PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 139, Number 2, February 2011, Pages 535–545 S 0002-9939(2010)10589-0 Article electronically published on August 18, 2010 INTEGRALS OF POWERS OF LOGGAMMA TEWODROS AMDEBERHAN, MARK W. COFFEY, OLIVIER ESPINOSA, CHRISTOPH KOUTSCHAN, DANTE V. MANNA, AND VICTOR H. MOLL (Communicated by Ken Ono) Abstract. Properties of the integral of powers of log Γ(x) from 0 to 1 are con- sidered. Analytic evaluations for the first two powers are presented. Empirical evidence for the cubic case is discussed. 1. Introduction The evaluation of definite integrals is a subject full of interconnections of many parts of mathematics. Since the beginning of Integral Calculus, scientists have developed a large variety of techniques to produce magnificent formulae. A partic- ularly beautiful formula due to J. L. Raabe [12] is 1 Γ(x + t) (1.1) log √ dx = t log t − t, for t ≥ 0, 0 2π which includes the special case 1 √ (1.2) L1 := log Γ(x) dx =log 2π. 0 Here Γ(x)isthegamma function defined by the integral representation ∞ (1.3) Γ(x)= ux−1e−udu, 0 for Re x>0. Raabe’s formula can be obtained from the Hurwitz zeta function ∞ 1 (1.4) ζ(s, q)= (n + q)s n=0 via the integral formula 1 t1−s (1.5) ζ(s, q + t) dq = − 0 s 1 coupled with Lerch’s formula ∂ Γ(q) (1.6) ζ(s, q) =log √ . ∂s s=0 2π An interesting extension of these formulas to the p-adic gamma function has recently appeared in [3]. Received by the editors February 23, 2010 and, in revised form, March 8, 2010. 2010 Mathematics Subject Classification. Primary 33B15. Key words and phrases. Integrals, transformations, loggamma, Hurwitz zeta function. c 2010 American Mathematical Society 535 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 536 T. AMDEBERHAN ET AL. Two of the current authors have investigated definite integrals involving the Hurwitz zeta function [6, 7]. As an unexpected corollary, the formula for the integral of log2 Γ(x), 1 2 2 2 γ π 1 4 2 − ζ (2) ζ (2) (1.7) L2 := log Γ(x) dx = + + γL1 + L1 (γ +2L1) 2 + 2 , 0 12 48 3 3 π 2π was produced. Here γ is Euler’s constant defined by n 1 (1.8) γ = lim − log n. n→∞ k k=1 The natural question addressed here is that of an analytic expression for the family of integrals 1 n (1.9) Ln := log Γ(x) dx, for n ∈ N, 0 that extends the values of L1 and L2 given above. Section 2 presents a direct approach to the evaluation of L1, very close in spirit to the original proof given by Raabe. The proof employs only elementary properties of the gamma function. Section 3 contains a new proof of the value of L2 based on the Fourier series expansion of log Γ(x). An expression for L3 remains an open question. The quest for such an expression is connected to a special kind of multiple zeta values known as Tornheim sums. The study of their relation with the value of L3 was begun in [8, 9]. Section 4 discusses the integrals 1 n n (1.10) Sn =(−1) log (sin πx)dx 0 that appear in the evaluation of L2. A notion of weight is introduced and a recur- rence for this family shows directly that Sn is a homogeneous form. The study of the loggamma integrals considered here has been motivated by our conjecture that Ln is a homogeneous form of weight n. This remains open for n ≥ 3. 2. A Riemann sum approach to the evaluation of L1 In this section we present an elementary evaluation of the formula for L1.This was originally obtained by J. L. Raabe [12] and it appears as entry 6.441.2inthe classical table [10]. Theorem 2.1. The integral L1 is given by 1 √ (2.1) log Γ(x) dx =log 2π. 0 Proof. Partition the interval [0,1] into n subintervals of length 1/n to produce 1 n 1 k (2.2) log Γ(x) dx = lim log Γ . n→∞ n n 0 k=1 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTEGRALS OF POWERS OF LOGGAMMA 537 On the other hand, assuming n is even, 1 n k 1 n k log Γ = log Γ n n n n k=1 ⎛k=1 ⎞ n/2 1 k k = log ⎝ Γ Γ 1 − ⎠ n n n ⎛k=1 ⎞ n/2 1 π = log ⎝ ⎠ n sin(πk/n) k=1 ⎛ ⎞1/n √ n/2 =logπ − log ⎝ sin(πk/n)⎠ . k=1 The reflection formula Γ(x)Γ(1 − x)=π/ sin πx for the gamma function has been employed in the third line. The classical trigonometric identity − n1 πk n sin = n 2n−1 k=1 now yields √ 1 n k 2π log Γ =log . n n (2n)1/2n k=1 Let n →∞to obtain the result. The case n odd is treated similarly. 3. The evaluation of L2 The expression for L2 given in (1.7) was obtained in [6] using integrals involving the Hurwitz zeta function ζ(z,s), defined in (1.4). Differentiate the identity 1 − − − 2Γ(1 z)Γ(1 z ) − − π(z z ) (3.1) ζ(z ,x)ζ(z,x) dx = 2−z−z ζ(2 z z )cos 0 (2π) 2 with respect to z and z and then set z = z = 0. The formula of Lerch (see [13], page 271), d √ (3.2) ζ(z,x) =logΓ(x) − log 2π, dz z=0 produces the result. In this section we provide a new proof of (1.7) based on the Fourier expansion of log Γ(x): 1 1 (3.3) log Γ(x)=L − log(2 sin πx)+ (γ +2L )(1 − 2x) 1 2 2 1 ∞ 1 log k + sin 2πkx. π k k=1 This expansion was given by E. Kummer [11]; the reader will find a detailed proof in [1]. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 538 T. AMDEBERHAN ET AL. Define 1 1 (3.4) g(x)=L − log(2 sin πx)+ (γ +2L )(1 − 2x), 1 2 2 1 ∞ 1 log k s(x)= sin 2πkx π k k=1 so that 1 1 1 2 2 (3.5) L2 = s (x) dx +2 s(x)g(x) dx + g (x) dx. 0 0 0 Each term in this sum is now considered separately. First term. The orthogonality of the trigonometric terms on [0, 1] yields 1 1 2 2 1 log k1 log k2 1 log k s (x) dx = 2 sin(2πk1x)sin(2πk2x) dx = 2 2 . 0 π k1 k2 0 π k k1,k2 k ∞ 1 log2 k Therefore s2(x) dx = ζ(2)/2π2 using = ζ(2). k2 0 k=1 Second term. The “cross term” in (3.5) reduces to ∞ 1 1 log k 1 2 g(x) s(x) dx = − sin(2πkx)log(2sinπx) dx π k 0 k=1 0 ∞ 2(γ +2L ) log k 1 − 1 x sin(2πkx) dx π k k=1 0 1 in view of the vanishing of sin(2πkx) dx =0, for k ≥ 1. Integration by parts 0 ∞ 1 1 log k yields x sin(2πkx) dx = − , converting the last series into = −ζ(2). 2πk k2 0 k=1 1 The evaluation sin(2πkx)log(2sinπx) dx = 0 appears as 4.384.1 in [10]. It 0 1 −ζ (2) follows that g(x)s(x) dx = 2 (γ +log2π). 0 2π 1 2 2 2 π 1 2 Third term. The last term in (3.5) is g (x) dx = L1 + + (γ +2L1) , 0 48 12 wherewehaveemployed 1 1 (3.6) log(2 sin πx) dx = (1 − 2x)log(2sinπx) dx =0 0 0 and 1 π2 (3.7) log2(2 sin πx) dx = . 0 12 The second integral in (3.6) is seen to vanish by using the change of variables t =1− x. The evaluation (3.7) is proven in Section 4. Every term in (3.5) has been evaluated, confirming (1.7). License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTEGRALS OF POWERS OF LOGGAMMA 539 Note 3.1. A second proof of (1.7) can be obtained from the Fourier expansion ∞ ∞ (3.8) log Γ(x)=a0 + an cos(2πnx)+ bn sin(2πnx), n=1 n=1 1 A+log n with a0 = L1,an = 2n and bn = πn , with A = γ +2L1. This appears in [6] (formulas (6.3) and (6.4)) and it follows directly from entries 6.443.1and6.443.3 in [10]. Parseval’s identity gives ∞ ∞ 1 1 (3.9) L = a2 + a2 + b2 , 2 0 2 n 2 n n=1 n=1 which produces (1.7). 4. A family of log-trigonometric integrals This section considers the family of integrals 1 k k (4.1) Sk =(−1) log (sin πx)dx 0 that appeared in the special cases k =1andk = 2 in the evaluation of L2 given in Section 3. These integrals were analyzed in [2], where the value − k k 1 ( 1) d Γ(α + 2 ) (4.2) Sk = √ π2k dα Γ(α +1) α=0 was employed to produce the exponential generating function ∞ k Γ 1−x x 1 2 (4.3) Sk = √ . k! π Γ 1 − x k=0 2 From there, the author derived the recurrence k −j k! (4.4) S = S log 2 + (1 − 2 )ζ(j +1) S − .