- cal constants, in terms of summations over the trigonometric integrals Si and Ci [8] (Propositions 2 and 3, Corollaries 7 and 9). The present work is in a sense a general- ization of those results. We present series representations of the Stieltjes (generalized Euler) constants that involve sums over certain generalized hypergeometric functions pFp+1. There are underlying connections to Si and Ci and certain logarithmic inte- grals of those functions, and our presentation includes some developments of special function theory. We let ζ(s)= ζ(s, 1) be the Riemann zeta function [11, 15, 21, 24], Γ the Gamma function, ψ = Γ′/Γ be the digamma function (e.g., [1]) with γ = −ψ(1) the Euler (k) constant, ψ be the polygamma functions [1], and pFq be the generalized hyperge- ometric function [2]. Although we concentrate on the Stieltjes constants γk = γk(1) corresponding to the Riemann zeta function, we briefly describe how the approach carries over to those for the Hurwitz zeta function. (See the discussion section.) The Stieltjes constants γk(a) [5, 6, 4, 7, 17, 18, 23, 25] arise in the regular part of the Laurent expansion of the Hurwitz zeta function ζ(s, a): n 1 ∞ (−1) ζ(s, a)= + γ (a)(s − 1)n, (1.1) s − 1 n! n nX=0 where γ0(a)= −ψ(a). 2 The Stieltjes constants may be expressed through the limit relation [3] (−1)n N lnn(k + a) lnn+1(N + a) γn(a)= lim − , n ≥ 0. (1.2) n! N " k + a n +1 # →∞ kX=0 Here, a∈ / {0, −1, −2,...}. The sequence {γk(a)}k 0 has rapid growth in magnitude ≥ with k for k large and changes in sign due to both k and a. Subsequences of the same sign of arbitrarily long length occur. For an asymptotic expression for these constants, even valid for moderate values of k, [16] (Section 2) may be consulted. The Stieltjes constants appear in applications including asymptotic analysis whether in computer science or high energy physics. Proposition 1. (a) 2 2 2 ∞ 1 π π n 5 2 2 γ1 =2 1+ (γ − 2)γ − − 3F4 1, 1, 1;2, 2, 2, ; −π n " 2 24 6 2 nX=1 1 1 + ln(2πn) γ − 1+ ln(2πn) + − γ, (1.3) 2 2 and (b) the slightly accelerated form 2 2 2 ∞ 1 π 5 1 π n 5 2 2 γ1 =2 1+ (γ − 2)γ − − − 3F4 1, 1, 1;2, 2, 2, ; −π n " 2 24 32π4 n4 6 2 nX=1 1 73 + ln(2πn) γ − 1+ ln(2πn) + − γ, (1.4) 2 144 and (c) 3 2 2 ∞ 2 γ π π 2 γ2 =1 − 2(γ + γ1)+2 2(1 − γ)+ γ + − + γ − ζ(3) ( 3 12 12 3 nX=1 2 2 π n 5 2 2 + 4F5 1, 1, 1, 1;2, 2, 2, 2, ; −π n 6 2 3 γ2 π2 1 1 −2 ln(2πn) 1 − γ + − + ln(2πn)(γ − 1)+ ln2(2πn) . (1.5) " 2 24 2 6 #) 3 Remark. The summand in (1.5) is O(n− ) as n →∞. All parts of the Proposition may be further accelerated in their rate of convergence. Proposition 2. The Stieltjes constant γj may be expressed as a summation over n of mathematical constants, terms lnk(2πn), with k =1,...,j, and the term ∞ 5 π2 n2 F 1, 1,..., 1;2, 2,..., 2, ; −π2n2 . (1.6) j+2 j+3 2 nX=1 The following section of the paper contains the proof of the Propositions. Section 3 contains various supporting and reference Lemmas. Some of these Lemmas present results of special function theory and may indeed be of occasional interest in them- selves. Certain logarithmic integrals of the Si and Ci functions are considered in the Appendix. Proof of Propositions We let P1(x)= B1(x−[x]) = x−[x]−1/2 be the first periodic Bernoulli polynomial, with {x} = x − [x] the fractional part of x. Being periodic, P1 has the Fourier series [1] (p. 805) ∞ sin 2πjx P (x)= − . (2.1) 1 πj jX=1 Proposition 1. (a) We take a = 1 in the representation s 1 s a− a − ∞ P1(x) ζ(s, a)= + − s s+1 dx, σ ≡ Re s> 0, (2.2) 2 s − 1 Z0 (x + a) so that 1 1 ∞ P1(x) ζ(s)= + − s s+1 dx, σ> −1, (2.3) 2 s − 1 Z1 x 4 and 1 ∞ P1(x) ∞ P1(x) ′ ζ (s)+ 2 = − s+1 dx + s s+1 ln x dx. (2.4) (s − 1) Z1 x Z1 x Taking s → 1 in (2.4) and using (1.1) gives ∞ P1(x) 1 γ1 = − 2 ln x dx + − γ. (2.5) Z1 x 2 Here, we have used the well known integral that also results from (2.3) P (x) ∞ j+1 (x − j − 1/2) ∞ 1 dx = dx 1 x2 j x2 Z jX=1 Z ∞ j +1 1 1 1 1 = ln − + = − γ. (2.6) " j ! 2 j +1 j !# 2 jX=1 From Lemmas 2, 3, and 6 we have 2 2 2 1 ∞ sin κx 1 π κ 5 κ 2 ln x dx =1+ (γ − 2)γ − − 3F4 1, 1, 1;2, 2, 2, ; − κ Z1 x 2 24 24 2 4 ! ln κ + (2γ − 2 + ln κ). (2.7) 2 Taking κ =2πn and using (2.1) and (2.5) so that 1 ∞ 1 ∞ sin(2πnx) 1 γ1 = ln x dx + − γ, (2.8) π n 1 x2 2 nX=1 Z gives the first part of the Proposition from (2.7). 4 (b) The summand of (1.3) is O(n− ) as n → ∞. By adding and subtracting 5 1 5 ζ(4) 1 6 32 n∞=1 n4 = 32 π4 = 288 we transform the summand to be O(n− ) as n →∞. P(c) From (2.4) we obtain 2 ∞ P1(x) ∞ P1(x) 2 ′′ ζ (s) − 3 =2 s+1 ln x dx − s s+1 ln x dx. (2.9) (s − 1) Z1 x Z1 x 5 Then from (1.1) ∞ P1(x) γ2 = 2 (2 − ln x) ln x dx, (2.10) Z1 x where by (2.5) the term ∞ P1(x) 2 2 ln x dx =1 − 2(γ + γ1). (2.11) Z1 x For the term P (x) 1 ∞ 1 sin(2πnx) ∞ 1 ln2 x dx = − ∞ ln2 x dx, (2.12) 1 x2 π n 1 x2 Z nX=1 Z we apply Lemmas 3, 4, and 6. We omit further details. Proposition 2. As may be proved from (2.2) by induction, for integers j ≥ 1 we have j k j (j) j 1 s j ln a (−1) s j ζ (s, a)=(−1) a − (j − k)! j k+1 + a− ln a k! (s − 1) − 2 kX=0 j ∞ P1(x) j 1 − +(−1) s+1 ln (x + a)[j − s ln(x + a)] dx. (2.13) Z0 (x + a) At a = 1 we have j (j) (−1) j! j ∞ P1(x) j 1 − ζ (s)= j+1 +(−1) s+1 ln x (j − s ln x) dx. (2.14) (s − 1) Z1 x Then by (1.1) we have ∞ P1(x) j 1 − γj = 2 ln x(j − ln x)dx Z1 x 1 ∞ 1 ∞ sin(2πnx) j 1 = − ln − x(j − ln x)dx. (2.15) π n 1 x2 nX=1 Z 6 Herein on the right side, j multiplies a contribution from γj 1. We then appeal to − Lemmas 2, 3, and 4 for logarithmic-sine integrals, and the result follows. Lemmas Herein the cosine integral is defined by cos t z cos t − 1 Ci(z) ≡ − ∞ dt = γ + ln z + dt. (3.1) Zz t Z0 t As usual, (w)n = Γ(w + n)/Γ(w) denotes the Pochhammer symbol. Lemma 1. (Hypergeometric form of the cosine integral) z2 3 z2 Ci(z)= γ + ln z − 2F3 1, 1;2, 2, ; − . (3.2) 4 2 4 ! Proof. This easily follows from the expression ℓ 2ℓ ∞ (−1) z Ci(z)= γ + ln z + . (3.3) 2ℓ(2ℓ)! Xℓ=1 Lemma 2. (a) For a =6 0, y Ci(az) 1 dz = γ ln(y/x)+ [ln2(ay) − ln2(ax)] Zx z 2 2 2 2 2 2 a 2 3 a y 2 3 a x − y 3F4 1, 1, 1;2, 2, 2, ; − − x 3F4 1, 1, 1;2, 2, 2, ; − , (3.4) 8 " 2 4 ! 2 4 !# and (b) for b>a> 0 b sin κx 2 ln x dx = κ {Ci(κb)(1 + ln b) − Ci(κa)(1 + ln a) − γ ln(b/a) Za x 2 2 2 2 2 1 2 2 κ 2 3 κ b 2 3 κ a + [ln (κa) − ln (κb)] + b 3F4 1, 1, 1;2, 2, 2, ; − − a 3F4 1, 1, 1;2, 2, 2, ; − 2 8 " 2 4 ! 2 4 !#) sin κa sin κb + (1 + ln a) − (1 + ln b).
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