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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 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 . (35) − 2 6 − 6 − Therefore, we obtain π π π π C1 = Cl2 +2ϕ + Cl2 2ϕ Cl2 2ϕ + Cl2 +2ϕ − 6 6 − − 2 − 2 1 2π 2π 5π 5π Cl2 +2ϕ + Cl2 2ϕ + Cl2 2ϕ Cl2 +2ϕ 2 3 − 3 6 − − 6 +Cl2(π +2ϕ). (36) For the combination C2 we have π π π π 2I2 = Cl2 +2ϕ Cl2 2ϕ + Cl2 2ϕ Cl2 +2ϕ , (37) 6 − 6 − 4 − − 4 2π 2π π π 2(I6+I7) = Cl2 +2ϕ Cl2 2ϕ +Cl2 2ϕ Cl2 +2ϕ , (38) − 3 − 3 − 2 − − 2 and π π π π 2(I3+I4 +I5) = Cl2 +2ϕ Cl2 2ϕ +Cl2 2ϕ Cl2 +2ϕ . (39) 4 − 4 − 2 − − 2 Therefore, we find π π π π 2C2 = Cl2 +2ϕ Cl2 2ϕ + Cl2 2ϕ Cl2 +2ϕ 6 − 6 − 2 − − 2 2π 2π 5π 5π 5Cl2 +2ϕ 5Cl2 2ϕ Cl2 2ϕ + Cl2 +2ϕ − 3 − − 3 − 6 − 6 π π 3π 3π +2 Cl2 +2ϕ Cl2 2ϕ + Cl2 +2ϕ + Cl2 2ϕ . (40) 4 − 4 − 4 − 4 8 By a combination of the quadriplication formula (16) and the duplication formula (14) we may write 1 π π 1 Cl2(4θ) = Cl2 θ + + Cl2 θ + Cl2(2θ). (41) 4 2 − 2 2 This enables other expressions for C1 and C2. Similarly, one may use the 6- and 12-fold multiplication formulas. In regard to the combination on the right side of (3), we comment on an obser- vation given previously [9]. We have 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 + , (42a) − 7 − 7 7 − 2 8 and 6π 2π 4π 3 √7 2 √7 p2(x)= x sin x + sin x + sin = x + x , (42b) − 7 7 7 2 − 8 we then have the factorization p1(x)p2(x) = T7(x)/64x. This invites questions as to whether scaled versions of these or other Chebyshev polynomials could be useful in developing identities underlying (3), (30), (31), or the like. Given the close relation of the Clausen function Cl2 and the dilogarithm function, one wonders if a set of ladder relations for the latter may be carried over to explain (3) and relations amongst the integrals In. In developing ladder relations, cyclotomic equations for the base have proven very useful. It would be of interest to see if Cl2 relations with θ7 could be discovered in this way. 9 We remark on using Kummer’s relation [14] (pp. 107, 254) to rewrite the right side of (3) in terms of the dilogarithm of complex argument. We have 1 7 iφ ZQ(√ 7) = Im Li2(Re ) b ln R , (43) 4 − 2 − h i where tan b R = . (44) sin φ + tan b cos φ Here, we may take φ = π/7 and b = 2π/7, or vice versa. Then by Proposition 2 of [9] we have the integral representation √7 ? 2π 4π 6π π ∞ ln y dy I7 = Cl2 + Cl2 Cl2 = 2 sin , 2 7 7 − 7 7 d y2 2y cos(π/7)+1 Z − (45) 1 where d = [2 cos(π/7) 1]− . − Finally, we use relations from [16] (Appendix A) and [19] to write a third evalua- 1 tion of the integral I . For this we introduce the angle θ 2tan− (√7/5) and the 7 75 ≡ Bloch-Wigner dilogarithm [17] D(z) = Im[Li (z)] + arg(1 z) ln z , (46) 2 − | | for which we have [16] (p. 246) 1 D(z)= [Cl (2θ) + Cl (2ω) Cl (2θ +2ω)], (47) 2 2 2 − 2 where θ = arg z and ω = arg (1 z¯). We note the interpretation that for z C, − ∈ the volume of the asymptotic simplex with vertices 0, 1, z, and in 3-dimensional ∞ 10 hyperbolic space is given by D(z) [16] (p. 271). We then rewrite the expression | | ([16], p. 384 or [19], p. 246) 4π2 1+ i√7 1+ i√7 ζQ(√ 7)(2) = ζ(2)L 7(2) = 2D + D − . (48) − − 21√7 " 2 ! 4 !# We apply (47), giving ? 8 1+ i√7 1+ i√7 I7 = L 7(2) = 2D + D − − 7√7 " 2 ! 4 !# 4 = [4Cl2(π θ7) Cl2(θ7) + Cl2(θ75) + Cl2(θ7 θ75)] . (49) 7√7 − − − In the case of D[(1 + i√7)/2] we used the duplication formula (14). In contrast to 1 (49), the expression in [9] for I involves θ tan− (√7/3). With the various analytic 7 + ≡ evaluations now known for I7 or L 7(2), we have enlarged the set of possible relations − amongst Cl2 values. From (10), (14), and (49) we obtain the conjecture Cl (3θ ) Cl (2θ ) =? Cl (θ ) + Cl (θ θ ). (50) 2 7 − 2 7 2 75 2 7 − 75 In fact, we have θ θ =2θ , and we conclude by proving (50), and thereby (49). 7 − 75 + We quickly show that both Cl2(3θ7) = Cl2(θ75) (51) and Cl (2θ )= Cl (θ θ ), (52) 2 7 − 2 7 − 75 for we have 3θ 2π = θ and θ π = θ . The latter relations require nothing 7 − 75 7 − − + more than the identity tan(x/2) = sin x/(1 + cos x). 11 We have similarly found many other angular pairs (θ , θ ) satisfying 3θ 2π = 1 2 1 − θ , immediately giving Cl (3θ ) = Cl (θ ). As these may be useful elsewhere ± 2 2 1 ± 2 2 [6, 16], we record several of them in the first Appendix. We also relegate to this Appendix a possibly new log trigonometric integral in terms of Cl2. In the second Appendix, we develop new series and integral representations of the Clausen function. Appendix A 1 1 We let θ 2tan− √k, and θ 2tan− √k/j. We find the relations k ≡ k,j ≡ 3θ 2π = θ , 3θ 2π = θ , (A.1) 2 − − 2,5 5 − 5,7 1 5√13 3θ11 2π = θ11,4, 3θ13 2π = 2tan− , (A.2) − − − − 19 ! and 1 8√91 1 2√91 3θ91,3 2π = 2tan− , 3θ91,5 2π = 2tan− , 3θ91,7 2π = θ91,28. − − 99 ! − 155 ! − − (A.3) 1 1 1 1 With θ 2tan− 3/2, θ 2tan− 5/3, θ 2tan− 13/3, θ 2tan− 7/3, 32 ≡ 53 ≡ 133 ≡ 73 ≡ q q q q we have 1 3 3 1 1 5 3θ32 2π = 2tan− , 3θ53 2π = 2tan− , (A.4) − − 7s2 − − 3s3 and 1 1 13 1 1 7 3θ133 2π = 2tan− , 3θ73 2π = 2tan− . (A.5) − 9s 3 − − 9s3 Based upon the trigonometric identity 3 + 4 cos θ +cos2θ = 2(1+cos θ)2, we have found the integral x ln(3 + 4 cos θ + cos2θ)dθ = x ln 2 + 4Cl2(π x), 0 x π. (A.6) Z0 − − ≤ ≤ 12 This obviously provides an integral expression for the Catalan constant G = k 2 k 0( 1) /(2k + 1) = Cl2(π/2) when x = π/2. ≥ − P The function Cl (t) for t (0, π) has its only maximum at t = π/3, when 2 ∈ Cl (π/3) 1.014941606409653625021. We mention that near this value Cl has a 2 ≃ 2 fixed point, Cl (y)= y for y 1.01447193895251725798414. 2 ≃ Related to the equality of expressions (10) and (49) for I7 we have the relation 1 3i√7 1+3i√7 2 2 6 Li2 − + Li2 = 3(π 2θ+) π " 8 ! 8 !# − − 2 2 1 2 2 = 3(θ θ ) π = 3[π tan− (3√7)] π . (A.7) 7 − + − − − Such relations follow readily from (21) as we have iθ iθ 2 2 6[Li (e ) + Li (e− )]=2π +3θ , 0 θ 2π. (A.8) 2 2 ≤ ≤ 13 Appendix B We have Proposition B1. We have for θ < π/n and n 1 an integer ≥ 2j 1 1 ∞ ζ(2j) (n 1) Cl (2nθ) Cl (2θ) = − θ2j+1 θ ln n. (B.1) 2 n 2 − 2 jπ2j (2j + 1) − jX=1 This result gives several Corollaries, including Corollary (i) n 1 2j 1 − 2π ∞ ζ(2j) (n 1) Cl 2θ + k = − θ2j+1 θ ln n, (B.2) 2 2 n jπ2j (2j + 1) − kX=1 jX=1 Corollary (ii) 1 1 Cl2(2nθ) Cl2(2θ) = θ ln n 2 n − − 2θ 1 dx + ∞ [sinh(nx) n sinh x] , (B.3) n 0 x2 − (eπx/θ 1) Z − Corollary (iii) 1 1 Cl2(2nθ) Cl2(2θ) = θ ln n 2 n − − 2 2 2 θ 1 nθ 1 θ π n θ + 2nθ tanh− 2θ tanh− + π ln − 2π " π ! − π ! π2 θ2 !# − ∞ 1 θ 1 1 nθ +2π tanh− tanh− P1(x)dx. (B.4) Z1 " πx ! − n πx!# In the last equation, P (x)= x [x] 1/2 is the first periodized Bernoulli polynomial. 1 − − Proof. The Proposition is based upon the relation [1] (p. 75) n sin x ∞ ζ(2j) ln = (n2j 1)x2j, x < π/n, (B.5) sin nx jπ2j − | | jX=1 14 wherein we have used the relation between ζ(2j) and the Bernoulli numbers B2j. (For more details, see the end of this Appendix.) With the series of (B.5) being bound- edly convergent, we may integrate term-by-term over any finite interval avoiding the singularity at x = π/n. Doing so, integrating over [0, θ], and using the first integral representation for Cl2 on the right side of (12) gives (B.1). Corollary (i) follows from the multiplication formula for Cl2 [14] (pp. 94, 253). Corollary (ii) uses a standard integral representation of the Riemann zeta function. With the interchange of summation and integration, with the integral being absolutely convergent, the Corollary follows. Corollary (iii) uses the representation for Re s> 1, − 1 1 P (x) ζ(s)= + s ∞ 1 dx. (B.6) 2 s 1 − 1 xs+1 − Z Again, the interchange of summation and integration is employed. Remarks. In connection with (B.5) we may note the relation with the Chebyshev polynomials Un of the second kind [12] (p. 1032), sin nφ Un 1(cos φ)= . (B.7) − sin φ For θ = π/4 and other values, Proposition B1 gives many relations involving the Catalan constant G. More generally, for θ a rational multiple of π, the results are expressible in terms of ψ′, the trigamma function [11, 13]. If we let 2j ∞ ζ(2j) (n 1) r(θ, n) − θ2j+1, θ π/n, (B.8) ≡ jπ2j (2j + 1) ≤ jX=1 15 we may write several simple examples: π 1 ψ (1/3) r , 2 = π (√3π + 6 ln 2) ′ , (B.9a) 3 " 18 − 4√3π # π 1 ψ (1/3) r , 3 = π (√3π + 9 ln 3) ′ , (B.9b) 3 " 27 − 6√3π # π 1 π r , 2 = G + ln 2, (B.10a) 4 −2 4 π 2 π r , 3 = G + ln 3, (B.10b) 4 −3 4 π 1 π r , 4 = G + ln 2, (B.10c) 4 −2 2 and π 1 ψ (1/3) r , 2 = π (2√3π + 9 ln 2) ′ , (B.11a) 6 "54 − 6√3π # π 1 ψ (1/3) r , 3 = π (√3π + 3 ln 3) ′ , (B.11b) 6 " 18 − 4√3π # π 1 7ψ (1/3) r , 4 = π (7√3π + 36 ln 2) ′ , (B.11c) 6 "108 − 24√3π # π 1 √3ψ (1/3) r , 5 = π (2√3π + 5 ln 5) ′ , (B.11d) 6 "30 − 10π # π 1 ψ (1/3) r , 6 = π (√3π + 3 ln 2 + 3 ln 3) ′ . (B.11e) 6 " 18 − 4√3π # Finally, we supply a derivation of (B.5). 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