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Alternative Evaluation of a Ln Tan Integral Arising in Quantum Field

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Alternative Evaluation of a Ln Tan Integral Arising in Quantum Field Alternative evaluation of a lntan integral arising in quantum field theory Mark W. Coffey Department of Physics Colorado School of Mines Golden, CO 80401 (Received 2008) November 9, 2008 Abstract A certain dilogarithmic integral I7 turns up in a number of contexts in- cluding Feynman diagram calculations, volumes of tetrahedra in hyperbolic geometry, knot theory, and conjectured relations in analytic number theory. We provide an alternative explicit evaluation of a parameterized family of inte- grals containing this particular case. By invoking the Bloch-Wigner form of the dilogarithm function, we produce an equivalent result, giving a third evaluation of I7. We also alternatively formulate some conjectures which we pose in terms of values of the specific Clausen function Cl2. arXiv:0810.5077v2 [math-ph] 15 Nov 2008 Key words and phrases Clausen function, dilogarithm function, Hurwitz zeta function, functional equation, duplication formula, triplication formula AMS classification numbers 33B30, 11M35, 11M06 1 The particular integral 24 π/2 tan t + √7 I7 ln dt, (1) ≡ 7√7 π/3 tan t √7 Z − occurs in a number of contexts and has received significant attention in the last several years [3, 4, 5, 6]. This and related integrals arise in hyperbolic geometry, knot theory, and quantum field theory [6, 7, 8]. Very recently [9] we obtained an explicit evaluation of (1) in terms of the specific Clausen function Cl2. However, much work remains. This is due to the conjectured relation between a Dirichlet L series and I7 [6], ? ∞ 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 (2) The ? here indicates that numerical verification to high precision has been per- formed but that no proof exists, the approximate numerical value of I being I 7 7 ≃ 1.15192547054449104710169. The statement (2) is equivalent to the conjecture, with 1 θ 2tan− √7, 7 ≡ 1 ? 1 7 2π 4π 6π [3Cl2(θ7) 3Cl2(2θ7) + Cl2(3θ7)] = ZQ(√ 7) = Cl2 + Cl2 Cl2 , 2 − 4 − 4 7 7 − 7 (3) relating triples of Clausen function values. Here, we alternatively evaluate I7 directly in terms of the left side of (3). In addition, we present another evaluation of I7, based upon a property of the Bloch-Wigner form of the dilogarithm function. We recall that the L series L 7(s) has occurred in several places before, including − hyperbolic geometry [19] and Dedekind sums of analytic number theory [2]. Let ζQ(√ p) denote the Dedekind zeta function of an imaginary quadratic field Q(√ p). − − 2 Then indeed we have [2, 19, 20] 1 1 ζQ(√ 7)(s)= 2 2 s (4) − 2 m,n∈Z (m + mn +2n ) (m,nX)=(0,0) 6 6 s ν ν = ζ(s)L 7(s)= ζ(s)7− ζ s, , (5) − 7 7 νX=1 ν where 7 is a Legendre symbol, ζ(s, a) is the Hurwitz zeta function, and ζ(s) = ζ(s, 1) is the Riemann zeta function. The series L 7(s) is an example of a Dirichlet L function corresponding to a real − character χ [here, modulo 7] with χ (k 1) = 1. Such L functions, extendable to k k − − the whole complex plane, satisfy the functional equation [20] 1 s s+1/2 sπ L k(s)= (2π) k− cos Γ(1 s)L k(1 s). (6) − π 2 − − − Owing to the relation Γ(1 s)Γ(s) = π/ sin(πs), this functional equation may also − be written in the form s s 1/2 πs L k(1 s) = 2(2π)− k − sin Γ(s)L k(s). (7) − − 2 − Integral representations are known for these L-functions [20, 10]. From the functional equation (6) we find ∂ k3/2 L k(s) = L k(2). (8) ∂s − 4π − s= 1 − In turn, we have k3/2 ζQ′ (√ k)( 1) = L k(2), (9) − − − 48π − where we used ζ( 1) = 1/12 and L k( 1) = 0. − − − − 3 We have Proposition 1. We have 4 I7 = [3Cl2(θ7) 3Cl2(2θ7) + Cl2(3θ7)] . (10) 7√7 − In fact, we treat integrals π/2 tan t + a I(a) ln dt, (11) ≡ Zπ/3 tan t a − and more general ones with varying limits. For (11), we assume that π/3 < ϕ = 1 tan− a <π/2. These other integrals permit us to explicitly write other conjectures directly in terms of linear combinations of specific Clausen function values. The Clausen function Cl2 can be defined by (e.g., [14, 16]) θ 1 t 1 x sin θ dx Cl2(θ) ln 2 sin dt = tan− (12) ≡ − Z0 2 Z0 1 x cos θ ! x − 1 ln x ∞ sin(nθ) = sin θ dx = . (13) − 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 [11, 13]. This Clausen function is odd and periodic, Cl (θ)= Cl ( θ), and Cl (θ) = Cl (θ +2π). It also satisfies the duplication 2 − 2 − 2 2 1 Cl (2θ) = Cl (θ) Cl (π θ), (14) 2 2 2 − 2 − triplication 1 2π 4π Cl2(3θ) = Cl2(θ) + Cl2 θ + + Cl2 θ + , (15) 3 3 3 and quadriplication 1 π 3π Cl2(4θ) = Cl2(θ) + Cl2 θ + + Cl2 (θ + π) + Cl2 θ + , (16) 4 2 2 4 formulas, as well as a more general multiplication formula [14]. We recall the specific relation 6 2π Cl j =0, (17) 2 7 jX=1 that arises as a special case of [14] (pp. 95, 253) n 1 − 2π Cl j =0. (18) 2 n jX=1 In (17), pairwise cancellation occurs, as Cl (θ)= Cl (2π θ). 2 − 2 − Further information on the special functions that we employ may readily be found elsewhere [15, 16, 18, 10]. In particular, with k ∞ z Li (z)= , z 1, (19) 2 k2 | | ≤ kX=1 or z ln(1 t) Li2(z)= − dt, (20) − Z0 t the dilogarithm function, we have the relation π2 1 Li (eiθ)= θ(2π θ)+ iCl (θ), 0 θ 2π. (21) 2 6 − 4 − 2 ≤ ≤ We omit discussion of further relations between the Clausen function Cl2 and the dilogarithm function. For the proof of Proposition 1 we repeatedly rely on [14] (pp. 227, 272) θ 1 1 1 ln(tan θ+tan ϕ)dθ = θ ln(cos ϕ) Cl2(2θ+2ϕ)+ Cl2(2ϕ) Cl2(π 2θ). (22) Z0 − − 2 2 − 2 − We may split the integral in (11), writing ϕ a + tan t π/2 tan t + a I(a)= ln dt + ln dt π/3 a tan t ϕ tan t a Z − Z − 5 π/2 ϕ π/2 = ln(a + tan t)dt ln(a tan t)dt ln(tan t a)dt. (23) Zπ/3 − Zπ/3 − − Zϕ − By the use of (22) we obtain for the first integral on the right side of (23) π/2 π 1 2π 1 π ln(a + tan t)dt = ln cos ϕ + Cl2 +2ϕ Cl2 (π +2ϕ) + Cl2 , Zπ/3 − 6 2 3 − 2 3 (24a) the second integral, ϕ π 1 π 1 π ln(a tan t)dt = ϕ ln cos ϕ Cl2 2 ϕ + Cl2 Cl2(π 2ϕ) , Zπ/3 − − − 3 −2 − 3 2 3 − − (24b) and the third integral, π/2 π 1 ln(tan t a)dt = ϕ ln cos ϕ = cot− a ln cos ϕ. (24c) Zϕ − − 2 − − This latter integral is readily obtained from (22) by taking a a so that simply → − tan ϕ tan ϕ. Then per (23) we have → − 1 2π 2π I(a)= Cl2 2ϕ + + Cl2 2ϕ Cl2(π +2ϕ). (25) 2 3 − 3 − Then we apply both the duplication formula (14) and the triplication formula (15) wherein Cl (θ +4π/3) = Cl (θ 2π/3) as Cl (θ) = Cl (θ 2π) by the 2π-periodicity 2 2 − 2 2 − of Cl2. We find 1 I(a)= [Cl (6ϕ) 3Cl (4ϕ) + 3Cl (2ϕ)] . (26) 6 2 − 2 2 1 When ϕ = tan− √7, the case (10) follows. We next present some reference integrals. We then apply them to write expressions for combinations of the integrals (n+1)π/24 tan t + √7 In ln dt, (27) ≡ nπ/24 tan t √7 Z − 6 where n 0 is an integer. We supplement (22) with ≥ y 1 ln(a tan t)dt = (y x) ln cos ϕ + [Cl2[2(ϕ y)] Cl2[2(ϕ x)]] Zx − − − 2 − − − 1 [Cl (π 2y) Cl (π 2x)] , (28) −2 2 − − 2 − where a = tan ϕ. We also have y tan t + a 1 ln dt = [Cl2(2x +2ϕ) Cl2(2x 2ϕ) + Cl2(2y 2ϕ) Cl2(2y +2ϕ)] , Zx tan t a 2 − − − − − (29) 1 with x<ϕ = tan− a<y. We now write expressions for the linear combinations C 2(I + I + I + I )+ I + I (I + I ) =0? , (30) 1 ≡ − 2 3 4 5 8 9 − 10 11 and C I + 3(I + I + I )+2(I + I ) 3I I =0? . (31) 2 ≡ 2 3 4 5 6 7 − 8 − 9 These relations have been detected with further PSLQ computations [5]. A similar conjecture for integrals In with increments nπ/60 has also been written [4] (p. 508). The latter linear combination may also be expressed in terms of Cl2 values, but here we concentrate on (30) and (31). We decompose the left side of (30) as indicated, and find for these contributions π π π π 2(I2 + I3 + I4 + I5) = Cl2 +2ϕ Cl2 2ϕ + Cl2 2ϕ Cl2 +2ϕ , 6 − 6 − 2 − − 2 (32) 1 where ϕ = tan− √7, 2π 3π 2π 3π 2I8 = Cl2 +2ϕ Cl2 +2ϕ + Cl2 2ϕ Cl2 2ϕ , (33) 3 − 4 − 3 − − 4 7 3π 3π 5π 5π 2I9 = Cl2 +2ϕ Cl2 2ϕ + Cl2 2ϕ Cl2 +2ϕ , (34) 4 − 4 − 6 − − 6 and 1 5π 5π I10 + I11 = Cl2(π +2ϕ)+ Cl2 +2ϕ Cl2 2ϕ .
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