The Laplace Transform of the Digamma Function: an Integral Due to Glasser, Manna and Oloa

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PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 136, Number 9, September 2008, Pages 3211–3221 S 0002-9939(08)09300-3 Article electronically published on April 30, 2008

THE LAPLACE TRANSFORM OF THE DIGAMMA FUNCTION: AN INTEGRAL DUE TO GLASSER, MANNA AND OLOA

TEWODROS AMDEBERHAN, OLIVIER ESPINOSA, AND VICTOR H. MOLL

(Communicated by Carmen C. Chicone)

Abstract. The deﬁnite integral 4 π/2 x2 dx M(a):= 2 2 −a π 0 x +ln (2e cos x) is related to the Laplace transform of the digamma function ∞ L(a):= e−asψ(s +1)ds, 0 by M(a)=L(a)+γ/a when a>ln 2. Certain analytic expressions for M(a) in the complementary range, 0

1. Introduction The classical table of integrals by I. S. Gradshteyn and I. M. Ryzhik [7] contains a large collection organized in sections according to the form of the integrand. In each section one finds a significant variation on the complexity of the integrals. For example, section 4.33–4.34, with the title Combinations of logarithms and exponentials, presents the elementary formula 4.331.1. For a>0, ∞ γ +lna (1.1) e−ax ln xdx= − , 0 a where γ is the Euler constant n 1 (1.2) γ = lim − ln n. n→∞ k k=1 Containedinthesamesectionarethemoreelaborate4.332.1and4.325.6: ∞ ln xdx 1 1 dx 2π 5 1 = ln ln = √ ln 2π − ln Γ . x −x − 2 − 0 e + e 1 0 x x x +1 3 6 6 The difficulty involved in the evaluation of a definite integral is hard to measure from the complexity of the integrand. For instance, the evaluation of Vardi’s integral, √ π/2 1 3 1 dx π Γ( 4 ) 2π (1.3) ln ln tan xdx= ln ln 2 = ln 1 , π/4 0 x 1+x 2 Γ( 4 )

Received by the editors July 23, 2007. 2000 Mathematics Subject Classiﬁcation. Primary 33B15. Key words and phrases. Laplace transform, digamma function. The work of the third author was partially funded by NSF-DMS 0409968.

c 2008 American Mathematical Society Reverts to public domain 28 years from publication 3211

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that appears as 4.229.7 in [7], requires a reasonable amount of number theory. The second integral form is 4.325.4, found in the section entitled Combinations of logarithmic functions of more complicated arguments and powers. The reader will find in [15] a discussion of this formula. It is a remarkable fact that combinations of elementary functions in the integrand often exhibit definite integrals whose evaluation is far from elementary. A system- atic study of the formulas in [7] has been initiated in the series [1, 2, 9, 10, 11, 12]. These papers are organized according to the combinations appearing in the integrand. Even the elementary cases, such as the combination of logarithms and rational functions discussed in [2], entail interesting results. The evaluation b ln tdt 1 − −n − 1 (1.4) n+1 = 1 (1 + b) ln b ln(1 + b) 0 (1 + t) n n − 1 n1 1 n − 1 − |s(j +1, 2)|bj, n(1 + b)n−1 j! j j=1 for b>0andn ∈ N, produces an explicit formula for the case where the rational function has a single pole. Here, |s(n, k)| are the unsigned Stirling numbers of the first kind, which count the number of permutations of n letters having exactly k cycles. The case of a purely imaginary pole, − x ln tdt 2n n1 tan−1 x + p (x) = n g (x)+p (x)lnx − k , (1 + t2)n+1 22n 0 n 2k +1 0 k=0 is expressed in terms of the rational function n 22j x (1.5) p (x)= , n 2j (1 + x2)j j=1 2j j and with x −1 −1 tan t (1.6) g0(x)=lnx tan x − dt. 0 t The special case x = 1 becomes − 1 ln tdt 2n n1 π + p (1) (1.7) = −2−2n G + 4 k , (1 + t2)n+1 n 2k +1 0 k=0 where ∞ (−1)k (1.8) G = (2k +1)2 k=0 is Catalan’s constant.Thevalues

k j 2 (1.9) pk(1) = 2j j=1 2j j do not admit a closed form (in the sense of [14]), but they do satisfy the three-term recurrence

(1.10) (2k +1)pk+1(1) − (3k +1)pk(1) + kpk−1(1) = 0.

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The study of deﬁnite integrals, where the integrand is a combination of powers, logarithms and trigonometric functions, was initiated by Euler [5], with the evaluation of π/2 7 (1.11) x ln(2 cos x) dx = − ζ(3) 0 16 and π/2 π (1.12) x2 ln(2 cos x) dx = − ζ(3), 0 4 which appear in his study of the Riemann zeta function at the odd integers. These type of integrals have been investigated in [8], [16]. The intriguing integral of D. and J. Borwein [3], π/2 11π 11π5 (1.13) x2 ln2(2 cos x) dx = ζ(4) = , 0 16 1440 was ﬁrst conjectured on the basis of a numerical computation by Enrico Au-Yueng while he was an undergraduate student at the University of Waterloo. A nice example of experimental mathematics in action.

Recently O. Oloa considered the integral 4 π/2 x2 dx (1.14) M(a):= , 2 2 −a π 0 x +ln (2e cos x) and the special value 4 π/2 x2 dx 1 (1.15) M(0) = = (1 + ln(2π) − γ) 2 2 π 0 x +ln (2 cos x) 2 is established in [13]. Oloa’s method of proof relies on the expansion ∞ x2 a a (1.16) = x sin 2x + (−1)n−1 n − n+1 x sin(2nx). 2 2 n! (n +1)! x +ln (2 cos x) n=1 Here 1 (1.17) an := (t)n dt, 0

where (t)n = t(t+1)···(t+n−1) is the Pochhammer symbol. The standard relation n k (1.18) (t)n = |s(n, k)|t k=1 gives n |s(n, k)| (1.19) a = . n k +1 k=1 M. L. Glasser and D. Manna [6] introduced the function ∞ (1.20) L(a):= e−asψ(s +1)ds, 0

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d where ψ(x)= dx ln Γ(x)isthedigamma function. After integrating by parts and making use of (1.1), one finds ∞ (1.21) L(a)=−γ − ln a + a e−at ln Γ(t) dt. 0 The main result in [6] gives a relation between M(a)andL(a). Theorem 1.1. If a>ln 2,then γ M(a)=L(a)+ . a That is, for a>ln 2, γ ∞ (1.22) M(a)= − γ − ln a + a e−at ln Γ(t) dt. a 0 The proof in [6] begins with the representation π/2 πΓ(ν +2) (1.23) cosν x cos ax dx = ν+1 ν a ν − a 0 2 (ν +1)Γ(1+ 2 + 2 )Γ(1+ 2 2 ) (3.631.9 in [7]). Differentiating with respect to a, evaluating at a = s, and using ψ(1) = −γ yields 2s+2 π/2 (1.24) ψ(s +1)= x coss x sin(sx) dx − γ. π 0 Replacing (1.24) into (1.20) produces ∞ π/2 γ 4 −a (1.25) L(a)+ = − Im xes(ln[2e cos x]−ix) dx ds. a π 0 0 The identity (1.22) follows from evaluating the s-integral as ∞ −a 1 (1.26) es(ln[2e cos x]−ix) ds = . − −a 0 ix ln [2e cos x] The authors of [6] produced a series expansion for M(a), which they recognize as a hypergeometric function in two variables, and state that this strongly suggests that for a general value of a, no further progress is possible.Thehypergeometric expression gives 1 1 (1.27) M(0) = 1 + t(1 − t) 3F2(1, 1, 2 − t;2, 3; 1) dt, 2 0 on which they invoke 2(1 − γ − ψ(t +1)) (1.28) F (1, 1, 2 − t;2, 3; 1) = 3 2 1 − t to give a new proof of (1.15). The graph of M(a) shown in Figure 1, obtained by the numerical integration of (1.14), has a well-defined cusp at a = ln 2. In this paper, analytic expressions for both branches of M(a) are provided. The region a>ln 2, determined in [6], has been reviewed in the present section. The corresponding expressions for 0

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2.0

1.5

1.0

0.5

0.0 0.0 0.5 1.0 1.5 2.0

Figure 1. The graph of M(a)for0≤ a ≤ 2

2. The case 0 ln 2, then ∞ s ln[2e−a cos x]+ix 1 (2.4) e ds = −a . 0 ix +ln[2e cos x] This implies π/2 ∞ 2 −a (2.5) M(a)= Im xes(ln[2e cos x]+ix) dx ds, π −π/2 0 where one uses the fact that the imaginary part of the integrand is an even function of x.Onemoreidentity,

−a −a 2ix (2.6) eisx · es ln[2e cos x] = es ln[e (1+e )], and the change of variables x → x/2, give the equality π ∞ 1 −a ix (2.7) M(a)= Im xes ln[e (1+e )] ds dx. 2π −π 0 Evaluating the s-integral yields π − 1 xdx (2.8) M(a)= Im −a ix . 2π −π ln [e (1 + e )]

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The formula 1 1 ut dt (2.9) = − ln u 0 u 1 now gives (2.1) from (2.8). Note 2.1. The proof outlined above is valid for a>ln 2, but (2.1) holds for a>0. Notation. Deﬁne b := ea − 1andlet0

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The left hand side of (2.14) reduces to −1/(k + 1) in view of the classical identity ∞ t +1 t (2.16) (−1)m−1 = . m + k +1 k +1 m=1 A simple evaluation of the right hand side in (2.14) also produces −1/(k +1). Therefore both sides of (2.14) are, up to a constant, the harmonic number Hk.The special case k = 0 shows that this constant vanishes. Continuing from (2.10), it follows that (2.17) ∞ j ∞ 1 (−1)j−k t ea 1 t M(a)=ea bj k + e−at bj ψ(j +1)dt j +1− k b j 0 j=0 k=0 0 j=1 ∞ ea 1 t − e−at bj ψ(t +1)dt ≡ M + M + M . b j 1 2 3 0 j=1

To simplify M1,observe ∞ j ∞ ∞ (−1)j−k t t (−1)ν bν bj k = bk j +1− k k ν +1 j=0 k=0 k=0 ν=0 ∞ ln(1 + b) t aeat = bk = . b k b k=0 −a Thus, M1 = a/(1 − e ).

The reduction of M2 employs the following result. Lemma 2.2. If 0

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Therefore, ln(1 − e−a) 1 ∞ e−at (2.23) M = + dt. 2 − −a − −a 1 e 1 e 1 t Finally, ⎛ ⎞ ∞ ea 1 t ea 1 M = − e−at ⎝ bj⎠ ψ(t +1)dt = − (1 − e−at)ψ(t +1)dt. 3 b j b 0 j=1 0 A direct computation shows that 1 ψ(t +1)dt = 0, and integration by parts gives 0 a 1 (2.24) M = e−at ln Γ(t +1)dt. 3 − −a 1 e 0 The identity ln Γ(t +1)=lnΓ(t)+lnt now yields a 1 1 M = e−at ln tdt+ e−at ln Γ(t) dt . 3 − −a 1 e 0 0 Replacing (2.17), (2.19) and (2.21) into (2.17) provides an expression for M(a): a γ ln(1 − e−a) (2.25) M(a)= + + − −a − −a 1 e a 1 e a ∞ a 1 + e−at ln tdt+ e−at ln Γ(t) dt. − −a − −a 1 e 0 1 e 0 The term γ/a comes from the index j = 0 in the sum (2.18). The main result presented here now follows from (1.1). This settles a conjecture of O. Oloa presented in [13]. Theorem 2.1. If 0 gamma function. Proof. Split up the ﬁrst integral in (2.25) and integrate by parts. The derivative of (2.1) at a = 0, the classical values 1 1 (2.26) ln Γ(t) dt = ln 2π 0 2 and 1 ζ (2) 1 − γ (2.27) t ln Γ(t) dt = 2 + ln 2π , 0 2π 6 12 obtained in [4], give π/2 x2 ln(2 cos x) dx 7π π ln 2π ζ(2) (2.28) = + − . 2 2 2 0 (x +ln (2 cos x)) 192 96 16π

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Further differentiation of (2.1) produces the evaluation of a family of integrals similar to (2.28). The integral in (2.1) can be expressed in an alternative form. Define ⎛ ⎞ n j (2.29) Λ(z) := lim ⎝ − ln n⎠ . n→∞ j2 + z2 j=1 Observe that Λ(0) = γ,soΛ(z) is a generalization of Euler’s constant. Lemma 2.3. Let a>0,c=1− e−a and define A := ln 2π + γ.Then ∞ 1 A(a − c) c a ln j (2.30) e−at ln Γ(t) dt = − Λ +2c . a2 2a 2π a2 +4π2j2 0 j=1 Proof. Expand the exponential into a MacLaurin series and use the value

n+1 1 1 2 n +1 (2k)! tn ln Γ(t) dt = (−1)k [Aζ(2k) − ζ(2k)] n +1 2k − 1 k(2π)2k 0 k=1 n √ 1 2 n +1 (2k)! ln 2π − (−1)k ζ(2k +1)+ n +1 2k 2(2π)2k n +1 k=1 given as (6.14) in [4]. Then interchange the resulting double sums.

The next corollary follows from the identity M(a)=L(a)+γ/a. Corollary 2.2. If 0

The reader will check that (2.32) is equivalent to the continuity of M(a)at a = ln 2. The identity (2.33) provides a proof of the next theorem, which in itself is worthy of singular (pun intended) interest. Theorem 2.2. The jump of M (a) at a =ln2is 4.

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3. Conclusions The integral 4 π/2 x2 dx M(a):= 2 2 −a π 0 x +ln (2e cos x)

satisﬁes γ ∞ (3.1) M(a)= + e−atψ(t +1)dt a 0

for a>ln 2 and γ a +ln(1− e−a)+Γ(0,a) 1 1 M(a)= + + e−atψ(t +1)dt − −a − −a a 1 e 1 e 0

for 0

References

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[15] I. Vardi. Integrals, an introduction to analytic number theory. Amer. Math. Monthly, 95:308– 315, 1988. MR935205 (89b:11073) [16] Z. Yue and K.S. Williams. Values of the Riemann zeta function and integrals involving θ θ log(2 sinh 2 ) and log(2 sin 2 ). Pac. Jour. Math., 168:271–289, 1995. MR1339953 (96f:11170) Department of Mathematics, Tulane University, New Orleans, Louisiana 70118 E-mail address: [email protected] Departmento de F´ısica, Universidad Tec.´ Federico Santa Mar´ıa, Valparaiso, Chile E-mail address: [email protected] Department of Mathematics, Tulane University, New Orleans, Louisiana 70118 E-mail address: [email protected]

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