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Evaluation of a Ln Tan Integral Arising in Quantum Field Theory

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Evaluation of a Ln Tan Integral Arising in Quantum Field Theory Evaluation of a lntan integral arising in quantum field theory Mark W. Coffey Department of Physics Colorado School of Mines Golden, CO 80401 (Received 2007) August 27, 2007 Abstract We analytically evaluate a dilogarithmic integral that is prototypical of volumes of ideal tetrahedra in hyperbolic geometry. We additionally obtain new representations of the Clausen function Cl2 and the Catalan constant G = Cl2(π/2), as well as new relations between sine and Clausen function values. arXiv:0801.0272v1 [math-ph] 1 Jan 2008 Key words and phrases Clausen function, trigamma function, polygamma function, dilogarithm function, Hurwitz zeta function AMS classification numbers 33B30, 33B15, 11M35 1 Introduction and statement of results In this paper, we provide a closed form evaluation of the integral 24 π/2 tan t + √7 I7 ln dt, (1.1) ≡ 7√7 π/3 tan t √7 Z − in terms of elementary mathematical constants and values of the Clausen function Cl2. This and related integrals originate in hyperbolic geometry and quantum field theory [7, 14]. The integral I7 arises in the analysis of the volume of ideal tetrahedra in hyperbolic space and is the simplest of 998 empirically investigated cases wherein the volume of a hyperbolic knot complement appears to be expressible in terms of a Dirichlet L series [5]. In particular, Borwein and Broadhurst have conjectured that ? ∞ 1 1 1 1 1 1 I7 = L 7(2) = + + . − "(7n + 1)2 (7n + 2)2 − (7n + 3)2 (7n + 4)2 − (7n + 5)2 − (7n + 6)2 # nX=0 (1.2) The ? here indicates numerical verification has been performed but that no proof exists. The integral (1.1) has been much discussed of late [4], and in fact it provides the opening equation of a very recent book [3]. The evaluation of I7 in the specific form of Eq. (1.2) remains to be completed–our evaluation is in terms of a certain other combination of Clausen function values. In addition to the Clausen value relations conjectured in Ref. [5], our work suggests further, presumably kindred, relations. Moreover, our work may provide a means of making progress in the other 997 cases linking a knot complement volume to an L series. The equality of expressions such as Eqs. (1.1) and (1.2) would tightly link topics in hyperbolic geometry with analytic 2 number theory. In the rest of this Introduction, we present two Lemmas, state our main result, and provide some background relations on the particular Clausen function Cl2. It seems worth while to note some of the numerous alternative expressions for the value L 7(2) and its companion Clausen value combination. We have − Lemma 1. Let ψ′ be the trigamma function and ζ(s, a) the Hurwitz zeta function (e.g, [1]). (a) We have 1 1 2 3 4 5 6 L 7(2) = ψ′ + ψ′ ψ′ + ψ′ ψ′ ψ′ (1.3a) − 49 7 7 − 7 7 − 7 − 7 1 1 2 3 2 2 π 2 2π 2 3π = 2 ψ′ + ψ′ ψ′ π csc + csc csc (1.3b) 49 7 7 − 7 − 7 7 − 7 2 1 2 3 2 3π 2 = ψ′ + ψ′ ψ′ + csc 4 π (1.3c) 49 7 7 − 7 7 − 1 1 2 3 4 5 6 = ζ 2, + ζ 2, ζ 2, + ζ 2, ζ 2, ζ 2, (1.4) 49 7 7 − 7 7 − 7 − 7 1 (1 + u u2 + u3 u4 u5) = − − − ln u du − 0 (1 u7) Z − 1 (1+2u + u2 +2u3 + u4) = 2 3 4 5 6 ln u du − Z0 (1 + u + u + u + u + u + u ) 1 u(1 + u u4 u5) =1 2 −3 −4 5 6 ln u du (1.5) − Z0 (1 + u + u + u + u + u + u ) (b) We have 2π 4π 6π 1 1 2 3 Cl2 + Cl2 Cl2 = 8 ψ′ + ψ′ ψ′ 7 7 − 7 56√7 7 7 − 7 π 3π π 3π 2π 5π 4π π2 csc2 + 2csc2 + +csc2 + csc2 + csc2 + csc2 csc2 . (1.6) − 14 7 7 14 7 14 − 7 3 In Proposition 1 we make use of some of the representations for Cl2 given in the following Lemma 2. Let a be a real number and set 2 1 1 a θ = cos− − . (1.7) 1+ a2 ! Then we have (a) ∞ u + a du Cl2(θ)= ln . (1.8) a u a 1+ u2 Z − Let H be the nth harmonic number, γ = ψ(1) the Euler constant, and ψ the n − digamma function. Then (b) j 1 1 ∞ ( 1) Hj Cl (θ) = 2cot− a ln 2 + − (1.9a) 2 a a2j (2j + 1) jX=1 j 1 1 1 1 1 ∞ ( 1) ψ(j) = cot− a(2 ln 2 + γ 2) + (2 γ) ln 1+ + − . (1.9b) − a − − a a2 a a2j (2j + 1) jX=1 (c) We have the integral representation 1 1 1 ∞ [y cot− ay cot− a] dy Cl2(θ) = 2cot− a ln 2 + 2a − . (1.10) 1 (1 y2) y Z − As an immediate consequence of Lemma 2(b) at a = 1 we obtain a harmonic number series representation for the Catalan constant G = Cl (π/2) = 1 1/32 + 2 − 1/52 1/72 + ... 0.9159655941: − ≃ Corollary 1. We have j π ∞ ( 1) G = ln 2 + − H . (1.11) 2 (2j + 1) j jX=1 This expression has been previously obtained in the very recent Ref. [6]. 4 1 Let θ = tan− (√7/3), ± ± √7+ √3 1 r73 sin θ r73 , ω = tan− ± , (1.12a) ≡ √7 √3 ± 1 r73 cos θ ! − − ± so that 1 1 2π ω = cot− (2√3 √7) = tan− √7 , + − − − 3 1 2√3 √7 1 ω = ω+ = tan− − , and ω+ ω = 2ω+ = 2cot− (2√3 √7). − − 5 ! − − − − − (1.12b) Proposition 1. We have 24 1 I7 = Cl2(θ+)+ [Cl2(2ω+) Cl2(2ω+ +2θ+)] . (1.13) 7√7 2 − The Clausen function Cl2 can be defined by (e.g., [13]) θ 1 t 1 x sin θ dx Cl2(θ) ln 2 sin dt = tan− ≡ − 0 2 0 1 x cos θ ! x Z Z − 1 ln x ∞ sin(nθ) = sin θ dx = . (1.14) − 0 x2 2x cos θ +1 n2 Z − nX=1 When θ is a rational multiple of π it is known that Cl2(θ) may be written in terms of the trigamma and sine functions [10, 8]. As we shall find useful, the imaginary part of the dilogarithm function Li2 of complex argument may be written in terms of Cl2: 1 Im Li (reiθ)= ω ln r + [Cl (2ω) Cl (2ω +2θ) + Cl (2θ)] , (1.15) 2 2 2 − 2 2 where 1 r sin θ ω = tan− . (1.16) 1 r cos θ ! − 5 Besides being periodic, Cl2(θ) = Cl2(θ +2π), the Clausen function satisfies the du- plication formula 1 Cl (2θ) = Cl (θ) Cl (π θ). (1.17) 2 2 2 − 2 − Further properties of Cl2 are given in [12] and [13] and Ref. [15] gives a very recent but necessarily incomplete review of the dilogarithm function with complex argument. We next provide the proofs of Lemmas 1 and 2 and Proposition 1 and finish with a discussion with further analytic results. Our proof of Proposition 1 shows how to make a ready generalization of the integral I7. Proofs of Lemmas 1 and 2, and expressions for the Catalan constant For Eqs. (1.3) and (1.4) of Lemma 1, we repeatedly apply the relation [9] (p. 944) between the specific Hurwitz zeta function ζ(2, a) and the trigamma function contained in q ∞ 1 1 p ( 1) 1 (q 1) p = ζ q, = − ψ − , (2.1) (7j + p)q 7q 7 7q (q 1)! 7 jX=0 − and use the reflection formula satisfied by the trigamma function 2 2 ψ′(1 x)= ψ′(x)+ π csc πx. (2.2) − − Equation (1.3c) follows from (1.3b) by the use of the identity π 2π 3π csc2 + csc2 + csc2 =8. (2.3) 7 7 7 For Eq. (1.5), we use the integral representation 1 ts 1e (a 1)t ζ(s, a)= ∞ − − − dt, Re s> 1, Re a> 0, (2.4) Γ(s) 0 et 1 Z − 6 t/7 where Γ is the Gamma function, and change variable to u = e− . For part (b) of Lemma 1, we first apply for p even and q odd, with q 3 the ≥ formula [8] (p. 329) q 1 pπ 1 − k 1 k p Cl2 = ψ′ 1 + ψ′ sin k π. (2.5) q ! −4q2 " − 2q ! 2 − 2q !# q kX=1 We then obtain 2π 4π 6π Cl2 + Cl2 Cl2 7 7 − 7 1 1 1 3 2 5 3 = ψ′ + ψ′ ψ′ + ψ′ ψ′ ψ′ 56√7 14 7 − 14 7 − 14 − 7 4 9 5 11 6 13 +ψ′ + ψ′ ψ′ + ψ′ ψ′ ψ′ , (2.6) 7 14 − 7 14 − 7 − 14 wherein we used the identity 2xπ 4xπ 6xπ √7 sin + sin sin = , (2.7) 7 7 − 7 ± 2 for x =1,..., 6 and the sign holds on the right side for x =3, 5, and 6. − In regard to Eq. (2.7), we observe that sin(2π/7), sin(4π/7), and sin(6π/7) ± ± ± are the nonzero roots of the Chebyshev polynomial T7(x). Indeed, if we write the cubic polynomials 2π 4π 6π 3 √7 2 √7 p1(x)= x sin x sin x + sin = x x + , (2.8a) − 7 − 7 7 − 2 8 and 6π 2π 4π 3 √7 2 √7 p2(x)= x sin x + sin x + sin = x + x , (2.8b) − 7 7 7 2 − 8 we then have the factorization p1(x)p2(x)= T7(x)/64x.
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