MATHEMATICS OF COMPUTATION Volume 83, Number 287, May 2014, Pages 1383–1395 S 0025-5718(2013)02755-X Article electronically published on July 29, 2013
SERIES REPRESENTATION OF THE RIEMANN ZETA FUNCTION AND OTHER RESULTS: COMPLEMENTS TO A PAPER OF CRANDALL
MARK W. COFFEY
Abstract. We supplement a very recent paper of R. Crandall concerned with the multiprecision computation of several important special functions and numbers. We show an alternative series representation for the Riemann and Hurwitz zeta functions providing analytic continuation throughout the whole complex plane. Additionally, we demonstrate some series representations for the initial Stieltjes constants appearing in the Laurent expansion of the Hur- witz zeta function. A particular point of elaboration in these developments is the hypergeometric form and its equivalents for certain derivatives of the incomplete Gamma function. Finally, we evaluate certain integrals including ζ(s) η(s) Res=c s ds and Res=c s ds,withζ the Riemann zeta function and η its alternating form.
1. Introduction and statement of results Very recently R. Crandall has described series-based algorithms for computing multiprecision values of several important functions appearing in analytic number theory and the theory of special functions [9]. These functions include the Lerch ∞ n s transcendent Φ(z,s,a)= n=0 z /(n + a) , and many of its special cases, and Epstein zeta functions. In the following we let Bn(x)andEn(x) be the Bernoulli and Euler polynomials, respectively, Γ the Gamma function, and Γ(x, y) the incomplete Gamma function (e.g., [1, 2, 13]). An example series representation from [9, (27)] is that for the Hurwitz zeta function ζ(s, a), ∞ ∞ 1 Γ[s, λ(n + a)] 1 B (a) λm+s−1 (1.1) ζ(s, a)= + (−1)m m , Γ(s) (n + a)s Γ(s) m! (m + s − 1) n=0 m=0 with free parameter λ ∈ [0, 2π). We complement the presentation of [9] with further representation of the Rie- mann zeta function ζ(s)=ζ(s, 1) and some results on the Stieltjes constants. The latter constants γn(a) (e.g., [4–7]) appear in the Laurent expansion ∞ 1 (−1)n (1.2) ζ(s, a)= + γ (a)(s − 1)n, s − 1 n! n n=0
Received by the editor April 2, 2012 and, in revised form, July 18, 2012 and August 27, 2012. 2010 Mathematics Subject Classification. Primary 11M06, 11M35, 11Y35, 11Y60. Key words and phrases. Riemann zeta function, Hurwitz zeta function, Stieltjes constants, Euler polynomials, Bernoulli polynomials, incomplete Gamma function, confluent hypergeometric function, generalized hypergeometric function.
c 2013 retained by the author 1383
License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1384 MARK W. COFFEY
where γ0(a)=−ψ(a), the Euler constant γ0 = γ = −ψ(1), and by convention one takes γk = γk(1). Here ψ =Γ/Γ denotes the digamma function (e.g., [1, 2, 13]). The Stieltjes constants may be expressed via the limit formula N lnk(n + a) lnk+1(N + a) γk(a) = lim − . N→∞ n + a k +1 n=0 We let η(s)=(1−21−s)ζ(s) denote the alternating Riemann zeta function, given − ∞ (−1)n 1 for Re s>0byη(s)= n=1 ns . The following series representation provides an analytic continuation of the Riemann zeta function to the whole complex plane. Proposition 1. Let |λ| <π.Then ∞ ∞ Γ(s, mλ) 1 E (0) λn+s (1.3) Γ(s)η(s)= (−1)m−1 + n . ms 2 n! (n + s) m=1 n=0 This representation holds in all of C. In particular, it delivers the special values j η(−j)=(−1) Ej(0)/2 for j ≥ 0. An alternative expression for the values En(0) −1 n+1 is [1, p. 805] En(0) = −2(n +1) (2 − 1)Bn+1 for n ≥ 1 in terms of Bernoulli numbers Bn = Bn(0).
Remark. The values En(0) = −En(1) for n ≥ 2evenarezerowhileE0(0) = 1. The signs of two adjacent values of odd indices are opposite. Proposition 1 has many implications. For example, for λ =1wehavethe following, with Lis denoting the polylogarithm function. Corollary 1. ∞ 1 E (0) (1.4a) ln 2 = ln(1 + e) − 1+ n 2 (n +1)! n=0 and ∞ 1 1 1 E (0) (1.4b) ζ(2) = ln(1 + e) − 1 − Li − + n . 2 2 e 2 n!(n +2) n=0 We apply the representation (1.1) for a later proposition. First, we verify some well-known properties of the Hurwitz zeta function from (1.1). Their proofs are given so as to elucidate how they follow from that series representation. Corollary 2. (a) (1.5) ζ(s, a)=ζ(s, a +1)+a−s, (b)
(1.6) ∂aζ(s, a)=−sζ(s +1,a), (c) for integers q ≥ 2, q−1 r (1.7) ζ s, =(qs − 1)ζ(s), q r=1 (d) for Re s<1, 1 (1.8) ζ(s, a)da =0. 0
License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use SERIES REPRESENTATION OF THE RIEMANN ZETA FUNCTION 1385
j − j By induction, (1.6) gives ∂aζ(s, a)=( 1) (s)jζ(s + j, a),where(z)n =Γ(z + n)/Γ(z) is the Pochhammer symbol.
We let pFq denote the generalized hypergeometric function (e.g., [1, 2, 13]). Derivatives of the incomplete Gamma function with respect to the first argument may be expressed in hypergeometric form. An example is given in the following. Proposition 2. ∂ ∞ Γ(1)(s, x) ≡ Γ(s, x)= ts−1 ln te−tdt ∂s (1.9) x xs = F (s, s; s +1,s+1;−x)+Γ(s)[− ln x + ψ(s)] + ln xΓ(s, x). s2 2 2 Later we provide a separate discussion for such derivatives. Proposition 3. Let Re a>0.Then (a) ∞ 1 (−1)m B (a) (1.10) −γ − ψ(a)=γ (a) − γ = e−aΦ , 1,a + m , 0 e m! m m=1 (b) ∞ ∞ γ2 ζ(2) Γ(0,n+ a) (−1)m B (a) (1.11) + γψ(a)+ − γ (a)= − m , 2 2 1 n + a m! m2 n=0 m=1
(c) γ3 ζ(2) γ2 ζ(2) ζ(3) γ (a) − − γ − + ψ(a)+γγ (a) − + 2 6 2 2 2 1 3 2 ∞ 1 γ2 ζ(2) 1 = − F (1, 1, 1; 2, 2, 2; −n −a)+ γ ln(n+a)+ + + ln2(n+a) 3 3 (a+n) 2 2 2 n=0
∞ (−1)m B (a) (1.12) + m . m! m3 m=1 Corollary 3. For λ ∈ [0, 2π), ∞ ln Γ(x)=γ(1 − x) − ln λ(x − 1) + {Γ[0,λ(n + x)] − Γ[0,λ(n +1)]} n=0 ∞ (−1)m + [B (x) − B ]λm−1. (m − 1)m! m m m=2 Alternative forms of the Bernoulli polynomial sums of Proposition 3 and of another sum in (1.11) are given in the following. Lemma 1. Let Re a>0.Then (a) ∞ (−1)m B (a) 1 (1.13) m = − e−a F (1,a; a +1;e−1)+a(γ + ψ(a)) . m! m a 2 1 m=1
License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1386 MARK W. COFFEY
In particular, ∞ ∞ B 1 e−n (1.14) − m = =1− ln(e − 1). m! m n m=1 n=1 (b) ∞ B (a) tm 1 (1.15) m = − eat F (1,a; a +1;et)+a(γ +ln(−t)+ψ(a)) . m! m a 2 1 m=1 (c) (1.16) ∞ B (a) zm 1 z dt m = − eat F (1,a; a +1;et)+a(γ +ln(−t)+ψ(a)) . m! m2 a 2 1 t m=1 0 (d) ∞ Γ(0,n+ a) 1 1/e ua−1 (1.17) = − F (1,a;1+a; u)du. n + a a ln u 2 1 n=0 0
2. Proof of Propositions Proposition 1. We apply the integration technique Crandall refers to as “Bernoulli splitting” and use the integral representation for Re s>0 (e.g., [1, p. 807]: ∞ ts−1 (2.1) Γ(s)η(s)= t dt. 0 e +1 Splitting the integral at λ, we use the generating function ∞ 2exz zn (2.2) = E (x) , |z| <π, ez +1 n n! n=0 for the integration on [0,λ). For the other integration we use a standard integral representation for the incomplete Gamma function, together with a geometric series expansion: ∞ ∞ ts−1 ∞ dt = (−1)m e−(m+1)tts−1dt et +1 λ m=0 λ (2.3) ∞ (−1)m = Γ[s, λ(m +1)]. (m +1)s m=0 The sum in (1.3) converges uniformly away from poles so it provides an analytic continuation to Re s ≤ 0. For Corollary 1 we recall that for n ≥ 0 [13, p. 941],
n xm (2.4) Γ(n +1,x)=n!e−x , m! m=0 ∞ n 2 and Li2(z)= n=1 z /n .
License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use SERIES REPRESENTATION OF THE RIEMANN ZETA FUNCTION 1387
Remarks. We have supplied details for Crandall’s Algorithm 1 for the η function. For the values En(0), their asymptotic form is a case of the following [11, 24.11.6], n+1 (n+1)/2 π sin(πx) ,neven, (−1) En(x) → 4(n!) cos(πx) ,nodd,
and one has the bounds for 0 − 2n − 4(2n 1)! 2 1 n > (−1) E − (x) > 0. π2n 22n − 2 2n 1 A similar treatment can be made of ∞ s−1 −(a−1)t t e −s a − a +1 t dt =2 Γ(s) ζ s, ζ s, 0 e +1 2 2 (2.5) ∞ (−1)n =Γ(s) , (n + a)s n=0 where Re a>0andRes>0. This yields, for |λ| <π, ∞ a a +1 Γ[s, λ(m + a)] 2−sΓ(s) ζ s, − ζ s, = (−1)m 2 2 (m + a)s m=0 (2.6) ∞ 1 E (1 − a) λn+s + n , 2 n! (n + s) n=0 again providing analytic continuation to all of C. Herethepolarpartisabsent, and this representation could be used to develop expressions for the differences of Stieltjes constants γk(a/2) − γk[(a +1)/2]. As an illustration of Proposition 1, Figure 1 plots the natural logarithm of the absolute value of the relative error in the value of η(−3/2), ln[|ηnum (−3/2) − η(−3/2)|/η(−3/2)], versus the truncation size N of the sums in (1.3), wherein ηnum(−3/2) is calculated from that equation, and the reference value of η(−3/2) has been computed from Mathematicac V8. The values of λ in this figure are 1, 1.75, and 2.5, with increasing accuracy obtained for decreasing λ.ForN =40 and λ = 1, the value of η(−3/2) from (1.3) is correct to 19 decimal digits. As Γ(s, y) ∼ e−yys−1 as y →∞, the rate of convergence is governed by the second sum in (1.3) with the Euler numbers En(0). m−1 Corollary 2. For part (a), use the property Bm(a +1)=Bm(a)+ma and the sum (by [13, p. 941]), ∞ am λm+s (2.7) (−1)m+1 = a−s[Γ(s, aλ) − Γ(s)]. m! (m + s) m=0 For part (b), use the derivative dΓ(α, x)/dx = −xα−1e−x and the functional re- lations αΓ(α, x)=Γ(α +1,x) − xαe−x and Γ(s)=Γ(s +1)/s. For part (c), use q−1 r 1−m − − m r=1 Bm q =(q 1)Bm and ( 1) Bm(1) = Bm. The following decomposi- tion, wherein the generating function (2.24) is employed, can then be used to verify License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1388 MARK W. COFFEY relative error N 4 8 12 16 20 24 28 32 36 40 −10 −20 −30 −40 Figure 1. Plot of the natural logarithm of the absolute value of therelativeerrorinη(−3/2) computed from (1.3). the stated property: qs ∞ ys−1 qsζ(s)= dy y − Γ(s) 0 e 1 qs λ/q ys−1 ∞ ys−1 = dy + dy y − y − Γ(s) 0 e 1 λ/q e 1 (2.8) ∞ 1 B λm+s−1 = q1−m m Γ(s) m! (m + s − 1) m=0 ∞ qs ∞ e−(n+1)qy(eqy − 1)ys−1 + dy. Γ(s) ey − 1 n=0 λ/q 1 ≥ For part (d), we first note that 0 Bm(a)da =0forallm 1, giving from (1.1), ∞ 1 1 Γ(s, λ(n + a)] λs−1 (2.9) Γ(s) ζ(s, a)da = da + . (n + a)s s − 1 0 n=0 0 Integrating by parts, 1 Γ(s, λ(n + a)] 1 1 d 1 da = − Γ(s, λ(n + a)] da s − s−1 0 (n + a) (s 1) 0 da (n + a) (2.10) 1 − s−1 −λn s−1 −λ(n+1) 1−s = − λ e + λ e + n Γ(s, λn) s 1 − (n +1)1−sΓ(s, λ(n +1)) . Then the sum in (2.9) telescopes to −λs−1/(s − 1) and (1.8) follows. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use SERIES REPRESENTATION OF THE RIEMANN ZETA FUNCTION 1389 Remark. More generally than part (c), we have the reciprocity relation ([7, Lemma 1]), q p−1 pr q s q + p qb (2.11) ζ s, − b = ζ s, − . q p p p r=1 =0 Here p ≥ 1andq ≥ 1 are integers, b ≥ 0, and min(p/q, q/p) >b. From (2.11) follows ([7, Corollary 3]), q − p−1 pr q 1 m q (2.12) B − b = B 1+( − b) . m q p m p r=1 =0 Proposition 2. First integrating by parts we have ∂ ∞ Γ(1)(s, x) ≡ Γ(s, x)= ts−1 ln te−tdt ∂s x ∞ ∂ ∞ Γ(s, t) (2.13) = − ln t Γ(s, t)dt = dt +lnx Γ(s, x). x ∂t x t We next recall a hypergeometric form of Γ(x, y), xα Γ(α, x)=Γ(α) − F (α, 1+α; −x) α 1 1 ∞ xα α (−x)j (2.14) = Γ(α) − , α (α + j) j! j=0 with 1F1 the confluent hypergeometric function. This expression may be integrated on a finite interval. For an improper integral extending to infinity we need some asymptotic information contained in the following. Lemma 2 (Asymptotic form of special pFp functions). As x →∞ (a) s2 (2.15) F (s, s; s +1,s+1;−x) ∼ e−x + x−ss2Γ(s)[ln x − ψ(s)], 2 2 x2 (b) 3 − ∼− −x s 3F3(s, s, s; s +1,s+1,s+1; x) e 3 (2.16) x s3 + x−s Γ(s)[ln2 x − 2ln xψ(s)+ψ2(s)+ψ(s)]. 2 Here ψ is the trigamma function. (a) We use a general procedure based upon the Barnes integral representation of pFq ([15, Section 2.3]). We have i∞ 1 y (2.17) 2F2 (s, s; s +1,s+1;−x)= Γ(−y)Γ(y +1)g(y)x dy, 2πi −i∞ where the path of integration is a Barnes contour, indented to the left of the origin but staying to the right of −s,and Γ2(y + s) Γ2(s +1) s2 (2.18) Γ(y +1)g(y)= = . Γ2(y + s +1) Γ2(s) (y + s)2 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1390 MARK W. COFFEY The contour can be thought of as closed in the right half-plane, over a semicircle of infinite radius. We then move the contour to the left of y = −s, picking up the residue there. We find − y 2 2 Γ( y)x s (2.19) s Res − = Γ(s)[ln x − ψ(s)]. y= s (y + s)2 xs This gives the algebraic part of the asymptotic form of the 2F2 function, that is, the leading portion. The higher order terms come from the exponential expansion of these functions, and they are infinite in number. The latter expansion may be determined according to ([15, Section 2.3, p. 57]). (b) proceeds similarly, where now − y 3 3 Γ( y)x s 2 2 (2.20) s Res − = Γ(s)[ln x − 2lnxψ(s)+ψ (s)+ψ (s)]. y= s (y + s)3 2xs We may integrate in (2.13) from x to b using (2.14), b s Γ(s, t) x − dt = 2 2F2(s, s; s +1,s+1; x) (2.21) x t s bs − F (s, s; s +1,s+1;−b)+Γ(s)(ln b − ln x). s2 2 2 Then with the lemma in hand, taking b →∞completes the proposition. Remark. The result (1.9) can be obtained directly from (2.14). For, since Γ(α)= Γ(α)ψ(α), we have ∞ ∂ 1 (−x)j Γ(α, x)=Γ(α)ψ(α)+lnx[Γ(α, x) − Γ(α)] + xα ∂α (α + j)2 j! j=0 ∞ ∂ xα (α)2 (−x)j = Γ(α, x)=Γ(α)ψ(α)+lnx[Γ(α, x) − Γ(α)] + j ∂α α2 (α +1)2 j! j=0 j ∂ xα = Γ(α, x)=Γ(α)ψ(α)+lnx[Γ(α, x) − Γ(α)] + F (α, α; α +1,α+1;−x). ∂α α2 2 2 Proposition 3. We take λ = 1 in (1.1) and, making use of Proposition 2 and related considerations, expand the function ∞ ∞ 1 Γ(s, n + a) (−1)m B (a) (2.22) Γ(s)ζ(s, a) − = + m s − 1 (n + a)s m! (m + s − 1) n=0 m=1 about s = 1, and apply the defining series (1.2). − x Corollary 3. Maintaining the free parameter λ, this follows from 1 [γ+ψ(a)]da = γ(1 − x) − ln Γ(x), the corresponding expression similar to (1.10), ∞ 1 (−1)m B (a) (2.23) − ln λ − γ − ψ(a)=e−λaΦ , 1,a + m λm, eλ m! m m=1 and the property Bm(a)da = Bm+1(a)/(m +1). To illustrate Corollary 3, Figure 2 shows a plot of the natural logarithm of the absolute value of the relative error in the value of ln Γ(7/4) versus N, the truncation size of the sums therein. Now λ = π and π/2, and the reference value of ln Γ(7/4) is computed from Mathematicac V8. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use SERIES REPRESENTATION OF THE RIEMANN ZETA FUNCTION 1391 relative error N 5 10 15 20 25 30 35 40 45 50 55 60 −20 −40 −60 −80 Figure 2. Plot of the natural logarithm of the absolute value of therelativeerrorinlnΓ(7/4) computed from Corollary 3. Lemma 1. For (a)–(c), repeatedly integrate the generating function ∞ B (a) zeaz (2.24) n zn = − 1, |z| < 2π. n! ez − 1 n=1 For (d), write ∞ ∞ ∞ Γ(0,n+ a) 1 ∞ dt 1 ∞ dv (2.25) = e−t = e−(n+a)v . n + a n + a t n + a v n=0 n=0 n+a n=0 1 Then interchange summation and integration and perform another change of vari- able. Remark. As Γ(0,x)=−Ei(−x), where Ei is the exponential integral having many integral representations, there are a multitude of ways of obtaining (1.17). 3. Discussion: Derivatives of the incomplete Gamma function Derivatives of the incomplete Gamma function are important for the computa- tional methods of [9], and they provide a starting point for many useful integrals. Geddes et al. [12] investigated these derivatives, introducing a function T (m, a, z), with Γ(a, z)=zT(2,a,z), m ≥ 1 an integer and initially |z| < 1, such that (3.1a) Γ(1)(a, x)=xT (3,a,x)+lnx Γ(a, x), (3.1b) Γ(2)(a, x)=ln2 x Γ(a, x)+2x[ln xT(3,a,x)+T (4,a,x)], License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1392 MARK W. COFFEY i − and more generally, with Pj = i!/(i j)!, dmΓ(a, x) ≡ Γ(m)(a, x)=lnm x Γ(a, x) dam (3.2) m−1 m−1 m−i−1 + mx Pi ln xT(3 + i, a, x). i=0 The function T satisfies the derivative relations dT (m, a, z) (3.3a) =lnzT(m, a, z)+(m − 1)T (m +1,a,z) da and dT (m, a, z) 1 (3.3b) = − [T (m − 1,a,z)+T (m, a, z)]. dz z One form of T is [12, (37)], m−1 1 s T (m, a, z)=−Res − − Γ(a − 1 − s)z s= 1 s +1 (3.4) ∞ (−1)iza+i−1 + , i!(−a − i)m−1 i=0 where a is neither zero or a negative integer. This can be obtained by expressing T in terms of the Meijer G-function and then using a contour integral for the latter function. Using [12, (38)], and the relation (a)i/(a +1)i = a/(a + i), as we have previously done, we may write − (−1)m−1 d m 2 T (m, a, z)= Γ(a − t)zt−1 (m − 2)! dt t=0 a−1 m−1 z (3.5) +(−1) − F − (a,a,...,a; a +1,a+1,...,a+1;−z). am−1 m 1 m 1 Proposition 2 and like considerations are in accord with this result. We note that Milgram [14] studied derivatives with respect to the order s of the s−1 exponential integral function Es(z)=z Γ(1 − s, z)byusingtheG-function and contour integral representation. 4. Certain zeta functions and other integrals Elsewhere we have considered second moment integrals of the Riemann zeta and other functions [8]. The following concerns the integrals ∞ ζ(c + it) ζ(s) (4.1) I(c) ≡ dt = −i ds, −∞ c + it Res=c s for integer p ≥ 2, ∞ ≡ ζ(c + it) − ζ(s) (4.2) Ip(c) p dt = i p ds, −∞ (c + it) Res=c s ∞ η(c + it) η(s) (4.3) Ia(c) ≡ dt = −i ds, −∞ c + it Res=c s License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use SERIES REPRESENTATION OF THE RIEMANN ZETA FUNCTION 1393 and ∞ Lic+it(x) Lis(x) (4.4) IL(c) ≡ dt = −i ds. −∞ c + it Res=c s Proposition 4. We have I(c)=−π for 0 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1394 MARK W. COFFEY |x| < 1, or alternatively that Li1(x)=− ln(1 − x) and the derivative property dLis(x)/dx =Lis−1(x)/x. So, for instance, for (4.1) with c>1, the sum on n in (4.5) may be performed, ∞ 1 ds 1 (4.7) ζ(s) = θ(1 − n)= . 2πi Res=c s 2 n=1 This gives Res=c ζ(s)(ds/s)=iπ.Forc<1 the contribution from the pole at s =1 is subtracted, πi − 2πi = −iπ. Similarly, for (4.4), when c<0, the stated formula for I (c)isobtainedfrom L 2πx x +1 πx − = πx . 1 − x x − 1 For (4.2), we repeatedly divide (4.5) by x and integrate with respect to x.This gives s − 1 x ds θ(x n) m x (4.8) s m+1 = ln . 2πi Res=σ n s m! n This relation follows by induction, and the base case is given by integration by parts. In detail for that case, s 1 x ds d − − − s 2 = θ(x n)lnx δ(x n)lnx dx 2πi Res=σ n s dx = θ(x − n)ln(x/n), wherein δ(x)=dθ(x)/dx is the Dirac delta function. Summing on n and then putting x = 1 then yields, for c>1, 1 ds ζ(s) m+1 =0. 2πi Res=c s m+1 Since Res[ζ(s)/s ]s=1 = 1, the stated result for 0 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use SERIES REPRESENTATION OF THE RIEMANN ZETA FUNCTION 1395 This line of development may be continued. For instance, an integration of (4.10) with respect to x now gives s+2 1 x ds 1 − 2 − (4.11) s = (x n) θ(x n). 2πi Res=σ n s(s +1)(s +2) 2 More generally, we have 1 xs+m ds 1 (4.12) = (x − n)mθ(x − n), s ··· 2πi Res=σ n s(s +1) (s + m) m! provable by induction on nonnegative integer m. This result may then be applied in a number of ways. Acknowledgment The author thanks R. Crandall for reading the manuscript, and in particular, for assisting with Section 4. References [1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, Washington, Na- tional Bureau of Standards (1964). [2] George E. Andrews, Richard Askey, and Ranjan Roy, Special functions, Encyclopedia of Mathematics and its Applications, vol. 71, Cambridge University Press, Cambridge, 1999. MR1688958 (2000g:33001) [3] Tom M. Apostol, Introduction to analytic number theory, Springer-Verlag, New York, 1976. Undergraduate Texts in Mathematics. MR0434929 (55 #7892) [4] M. W. Coffey, Series representations for the Stieltjes constants, arXiv:0905.1111 (2009), to appear in Rocky Mtn. J. Math. [5]MarkW.Coffey,Addison-type series representation for the Stieltjes constants,J.Num- ber Theory 130 (2010), no. 9, 2049–2064, DOI 10.1016/j.jnt.2010.01.003. MR2653214 (2011f:11111) [6] M. W. Coffey, Series representations of the Riemann and Hurwitz zeta functions and series and integral representations of the first Stieltjes constant, arXiv:1106.5147 (2011). [7] M. W. Coffey, Parameterized summation relations for the Stieltjes constants, arXiv:1002.4684 (2010). [8] M. W. Coffey, Evaluation of some second moment and other integrals for the Riemann, Hurwitz, and Lerch zeta functions, arXiv:1101.5722 (2011). [9] R. E. Crandall, Unified algorithms for polylogarithm, L-series, and zeta variants, preprint (2012). http://www.perfscipress.com/papers/universalTOC25.pdf [10] Richard Crandall and Carl Pomerance, Prime numbers, A computational perspective, Springer-Verlag, New York, 2001. MR1821158 (2002a:11007) [11] http://dlmf.nist.gov/ [12] K. O. Geddes, M. L. Glasser, R. A. Moore, and T. C. Scott, Evaluation of classes of definite integrals involving elementary functions via differentiation of special functions, Appl. Algebra Engrg. Comm. Comput. 1 (1990), no. 2, 149–165, DOI 10.1007/BF01810298. MR1325519 (97e:33005) [13] I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Academic Press, New York (1980). [14] M. S. Milgram, The generalized integro-exponential function,Math.Comp.44 (1985), no. 170, 443–458, DOI 10.2307/2007964. MR777276 (86c:33024) [15] R. B. Paris and D. Kaminski, Asymptotics and Mellin-Barnes integrals, Encyclopedia of Mathematics and its Applications, vol. 85, Cambridge University Press, Cambridge, 2001. MR1854469 (2002h:33001) [16] E. T. Whittaker and G. N. Watson, A course of modern analysis, An introduction to the general theory of infinite processes and of analytic functions; with an account of the princi- pal transcendental functions; Reprint of the fourth (1927) edition, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 1996. MR1424469 (97k:01072) Department of Physics, Colorado School of Mines, Golden, Colorado 80401 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use