Some Infinite Series Involving the Riemann Zeta Function

Home , Lerch zeta function, Reflection formula

International Journal of Mathematics and M Computer Science, 7(2012), no. 1, 11–83 CS

Some inﬁnite series involving the Riemann zeta function

Donal F. Connon

Elmhurst Dundle Road Matﬁeld, Kent TN12 7HD, United Kingdom

(Received May 16, 2010, Accepted July 12, 2012) Abstract This paper considers some inﬁnite series involving the Riemann zeta function. Some examples are set out below ∞ n n − kxk 1 ( 1) s x, s ,y − x x, s, y n k k y s = Φ( +1 ) log Φ( ) n=0 +1k=0 ( + ) ∞ n n k − n 2 1 − 1−s ζ s ( 1) k k s = 1 2 ( ) n=0 k=0 ( +1) 2 ∞ n n − k − k 1 ( 1) ( 1) π πy n+1 k k y + k − y = cosec( ) n=0 2 k=0 + +1

∞ 2n 2n u n −1 2 sin 2 u2 u u u u − ζ n3 n =4 log(2 sin )+2Cl3(2 )+4 Cl2(2 ) 2 (3) n=1

∞ 4 7 1 24n [n!] ζ(3) − πG = n 3 n 2 4 2 n=0 (2 +1) [(2 )!]

where Φ(x, s, y) is the Hurwitz-Lerch zeta function and Cln(t) are the Clausen functions.

Key words and phrases: Inﬁnite series, Riemann zeta Function. AMS (MOS) Subject Classiﬁcations: 40A05, 11M32. 12 Donal F. Connon

1 Some Hasse-type series

It was shown in [21] that ∞ n k k y 1 n (−1) y log y = Lis(y) − Lis−1(y)(1) − k s−1 − s 1 n=0 n +1 k=0 (k +1) s 1 where Lis(y) is the polylogarithm function [39]

∞ yn Lis(y)= s n=1 n and, with y = 1, this becomes the formula originally discovered by Hasse [32] in 1930 ∞ n k 1 1 n (−1) = Lis(1) = ζ(s)(2) − k s−1 s 1 n=0 s +1 k=0 (k +1)

∞ s−1 −t The gamma function is deﬁned as Γ(s)= 0 t e dt, s>0. Using the substitution t =(k + y)u,weobtain

∞ 1 1 s−1 −(k+y)u s = u e du (3) (k + y) Γ(s) 0 We now consider the ﬁnite sum set out below n k n x Sn(x, y)= (4) k s k=0 (k + y) and combine (3) and (4) to obtain

n k n ∞ n x n k 1 s−1 −(k+y)u Sn(x, y)= = x u e du k s k k=0 (k + y) k=0 Γ(s) 0 ∞ n 1 s−1 n k −(k+y)u = u x e du k Γ(s) 0 k=0 Some inﬁnite series involving the Riemann zeta function 13

∞ n 1 s−1 −yu n −u k = u e xe du k Γ(s) 0 k=0 Using the binomial theorem this becomes

∞ 1 s−1 −yu −u n Sn(x, y)= u e 1+xe du (5) Γ(s) 0 Making the summation, we see that

∞ n k ∞ ∞ n n x 1 n s−2 −yu −u n t = t u e 1+xe du k s n=0 k=0 (k + y) Γ(s) n=0 0

The geometric series gives us for |t (1 + xe−u)|< 1

∞ n 1 t xe−u 1+ = − −u n=0 1 t (1 + xe ) and we then have ∞ n k ∞ s−1 −yu n n x 1 u e t = du (6) k s − −u n=0 k=0 (k + y) Γ(s) 0 1 t (1 + xe ) We now integrate (6) with respect to t

∞ n+1 n k w ∞ s−1 −yu w n x 1 u e = dt du k s − −u n=0 n +1 k=0 (k + y) Γ(s) 0 0 1 t(1 + xe )

∞ s−1 −yu − −u − 1 u e log [1 w(1 + xe )] = −u du Γ(s) 0 1+xe and obtain 14 Donal F. Connon

∞ n+1 n k ∞ s−1 −(y−1)u −u w n x 1 u e log [1 − w(1 + xe )] = − du k s u n=0 n +1 k=0 (k + y) Γ(s) 0 e + x (7) When w =1andx →−x we get

∞ n k k ∞ s−1 −(y−1)u −u 1 n (−1) x 1 u e log [xe ] = − du (8) k s u − n=0 n +1 k=0 (k + y) Γ(s) 0 e x

∞ s −(y−1)u ∞ s−1 −(y−1)u 1 u e − log x u e = u du u du Γ(s) 0 e − x Γ(s) 0 e − x We see from (3) that

k k ∞ x x s−1 −(k+y)u s = u e du (k + y) Γ(s) 0 and we have the summation

∞ k ∞ ∞ x 1 s−1 −yu k −ku s = u e x e du k=0 (k + y) Γ(s) 0 k=0

1 ∞ us−1e−yu = −u du Γ(s) 0 1 − xe We therefore obtain the well-known formula [43, p. 121] for the Hurwitz- Lerch zeta function Φ(x, s, y)

∞ xk 1 ∞ us−1e−(y−1)u x, s, y du Φ( )= s = u − (9) k=0 (k + y) Γ(s) 0 e x and with y =1weobtain

∞ k ∞ k x 1 x Lis(x) Φ(x, s, 1) = s = s = k=0 (k +1) x k=1 k x Some inﬁnite series involving the Riemann zeta function 15 giving us [46, p. 280]

x ∞ us−1 Lis(x)= u du (10) Γ(s) 0 e − x Reference to (8) then shows that

∞ n k k 1 n (−1) x = s Φ(x, s +1,y) − log x Φ(x, s, y) (11) k s n=0 n +1 k=0 (k + y) With x = 1 we obtain Hasse’s formula (2)

∞ n k 1 n (−1) = s Φ(1,s+1,y)=sζ(s +1,y) k s n=0 n +1 k=0 (k + y) With y = 1 equation (11) becomes

∞ n k k 1 n (−1) x = s Φ(x, s +1, 1) − log x Φ(x, s, 1) k s n=0 n +1 k=0 (k +1) or equivalently ∞ n k k 1 n (−1) x s 1 = Lis+1(x) − log xLis(x) (12) k s n=0 n +1 k=0 (k +1) x x which corresponds with (1). A closed form expression may be obtained for example with x =1/2ands =2. We see from (6) that

∞ n k ∞ s−1 −yu n n x 1 u e t = du k s − − −u − n=0 k=0 (k + y) (1 t)Γ(s) 0 1 e tx/(1 t)

1 ∞ us−1e−(y−1)u = u du (1 − t)Γ(s) 0 e − tx/(1 − t) 16 Donal F. Connon and hence referring to (9) we have ∞ n k n n x 1 tx t = Φ ,s,y (13) k s − − n=0 k=0 (k + y) 1 t 1 t With t =1/2weobtain ∞ n k 1 n x =Φ(x, s, y) (14) n+1 k s n=0 2 k=0 (k + y) which is clearly an example of Euler’s transformation of series [33, p. 244]. tx Letting w = 1−t , (13) may be represented by ∞ n n k w n x w + x = Φ(w, s, y) (15) k s n=0 w + x k=0 (k + y) x and with x = −1 this becomes ∞ n n k −w n (−1) =(1− w)Φ (w, s, y) (16) − k s n=0 1 w k=0 (k + y) as previously reported by Guillera and Sondow [57]. Letting w =1andx = −1 in (15) results in another identity due to Hasse ∞ n k 1 n (−1) =Φ(−1,s,y)=ζa (s, y) (17) n+1 k s n=0 2 k=0 (k + y) where ζa (s, y) is the alternating Hurwitz-Lerch zeta function

∞ (−1)k ζa (s, y)= s k=0 (k + y) With y =1wehave ∞ n k 1 n (−1) = ζa (s, 1) = ζa (s) (18) n+1 k s n=0 2 k=0 (k +1) Some inﬁnite series involving the Riemann zeta function 17

Letting w = x = 1 in (15) results in ∞ n 1 n 1 =Φ(1,s,y)=ζ (s, y) (19) n+1 k s n=0 2 k=0 (k + y) Letting s =1andy =1/2 in (17) results in ∞ n k 1 n (−1) = ζa (1, 1/2) n k n=0 2 k=0 2k +1 We have from [15, p. 523] π ζa (1, 1/2) = 2 and we therefore obtain ∞ n k 1 n (−1) π = n k n=0 2 k=0 2k +1 2 which was also determined in a diﬀerent manner in equation (8.11e) in [22]. Since [15, p. 523]

ζa (1,y)+ζa (1, 1 − y)=π cosec(πy) we have for 0

∞ n+1 n k v −u ∞ s−2 −(y−1)u v n x 1 log [1 − w(1 + xe )] u e = − dw du 2 k s u n=0 (n +1) k=0 (k + y) Γ(s) 0 w 0 e + x

Since 18 Donal F. Connon

log(1 − ax) dx = −Li2(ax) x we have

∞ n+1 n k ∞ s−1 −(y−1)u −u v n x 1 u e Li2 [v(1 + xe )] = du 2 k s u n=0 (n +1) k=0 (k + y) Γ(s) 0 e + x (20) Further operations of the same kind will result in

∞ n+1 n k ∞ s−1 −(y−1)u −u v n x 1 u e Lir [v(1 + xe )] = du r k s u n=0 (n +1) k=0 (k + y) Γ(s) 0 e + x (21) With t = −1 in (6) we obtain ∞ n k ∞ s−1 −yu n n x 1 u e (−1) = du k s −u n=0 k=0 (k + y) Γ(s) 0 2+xe and letting x →−x we get ∞ n k k ∞ s−1 −yu n n (−1) x 1 u e (−1) = du k s − −u n=0 k=0 (k + y) Γ(s) 0 2 xe With x =2wehave

∞ n k k ∞ s−1 −yu ∞ s−1 −(y−1)u n n (−1) 2 1 u e 1 u e (−1) = du = du k s − −u u − n=0 k=0 (k + y) 2Γ(s) 0 1 e 2Γ(s) 0 e 1

Hence we obtain using (9) ∞ n k k n n (−1) 2 1 (−1) = Φ(1,s,y) (22) k s n=0 k=0 (k + y) 2 Some inﬁnite series involving the Riemann zeta function 19 and with s = n +1andy = 1 this becomes ∞ n k k n n (−1) 2 1 (−1) = ζ(n + 1) (23) k n+1 n=0 k=0 (k +1) 2 This identity, in the case where n is a positive integer, was recently reported by Alzer and Koumandos [6] where it was derived in a very diﬀerent manner. With x = −2wehave ∞ n k ∞ s−2 −yu n n 2 1 u e (−1) = du k s u n=0 k=0 (k + y) 2Γ(s) 0 e +1 1 = Φ(−1,s,y) 2 we see that with y =1 1 1 = Φ(−1,s,1) = ζa(s) 2 2 Hence we have ∞ n k n n 2 1 2−s (−1) = 1 − 2 ζ(s) (24) k s n=0 k=0 (k +1) 2 1 With t = x = 2 in (6) we obtain ∞ n ∞ s−2 −yu 1 n 1 2 u e = du n k k s u − n=0 2 k=0 2 (k + y) Γ(s) 0 e 1

= 2Φ(1,s,y)=2ζ(s, y) and therefore we have with y =1 ∞ n 1 n 1 ζ(s)= (25) n+1 k k s n=0 2 k=0 2 (k +1) With t = −1, x =1/2andy =1in(6)wehave 20 Donal F. Connon

∞ n ∞ s−1 n n 1 1 u (−1) = du = −2Lis(−1/4) k k s u 1 n=0 k=0 2 (k +1) 2Γ(s) 0 e + 4 (26)

2 Some series involving the central binomial numbers

From, for example, Knopp’s book [33, p. 271] we have the well-known Maclaurin expansion ∞ 2 −1 2 1 [n!] 2n − ≤ ≤ sin y = 2 (2y) , 1 y 1 (27) 2 n=1 n (2n)! and therefore upon letting x =sin−1 y we get

∞ 2 2 1 [n!] 2n 2n − ≤ ≤ x = 2 2 sin x, π/2 x π/2 (28) 2 n=1 n (2n)! which was known by Euler [34]. Using B¨urmann’s theorem, it is an exercise in Whittaker & Watson [46, p. 130] to prove

∞ 2 2 2 1 4 2.4 1 6 2n x =sin x +( ) sin x +( ) sin x + ... = An sin x 3 2 3.5 3 n=1 where (2n − 2)!! 1 An = . (2n − 1)!! n Using the deﬁnitions of the double factorials

(2n +1)! (2n)!! = 2.4...(2n)=2nn!, (2n +1)!!=1.3.5...(2n +1)= 2nn! Some inﬁnite series involving the Riemann zeta function 21 weseethisisthesameastheaboveidentity. We now multiply equation (28) by cot x and integrate to obtain

t ∞ 2 t 2 1 [n!] 2n 2n−1 x cot xdx= 2 2 sin x cos xdx 0 2 n=1 n (2n)! 0 which, for −π/2 ≤ x ≤ π/2, results in [12, p. 234] t ∞ 2n 2n −1 2 1 2 sin t 2n x cot xdx= (29) 3 n 0 4 n=1 n 2n (2n)! where = is the central binomial coeﬃcient. n [n!]2 This integral also appears in [12, p. 234]. Using integration by parts we have t t t 2 2 x cot xdx= x log sin x − 2 x log sin xdx 0 0 0 Since x2 log sin x = x2 log sin x + x2 log x we see that lim x2 log sin x =0and x x→0 thus

t t x2 cot xdx= t2 log sin t − 2 x log sin xdx 0 0 Hence we get for 0 ≤ t ≤ π/2 t ∞ 2n 2n −1 1 2 1 2 sin t 2n x log sin xdx= t log sin t − (30) 3 n 0 2 8 n=1 n According to Ayoub [8, p. 1084], Euler returned to the zeta function, for what appears to be the last time, in 1772 in a paper entitled “Exercitationes Analyticae” [26]. Notwithstanding that by this time Euler had been blind for six years, through what Ayoub describes as “a striking and elaborate scheme”, he was able to prove that 22 Donal F. Connon

π 2 2 7 π x log sin xdx= ζ(3) − log 2 (31) 0 16 8 A very elementary proof of (31) is given in equation (6.20a) in [22] where we used the basic identity

b ∞ b p(x)cotxdx=2 p(x)sin2nx dx (32) a n=1 a which, as shown in [22], is valid for a wide class of suitably behaved functions. Speciﬁcally we require that p(x) is a twice continuously diﬀerentiable function. It should be noted that in the above formula we require either (i) both sin(x/2) and cos(x/2) have no zero in [a, b] or (ii) if either sin(a/2) or cos(a/2) is equal to zero then p(a) must also be zero. Condition (i) is equivalent to the requirement that sin x has no zero in [a, b]. For example, letting p(x)=x2 in (32) we have for −π

t ∞ t x2 cot xdx=2 x2 sin 2nx dx 0 n=1 0 ∞ ∞ ∞ t 1 cos 2nx cos 2nx sin 2nx = − x2 + x 2 n3 n n2 n=1 n=1 n=1 0 This gives us

t ∞ ∞ ∞ 2 1 cos 2nt − 2 cos 2nt sin 2nt − 1 x cot xdx= 3 t + t 2 ζ(3) (33) 0 2 n=1 n n=1 n n=1 n 2 and with t = π/2weobtain

π ∞ n 2 ∞ n 2 − − 2 1 ( 1) − π ( 1) − 1 x cot xdx= 3 ζ(3) 0 2 n=1 n 4 n=1 n 2 Hence using the alternating zeta function Some inﬁnite series involving the Riemann zeta function 23

∞ − n+1 ( 1) − 1−s ζa(s)= s =(1 2 )ζ(s) n=1 n we see that

π 2 2 7 π x2 cot xdx= − ζ(3) + log 2 (34) 0 8 4 Another elementary evaluation of this integral has recently been provided by Fujii and Suzuki [28]. Using the substitutions y =sinx and z =tanx we may also note that

n t sin t sin−1 y xn cot xdx= dy 0 0 y

n t xn tan t (tan−1 z) 2 dx = dz 0 sin 2x 0 z We also see from (29) and (33) that for −π/2 ≤ t ≤ π/2

∞ 2n 2n −1 ∞ ∞ ∞ 1 2 sin t 2n 1 cos 2nt 2 cos 2nt sin 2nt 1 = −t +t − ζ(3) 3 n 3 2 4 n=1 n 2 n=1 n n=1 n n=1 n 2 (35) and with t = π/2 we immediately see that ∞ 2n −1 2 2 2 2 2n ζ(3) = π log 2 − (36) 3 n 7 7 n=1 n which is in agreement with Sherman’s compendium of formulae [42]. We have from [12, p. 198]

−1 22n 2n B(n, 1/2) = (37) n n where the beta function is deﬁned by 24 Donal F. Connon

1 u−1 B(u, v)= (1 − t) tv−1dt 0 This then gives us

−1 1 n 22n 2n (1 − t) = √ dt n n 0 (1 − t) t and we have the summation involving the dilogarithm function

∞ 2n −1 1 2 2n Li2 (1 − t) = √ dt 3 n − n=1 n 0 (1 t) t It may be noted that the Wolfram Mathematica Online Integrator evaluates this integral in terms of polylogarithms of order 2 and 3 and hence this will not provide us with any new information regarding the nature of ζ(3). It should be easier to obtain a simpler version of the output by using integration by parts, noting that

Li2 (1 − t) 2 √ dt =2 Li2 1 − u du t and

2 2 Li2 1 − u du = uLi2 1 − u −Li2 (1 − u)−Li2 (−u)−log u log(1+u)−2u log u−2u

We now attempt to use (37) in (29) t ∞ 2n 2n −1 2 1 2 sin t 2n x cot xdx= 3 n 0 4 n=1 n and this gives us

∞ 2n 2n −1 1 2 2 sin t 2n Li2 (1 − t)sin t = √ dt 3 n − n=1 n 0 (1 t) t Some inﬁnite series involving the Riemann zeta function 25

Following an idea used by Batir [9] we now substitute an integral for the dilogarithm function; we have

(−1)n−1 1 z logn−1 u 1 z log u Lin(z)= du, Li2(z)=− du (38) (n − 1)! 0 1 − zu 0 1 − zu and thus

1 2 1 1 2 Li2 (1 − t)sin t √ − √ sin t log u dt = 2 dt du 0 (1 − t) t 0 0 t 1 − u (1 − t)sin t

However, the Wolfram Mathematica Online Integrator cannot evaluate this integral. The integral (38) was also used by Batir [9] to derive (107) using his complex double integral

π ∞ 2n 4 k−1 arcsin(x sin y/4) x [n!] (−1) 2 1 = φ logk−3[4 sin φ/x sin y]dφdy k 2 − n=1 (2n) [(2n)!] (k 3)! 0 sin y 0 π 1 Since sin 6 = 2 we see from (29) that ∞ −1 π 1 2n 6 = −8 x log [2 sin x] dx (39) 3 n n=1 n 0 which I ﬁrst came across in van der Poorten’s 1979 paper “Some wonderful formulae...anintroductiontoPolylogarithms” [45]. In passing, we note from [12, p.122] that

∞ 2n −1 t −1 2 (2t) 2n sin x 2 3 2 =4 dx =2t 4F3 {1, 1, 1, 1}, , 2, 2 ,t 3 n n=1 n 0 x 2 in terms of the hypergeometric functions. 26 Donal F. Connon

We now multiply (28) by x cot x and integrate

t ∞ 2 t 3 1 [n!] 2n 2n−1 x cot xdx= 2 2 x sin x cos xdx 0 2 n=1 n (2n)! 0 Integration by parts yields

t t 1 t x sin2n−1 x cos xdx= sin2n t − sin2n xdx 0 2n 2n 0 and therefore we have

t ∞ 2n 2n −1 ∞ 2n −1 t 3 1 2 sin t 2n 1 2 2n 2n x cot xdx= t − sin xdx 3 n 3 n 0 4 n=1 n 4 n=1 n 0 (40) With t = π/2weobtain

π π ∞ 2n −1 ∞ 2n −1 2 3 π 2 2n 1 2 2n 2 2n x cot xdx= − sin xdx 3 n 3 n 0 8 n=1 n 4 n=1 n 0 and using [12, p. 195]

π 2 2n (2n)! π 1 2n π sin xdx= 2n 2 = 2n (41) 0 2 (n!) 2 2 n 2 we have

π ∞ 2n −1 ∞ 2 3 π 2 2n π 1 x cot xdx= − 3 n 3 0 8 n=1 n 8 n=1 n Using (36) we obtain

π 3 2 π 9π x3 cot xdx= log 2 − ζ(3) (42) 0 8 16 As another example, letting p(x)=x3 in (32) we have for −π

t ∞ t x3 cot xdx=2 x3 sin 2nx dx 0 n=1 0 ∞ ∞ ∞ ∞ t 3 sin 2nx 3 sin 2nx cos 2nx 3 cos 2nx = x2 − − x3 + x 2 n2 4 n4 n 2 n3 n=1 n=1 n=1 n=1 0

∞ ∞ ∞ ∞ 3 2 sin 2nt − 3 sin 2nt − 3 cos 2nt 3 cos 2nt = t 2 4 t + t 3 2 n=1 n 4 n=1 n n=1 n 2 n=1 n

Using (40) together with (90) we see that

t x3 cot xdx = 0 ∞ 2n 2n −1 ∞ 2n −1 n j 1 2 sin t 2n 1 2 2n 1 2n (−1) 2n t − t + sin 2jt 3 n 3 n 2n n n − j 4 n=1 n 4 n=1 n 2 j=1 j

∞ 2n 2n −1 ∞ ∞ −1 n j 1 2 sin t 2n t 1 1 1 2n (−1) 2n = t − − sin 2jt 3 n 3 3 n n − j 4 n=1 n 4 n=1 n 4 n=1 n j=1 j

This gives us

∞ 2n 2n −1 ∞ −1 n j 1 2 sin t 2n t 1 1 2n (−1) 2n t − ζ(3)− sin 2jt 3 n 3 n n − j 4 n=1 n 4 4 n=1 n j=1 j

∞ ∞ ∞ ∞ 3 2 sin 2nt − 3 sin 2nt − 3 cos 2nt 3 cos 2nt = t 2 4 t + t 3 2 n=1 n 4 n=1 n n=1 n 2 n=1 n and upon substituting (35) we have for −π/2 ≤ t ≤ π/2 28 Donal F. Connon

∞ −1 n j 1 2n (−1) 2n sin 2jt 3 n n − j n=1 n j=1 j ∞ ∞ ∞ sin 2nt − cos 2nt − 2 sin 2nt − =3 4 4t 3 2t 2 3tζ(3) n=1 n n=1 n n=1 n (43)

It was shown in equation (4.3.168a) of [19] that for 0 ≤ t< 1

t πx2 cot πx dx 0 ζ(3) = −[ζ(−2,t)+ζ(−2, 1−t)]+2t[ζ(−1,t)−ζ(−1, 1−t)]+t2 log(2 sin πt)− 2π2 (44) ∂ where ζ (s, t)= ∂sζ(s, t) is a derivative of the Hurwitz zeta function. Using (29) we see that for 0 ≤ t ≤ 1/2

t πt ∞ 2n 2n −1 2 1 2 1 2 sin πt 2n πx cot πx dx = u cot udu= 2 2 3 n 0 π 0 4π n=1 n (45) and hence we see from (44) and (45) that for 0 ≤ t ≤ 1/2 ∞ 2n 2n −1 1 2 sin πt 2n 2 3 n 4π n=1 n ζ(3) = −[ζ(−2,t)+ζ(−2, 1−t)]+2t[ζ(−1,t)−ζ(−1, 1−t)]+t2 log(2 sin πt)− 2π2 (46) I initially thought of letting t → 1 − t in (46) with the hope that Some inﬁnite series involving the Riemann zeta function 29

∞ 2n 2n −1 1 2 sin πt 2n = −[ζ (−2,t)+ζ (−2, 1−t)]−2(1−t)[ζ (−1,t)−ζ (−1, 1−t)] 2 3 n 4π n=1 n

ζ(3) +(1 − t)2 log(2 sin πt) − 2π2 and subtraction of these two equations appeared to indicate that for 1 ζ(−1,t) − ζ(−1, 1 − t)= (1 − 2t) log(2 sin πt) 2 We note from (55) below that 1 3 1 G ζ −1, − ζ −1, = Cl2(π/2) = 4 4 2π 2π and apparently this would lead to a closed-form expression for Catalan’s constant. Unfortunately, the analysis is incorrect because the restriction that 0 ≤ t ≤ 1/2 prevents us from letting t → 1 − t in (46). We may also express (44) in terms of the Barnes multiple gamma functions Γn(x) defined in [43, p. 24]. Adamchik [5] has shown that for (x) > 0 n n ζ (−n, x) − ζ (−n)=(−1) k!Qk,n(x)logΓk+1(x) (47) k=0 where Qk,n(x) are polynomials defined by n n−j n j Qk,n(x)= (1 − x) j k j=k j and are the Stirling subset numbers defined by k j n − 1 n − 1 n 1,n=0 = k + , = k k k − 1 0 0,n=0

Particular cases of (47) are 30 Donal F. Connon

ζ (−1,x) − ζ (−1) = x log Γ(x) − log G(x + 1) (48)

2 ζ (−2,x) − ζ (−2) = 2 log Γ3(x)+(3− 2x)logG(x) − (1 − x) log Γ(x) and letting x → 1 − x we see that

ζ (−1, 1 − x) − ζ (−1) = (1 − x) log Γ(1 − x) − log Γ(1 − x) − log G(1 − x)

= −x log Γ(1 − x) − log G(1 − x)

2 ζ (−2, 1 − x)−ζ (−2) = 2 log Γ3(1−x)+(1+2x)logG(1−x)−x log Γ(1−x)

Hence we have

G(1 − x) ζ (−1,x) − ζ (−1, 1 − x)=x log[Γ(x)Γ(1 − x)] + log (49) G(1 + x) which we have previously derived in [19]. 1 Letting t = 2 in (46) we obtain ∞ 2n −1 1 2 2n 1 1 ζ(3) = −2ζ −2, + log 2 − 2 3 n 2 4π n=1 n 2 4 2π and using (36) we obtain the known result 1 3ζ(3) ζ −2, = (50) 2 16π2 which is also derived, for example, in equation (4.3.168d) of [19]. 1 When t = 4 in (46) we get Some inﬁnite series involving the Riemann zeta function 31

∞ n −1 1 2 2n 2 3 n 4π n=1 n 1 3 1 1 3 1 ζ(3) = − ζ −2, + ζ −2, + ζ −1, − ζ −1, + log 2− 4 4 2 4 4 32 2π2 (51) We now refer to Adamchik’s result [2]

(2n +1)! ζ (−2n − 1,t) − ζ (−2n − 1, 1 − t)= Cl2n+2(2πt) (52) (2π)2n+1

n (2n)! ζ (−2n, t)+ζ (−2n, 1 − t)=(−1) Cl2n+1(2πt) (53) (2π)2n where Cln(t) is the Clausen function deﬁned by

∞ ∞ cos kt sin kt Cl2n+1(t)= 2n+1 Cl2n(t)= 2n (54) k=1 k k=1 k This is also derived in equation 4.3.167 of [19]. We then see that (as noted by Adamchik [2]) 1 3 1 G ζ −1, − ζ −1, = Cl2(π/2) = (55) 4 4 2π 2π where G is Catalan’s constant. Using the formula [18, equation 5.11] (where I have corrected a misprint)

1 − 22n Cl2n+1(π/2) = ζ(2n + 1) (56) 24n+1 this then gives us for t =1/4

2n 1 3 n (2n)! n 1 − 2 (2n)! ζ −2n, +ζ −2n, =(−1) Cl2n+1(π/2) = (−1) ζ(2n+1) 4 4 (2π)2n 24n+1 (2π)2n 32 Donal F. Connon and in particular we have for n =1 1 3 3ζ(3) ζ −2, + ζ −2, = 4 4 64π2 We therefore have ∞ n −1 2 2 2n 35ζ(3) π = − + log 2 + πG (57) 3 n n=1 n 16 8 in agreement with equation 3.167 of Sherman’s paper [42]. We note that this also concurs with equation (6.69o) of [22] and also [18] where, by entirely diﬀerent methods, it is shown that

π 2 4 35 π πG x2 cot xdx = − ζ(3) + log 2 + (58) 0 64 32 4 1 With t = 6 in (46) we obtain ∞ −1 1 1 2n 2 3 n 4π n=1 n 1 5 1 1 5 ζ(3) = − ζ −2, − ζ −2, + ζ −1, − ζ −1, − 6 6 3 6 6 2π2 (59) We see from (53) that 1 5 1 π ζ −2, + ζ −2, = − Cl3 6 6 2π2 3 and from Lewin’s monograph [39, p. 198] we have π 1 −2n −2n Cl2n+1 = (1 − 2 )(1 − 3 )ζ(2n +1) 3 2 which gives us π 1 Cl3 = ζ(3) 3 4 Some inﬁnite series involving the Riemann zeta function 33

This results in 1 5 ζ(3) ζ −2, + ζ −2, = − (60) 6 6 8π2 We also have from (52) 1 5 1 π ζ −1, − ζ −1, = Cl2 6 6 2π 3 and from (59) we obtain

∞ −1 1 1 2n 1 5 1 π ζ(3) = − ζ −2, − ζ −2, + Cl2 − 2 3 n 2 4π n=1 n 6 6 6π 3 2π (61) − 1 Accordingly we have two simultaneous equations involving ζ 2, 6 and − 5 ζ 2, 6 (unfortunately (2.27) contains two other unknown constants; in this regard see (67)). We have the well-known Hurwitz’s formula for the Fourier expansion of the Riemann zeta function ζ(s, t) as reported in Titchmarsh’s treatise [43b, p. 37]

∞ ∞ − πs cos 2nπt πs sin 2nπt ζ(s, t)=2Γ(1 s) sin 1−s +cos 1−s (62) 2 n=1 (2πn) 2 n=1 (2πn) where (s) < 0and0functional equation for ζ(s). Letting s → 1 − s we may write this as

∞ ∞ − πs cos 2nπt πs sin 2nπt ζ(1 s, t)=2Γ(s) cos s +sin s (63) 2 n=1 (2πn) 2 n=1 (2πn)

Using (63) it is easily shown that 34 Donal F. Connon

∞ − − − − − log n cos 2nπt 1 2ζ ( 1,t) B2(t)(1 γ log(2π)) = 4 2 + Cl2(2πt) n=1 (2πn) 2π (64)

∞ − − 1 − 1 − 1 − log n sin 2nπt− 1 ζ ( 2,t) B3(t) γ log(2π) = 4 3 2 Cl3(2πt) 2 3 2 n=1 (2πn) (2π) (65) With t =1/6 in (65) we obtain

∞ − 1 − 5 1 − 1 − 1 − log n sin(nπ/3) − ζ(3) ζ 2, γ log(2π) = 4 3 2 6 216 2 3 2 n=1 (2πn) 16π and this indicates the complexities involved in determining a closed form expression for 1 ζ −2, . 6 Diﬀerentiating (73) gives us ∞ √ −s − −s − log n sin(nπ/3) 3 1 − 3 log 3 s = 3 ζ (s) ζ(s) n=1 n 2 2

√ 1 1 1 1 + 3 6−s ζ s, + ζ s, − 6−s log 6 ζ s, + ζ s, 6 3 6 3 so that with s =3wehave ∞ √ −3 − −3 − log n sin(nπ/3) 3 1 − 3 log 3 3 = 3 ζ (3) ζ(3) n=1 n 2 2 Some inﬁnite series involving the Riemann zeta function 35

√ 1 1 1 1 + 3 6−3 ζ 3, + ζ 3, − 6−3 log 6 ζ 3, + ζ 3, 6 3 6 3

With t =1/4 in (64) we obtain

∞ − 1 1 − − − log n cos(nπ/2) 1 π 2ζ 1, + [1 γ log(2π)] = 4 2 + Cl2 4 48 n=1 (2πn) 2π 2

∞ − 1 1 − − − − n log 2n G 2ζ 1, + [1 γ log(2π)] = 4 ( 1) 2 + 4 48 n=1 (4πn) 2π

∞ − n+1 ∞ log 2 ( 1) 1 − n+1 log n G = 2 2 + 2 ( 1) 2 + 4π n=1 n 4π n=1 n 2π

log 2 1 G = ζa(2) − ζ (2) + 4π2 4π2 a 2π 1 1 1 Since ζa(2) = 2 ζ(2) and ζa(2) = 2 ζ(2) log 2 + 2 ζ (2) we have 1 1 1 G 2ζ −1, + [1 − γ − log(2π)] = − ζ(2) + 4 48 8π2 2π It is easily found from the functional equation for the Riemann zeta function that

1 1 ζ (−1) = (1 − γ − log(2π)) + ζ(2) 12 2π2 and hence we have (as originally determined by Adamchik [2]) 1 G 1 ζ −1, = − ζ (−1) (66) 4 4π 8 Ghusayni [29] showed in 1998 that 36 Donal F. Connon

∞ ∞ −1 π sin (nπ/3) 3 1 2n ζ(3) = − (67) 2 3 n 2 n=1 n 4 n=1 n Later in 2000, Ghusayni [30], having noted an earlier paper [31], reported that √ ∞ sin(nπ/3) 3 1 1 − 1 − 1 1 1 − 2 = 2 + 2 2 2 + 2 + 2 ... (68) n=1 n 2 1 2 4 5 7 8 √ 3 1 1 2 = ψ(1) − π2 2 3 3 9 and Ghusayni concluded that √ √ ∞ ∞ −1 3 3 3 3 1 3 1 2n ζ(3) = − π + π − (69) − 2 3 n 18 4 n=1 (3n 2) 4 n=1 n It is shown in [30] that

1 1 1 − 1 − 1 1 1 − − 2 2 − log x 2 + 2 2 2 + 2 + 2 ... = π 2 3 dx 1 2 4 5 7 8 27 0 1+x Mathematica evaluates this integral as

1 log x − 2 − 1 5 324 3 dx = 8π 6PolyGamma 1, + 3PolyGamma 1, 0 1+x 6 6

1 2 −3PolyGamma 1, + 6PolyGamma 1, 3 3

The Mathworld website for the central binomial coefficient reports the following formula which was experimentally obtained by Plouffe [41] (and in fact determined analytically in 1985 by Zucker [49]) Some infinite series involving the Riemann zeta function 37

∞ −1 √ 1 2n 1 (1) 1 (1) 2 4 = π 3 ψ − ψ − ζ(3) (70) 3 n n=1 n 18 3 3 3

We have the reﬂection formula [43, p. 14]

ψ(1 − x) − ψ(x)=π cot πx and diﬀerentiation gives us

dk ψ(k)(1 − x)+(−1)(k+1)ψ(k)(x)=(−1)kπ cot πx dxk and therefore we see that 1 2 π 4 ψ(1) + ψ(1) = π2/ sin2 = π2 3 3 3 3 We then obtain ∞ −1 √ √ 1 2n 1 (1) 1 4 2 3 = π 3 ψ − ζ(3) − π 3 (71) 3 n n=1 n 9 3 3 27 We have p p ψ(k)( )=(−1)k+1k!ζ(k +1, ) q q and therefore we see that ∞ 1 1 1 ψ ( )=ζ(2, )=9 2 3 3 n=0 (3n +1) or, equivalently, changing the order of summation ∞ 1 1 ψ ( )=9 − 2 3 n=1 (3n 2) Hence we see that (70) is equivalent to Ghusayni’s result (69) 38 Donal F. Connon

√ √ ∞ −1 3 3 3 (1) 1 3 1 2n ζ(3) = − π + πψ − 3 n 18 12 3 4 n=1 n We now refer back to (35) which is valid for −1/2 ≤ t ≤ 1/2

∞ 2n 2n −1 ∞ ∞ ∞ 1 2 sin πt 2n 1 cos 2nπt 2 2 cos 2nπt sin 2nπt 1 = −π t +πt − ζ(3) 3 n 3 2 4 n=1 n 2 n=1 n n=1 n n=1 n 2 so that with t =1/6wehave

∞ −1 ∞ ∞ ∞ 1 1 2n 1 cos(nπ/3) 1 2 cos(nπ/3) 1 sin(nπ/3) 1 = − π + π − ζ(3) 3 n 3 2 4 n=1 n 2 n=1 n 36 n=1 n 6 n=1 n 2

Lewin [39] and Srivastava and Tsumura [43, p. 293] reported for Re(s)> 1 ∞ cos(nπ/3) 1 1−s − 1−s − 1−s s = (6 3 2 +1)ζ(s) (72) n=1 n 2

∞ √ −s sin(nπ/3) 3 − 1 −s 1 1 s = 3 ζ(s)+6 ζ s, + ζ s, (73) n=1 n 2 6 3

Hencewehavewiths =3ands = 2 respectively

∞ cos(nπ/3) 1 3 = ζ(3) (74) n=1 n 3

∞ √ −2 sin(nπ/3) 3 − 1 −2 1 1 2 = 3 ζ(2) + 6 ζ 2, + ζ 2, (75) n=1 n 2 6 3

We have the Fourier series [44, p. 148] for 0

∞ cos 2nπt = − log(2 sin πt) n=1 n Using (75) we then obtain Ghusayni’s result

√ −2 ∞ −1 π 3 − 1 −2 1 1 3 1 2n ζ(3) = 3 ζ(2) + 6 ζ 2, + ζ 2, − 3 n 2 2 6 3 4 n=1 n

Srivastava and Tsumura [43, p. 293] have also reported for Re(s)> 1

∞ cos(2nπ/3) 1 1−s − s = (3 1)ζ(s) (76) n=1 n 2 ∞ √ −s sin(2nπ/3) 3 − 1 −s 1 s = 3 ζ(s)+3 ζ s, (77) n=1 n 2 3 ∞ cos(nπ/2) −s 1−s − s =2 (2 1)ζ(s) (78) n=1 n ∞ sin(nπ/2) −s − 1−2s 1 s =(2 1)ζ(s)+2 ζ s, (79) n=1 n 4 and the relevant values may be easily inserted in (35). Reference should also be made to the paper by Koyama and Kurokawa [34]. More generally we have from (46), (52) and (53)

∞ 2n 2n −1 2 sin πt 2n 2 2 =4π t log(2 sin πt)+2Cl3(2πt)+4πt Cl2(2πt)−2ζ(3) 3 n n=1 n (80) or equivalently 40 Donal F. Connon

∞ 2n 2n −1 2 sin u 2n 2 =4u log(2 sin u) + 2Cl3(2u)+4u Cl2(2u) − 2ζ(3) 3 n n=1 n (81) Having recently seen a paper by Bradley et al. [14], I noted that this is equivalent to the identity previously discovered by Zucker [49] in 1985 (and I thank John Zucker for subsequently sending me a reprint of his original paper). With u = π/6andu = π/4 we obtain respectively ∞ −1 1 2n 2π = 2Cl3(π/3) + Cl2(π/3) − 2ζ(3) (82) 3 n n=1 n 3 ∞ n −1 2 2 2n π = log 2 + 2Cl3(π/2) + πCl2(π/2) − 2ζ(3) (83) 3 n n=1 n 8

Then using (56) we see that 3 Cl3(π/2) = − ζ(3) 32 and hence from (83) we obtain another derivation of (57) ∞ n −1 2 2 2n 35ζ(3) π = − + log 2 + πG 3 n n=1 n 16 8 pπ We also have a number of formulae involving Cl2 q in Browkin’s paper in Lewin’s survey [59, p. 244], including for example π 5π 4 4 π Cl2 +Cl2 = G= Cl2 6 6 3 3 2 It is easily seen from the deﬁnition of the Clausen function (54) that

Cl2n(π)=Cl2n(2π)=0 Some inﬁnite series involving the Riemann zeta function 41

−2n Cl2n+1(π)=(2 − 1)ζ(2n +1)=−ζa(2n +1)

Cl2n+1(2π)=ζ(2n +1) We also have

Cl2(π/2) = G = −Cl2(3π/2)

1 Cl2(2x)=Cl2(x) − Cl2(π − x) 2 which implies that 2π 2 π Cl2 = Cl2 3 3 3 The Clausen function may be expressed in closed form in at least three other cases and from Lewin’s book [39, p. 198] we have π −2n−1 −2n Cl2n+1 = −2 (1 − 2 )ζ(2n +1) 2 π 1 −2n −2n Cl2n+1 = (1 − 2 )(1 − 3 )ζ(2n +1) 3 2 2π 1 −2n Cl2n+1 = − (1 − 3 )ζ(2n +1) 3 2 For example, we see from the deﬁnition that π 1 1 1 Cl2n+1 = − + − + ... 2 22n+1 42n+1 62n+1 1 1 1 1 = − − + ... = − ζa(2n +1) 22n+1 12n+1 22n+1 22n+1

We then see Cl3(π/3) = ζ(3)/3 and from (82) that 42 Donal F. Connon

∞ −1 1 2n 2π 4 = Cl2(π/3) − ζ(3) 3 n n=1 n 3 3 and comparing this with (71) we obtain √ (1) 1 2 2 2 3Cl2(π/3) = ψ − π (84) 3 3 as previously derived by Fettis [55]. More generally, we have [39a, p. 358]

2 (q−1)/2 (1) p π 2 πp 2mπp 2mπp ψ = cosec +2q sin Cl2 (85) q 2 q m=1 q q and a particular case of this is (84). Another example is √ √ 1 (1) 1 2 Cl2(π/6) = 3 ψ +16G − 2π / 3 (86) 24 3 Using PSLQ, Bailey et al. [50] discovered experimentally that

2π 4π 6π 6Cl2 (α) − 6Cl2 (2α) + 2Cl2 (3α) = 7Cl2 + 7Cl2 − 7Cl2 7 7 7 √ (87) where α =2tan−1 7. It appears that there must be a connection with (85) in the case where q =7. See also Coﬀey’s recent paper [54]. Integrating (80) results in ∞ 2n −1 x 2 2n 2n sin udu (88) 3 n n=1 n 0

x 4 3 2 = x log 2 + 4 u log sin udu+ 2Cl4(2x) − 2x Cl3(2x) − 2xζ(3) 3 0 Some inﬁnite series involving the Riemann zeta function 43

x sin 2kx − x cos 2kx In the above we used the fact that 0 u sin 2ku du = 4k2 2k and this results in

x 1 1 u Cl2(2u) du = Cl4(2x) − x Cl3(2x). 0 4 2 When x = π/2weget

π ∞ 2n −1 ∞ 2n −1 ∞ 2 2n 2 2n 2 2n π 1 2n π 1 π sin udu= = = ζ(3) 3 n 3 n 2n n 3 n=1 n 0 n=1 n 2 2 2 n=1 n 2

3 π π 2 2 = log 2 + 4 u log sin udu− π Cl3(π) − πζ(3) 6 0 We see that

∞ − k ( 1) − −3 Cl3(π)= 3 = ζa(3) = ζ(3) k=1 k 4 which gives us

3 π π π 2 π ζ(3) = log 2 + 4 u2 log sin udu− ζ(3) 2 6 0 4 and we therefore obtain a new derivation of Euler’s integral

π 3 2 π 3π u2 log sin udu= − log 2 + ζ(3) (89) 0 24 16 Wiener [47] has shown that

x n j 2n 1 2n (−1) 2n sin udu= x + sin 2jx (90) 2n n n − j 0 2 j=1 j and we therefore obtain from (88) 44 Donal F. Connon

∞ −1 n j 1 2n (−1) 2n sin 2jx (91) 3 n n − j n=1 n j=1 j

x 4 3 2 = x log 2 + 4 u log sin udu+ 2Cl4(2x) − 2x Cl3(2x) − 3xζ(3) 3 0

Using integration by parts we have x x x 3 3 2 u cot udu= u log sin u − 3 u log sin udu 0 0 0 x = x3 log sin x − 3 u2 log sin udu 0 Using (32) we have

x ∞ x u3 cot udu=2 u3 sin 2nu du 0 n=1 0 ∞ ∞ ∞ ∞ x 3 sin 2nu 3 sin 2nu 1 cos 2nu 3 cos 2nu = u2 − − u3 + u 4 n2 8 n4 2 n 4 n3 n=1 n=1 n=1 n=1 0

∞ ∞ ∞ ∞ 3 2 sin 2nx − 3 sin 2nx − 1 3 cos 2nx 3 cos 2nx = x 2 4 x + x 3 4 n=1 n 8 n=1 n 2 n=1 n 4 n=1 n

3 2 3 1 3 3 = x Cl2(2x) − Cl4(2x) − x Cl1(2x)+ x Cl3(2x) 4 8 2 4 Therefore we may write (91) as ∞ −1 n j 1 2n (−1) 2n sin 2jx (92) 3 n n − j n=1 n j=1 j Some inﬁnite series involving the Riemann zeta function 45

1 3 4 3 − − = 3 x log sin x + 3 x log 2 + 2Cl4(2x) 2x Cl3(2x) 3xζ(3)

− 1 2 1 1 3 − 1 4 x Cl2(2x)+ 8 Cl4(2x)+ 3 x Cl1(2x) 4 x Cl3(2x)

1 3 4 3 17 9 1 2 1 3 = x log sin x+ x log 2+ Cl4(2x)− x Cl3(2x)− x Cl2(2x)+ x Cl1(2x)−3xζ(3) 3 3 8 4 4 3 Diﬀerentiating (28) gives us ∞ 2n 2n−1 −1 1 2 sin x cos x 2n x = (93) n 2 n=1 n and with x → πx this becomes ∞ 2n 2n−1 −1 1 2 sin πx cos πx 2n πx = (94) n 2 n=1 n We now multiply this across by cot πx and integrate to obtain

t ∞ 2n −1 t 1 2 2n 2n−2 2 πx cot πx dx = sin πx cos πx dx n 0 2 n=1 n 0 ∞ 2n −1 t 1 2 2n 2n−2 2n = sin πx − sin πx dx n 2 n=1 n 0 The Wolfram Mathematica Online Integrator evaluates the above integral in 3 3 − 5 2 terms of the hypergeometric function 3F2 2 , 2 , n; 2 ;cos πt . Alternatively, using integration by parts, we get

t t sin2n−2 πx cos2 πx dx = sin2n−2 πx cos πx cos πx dx 0 0 2n−1 t t sin πx 1 2n =cosπx + sin πx dx (2n − 1)π 0 (2n − 1) 0 46 Donal F. Connon and with t =1/2wehave

1 1 2 1 2 sin2n−2 πx cos2 πx dx = sin2n πx dx. 0 2n − 1 0 Since

1 π 2 1 2 sin2n πx dx = sin2n xdx 0 π 0 we have

π 2 2n (2n − 1)!! π 1 2n π sin xdx= = 2n 0 (2n)!! 2 2 n 2 Therefore we see that

1 2 2n−2 2 1 1 2n sin πx cos πx dx = 2n+1 0 2n − 1 2 n Hence we obtain

1 ∞ 2 1 1 πx cot πx dx = − 0 4 n=1 n(2n 1) Clearly

∞ ∞ 1 1 1 = − − − 1 n=1 n(2n 1) n=1 n 2 n Since the digamma function may be represented by [43, p. 14]

∞ 1 1 1 ψ(a)=−γ − − − a n=1 n + a n we see that

∞ 1 1 1 − = − ψ − − γ +2 − 1 n=1 n 2 n 2 Some inﬁnite series involving the Riemann zeta function 47

We know that [43, p. 22] 1 ψ − =2− γ − 2log2 2 and we therefore deduce that

∞ 1 1 − =2log2 − 1 n=1 n 2 n Hence we have the well-known integral

1 2 log2 x cot πx dx = 0 2π We now divide (93) by sin x and integrate to obtain t ∞ 2n 2n−1 −1 x 1 2 sin t 2n dx = − n 0 sin x 2 n=1 n(2n 1) Alternatively, with n → n + 1 in the summation this may be written as t ∞ 2n 2n+1 −1 x 2 sin t 2n dx = (95) 2 n 0 sin x n=0 (2n +1) We recall from [22] that

∞ b p(x) b dx =2 p(x)sin(2n +1)xdx (96) a sin x n=0 a where p(x) is a twice continuously diﬀerentiable function. It should be noted that in the above formula we require either (i) both sin(x/2) and cos(x/2) have no zero in [a, b] or (ii) if either sin(a/2) or cos(a/2) is equal to zero then p(a) must also be zero. Condition (i) is equivalent to the requirement that sin x has no zero in [a, b]. Letting p(x)=x in (96) we get 48 Donal F. Connon

t ∞ ∞ x − cos(2n +1)t sin(2n +1)t dx = 2t +2 2 (97) 0 sin x n=0 2n +1 n=0 (2n +1) and hence we obtain

∞ 2n 2n+1 −1 ∞ ∞ 2 sin t 2n cos(2n +1)t sin(2n +1)t = −t + (98) 2 n 2 n=0 (2n +1) n=0 2n +1 n=0 (2n +1)

In particular, we have from (97)

π ∞ n 2 x (−1) dx =2 2 =2G (99) 0 sin x n=0 (2n +1) and we then determine that ∞ 2n −1 1 2 2n G = (100) 2 n 2 n=0 (2n +1) in agreement with Sherman’s formula [42]. Using integration by parts Bradley [53] showed that

z z u sec2 u log tan udu− z log tan z = − du 0 0 tan u

2z −1 udu = 2 4 0 tan(u/2) cos (u/2)

1 2z u = − du 2 0 sin u

sin(2z) −1 − 2√x sin x dx = 2 2 0 1 − x x Diﬀerentiating (27) gives us Some inﬁnite series involving the Riemann zeta function 49

−1 −1 ∞ (2x)2n 2n 2x√sin x = for |x| < 1 and integrating term by term 1−x2 n=1 n n results in

z ∞ 2n+1 −1 1 (2 sin 2z) 2n log tan udu− z log tan z = − 2 n 0 4 n=0 (2n +1) for 0 ≤ z ≤ π/4. Therefore we obtain (95) again z ∞ 2n 2n+1 −1 u 2 sin z 2n du = 2 n 0 sin u n=0 (2n +1) We note from (102) below that

∞ t x t sin(2n +1)t dx = t log tan +2 2 0 sin x 2 n=0 (2n +1) and hence we have

∞ ∞ 2n 2n+1 −1 sin 2(2n +1)t 2 sin t 2n t log tan t + = (101) 2 2 n n=0 (2n +1) n=0 (2n +1)

Reference to [44, p. 149] shows that for 0

∞ cos(2n +1)t 1 t = − log tan n=0 2n +1 2 2 and for 0 ≤ t ≤ π

∞ t sin(2n +1)t −1 x 2 = log tan n=0 (2n +1) 2 0 2 Referring to (97) we have 50 Donal F. Connon

∞ t x t sin(2n +1)t dx = t log tan +2 2 (102) 0 sin x 2 n=0 (2n +1) and noting the Clausen function

t ∞ − u sin nt Cl2(t)= log[2 sin( )]du = 2 0 2 n=1 n we see that

∞ − ∞ − sin n(π t) − n sin nt Cl2(π t)= 2 = ( 1) 2 n=1 n n=1 n ∞ − sin(2n +1)t Cl2(t)+Cl2(π t)=2 2 n=0 (2n +1) Hence, as shown by Lewin [39, p. 255] we have t x t dx = t log tan +Cl2(t)+Cl2(π − t) (103) 0 sin x 2 More generally, we have

∞ ∞ ∞ 2n 2n+1 −1 sin(2n +1)t cos(2n +1)t 2 sin t 2n − t = (104) 2 2 n n=0 (2n +1) n=0 2n +1 n=0 (2n +1) t = t log tan +Cl2(t)+Cl2(π − t) 2 See also the recent paper by Cho et al. [16] who consider the integrals t xm 0 sin x dx and, in particular,

π √ √ 3 x π 3 3 1 dx = − log 3 − π2 + ζ 2, 0 sin x 6 9 18 6 Some inﬁnite series involving the Riemann zeta function 51

Such integrals may be easily evaluated using the method set out in [22] (and we have the added advantage that the upper end of the interval of integration does not need to be restricted to the form π/p where p is an integer). We have

t t t sin2n+1 xdx= sin2n x sin xdx= (1 − cos2 x)n sin xdx 0 0 0 t n n k 2k = (−1) cos x sin xdx k 0 k=0 and we therefore obtain

t n k n k 2k+1 2n+1 n (−1) n (−1) cos t sin xdx= − k k 0 k=0 2k +1 k=0 2k +1

With t = π/2weget

π n k 2 2n+1 n (−1) sin xdx= k 0 k=0 2k +1 as compared with the frequently quoted form of the Wallis integral formula

π 2 (2n)!! sin2n+1 xdx= 0 (2n +1)!! We therefore have the following identity which is reported in [33, p. 270] n n 2 n k 1 (2n)!! [2 n!] (−1) = = (105) k k=0 2k +1 (2n +1)!! (2n +1)! so that

π n 2 2 [2 n!] sin2n+1 xdx= (106) 0 (2n +1)! 52 Donal F. Connon

We recall (5) n k ∞ n x 1 s−1 −yu −u n = u e 1+xe du k s k=0 (k + y) Γ(s) 0 and note that

n n ∞ n k 1 1 n k 1 1 −u/2 −u n (−1) = (−1) = e 1 − xe du k k 1 k=0 2k +1 2 k=0 k + 2 2 0

Integrating (104) gives us ∞ ∞ ∞ 1 − cos(2n +1)t − sin(2n +1)t 2 3 2 3 t 2 n=0 (2n +1) n=0 (2n +1) n=0 (2n +1) ∞ 2n −1 t 2 2n 2n+1 = sin xdx 2 n n=0 (2n +1) 0 and with t = π/2wehave

π ∞ n+1 ∞ 2n −1 7 π (−1) 2 2n 2 2n+1 ζ(3) + = sin xdx 2 2 n 4 2 n=0 (2n +1) n=0 (2n +1) 0 ∞ 2n −1 n 2 2 2n [2 n!] = 2 n n=0 (2n +1) (2n +1)! Hence we obtain

∞ 4n 4 7 − 1 2 [n!] ζ(3) πG = 3 2 (107) 4 2 n=0 (2n +1) [(2n)!] which was also previously obtained by Batir [9]. We now multiply (28) by sin x and integrate to obtain Some inﬁnite series involving the Riemann zeta function 53

π ∞ 2 π 2 2 1 [n!] 2n 2 2n+1 x sin xdx= 2 2 sin xdx 0 2 n=1 n (2n)! 0 and therefore we see that

∞ 4n 4 − 1 2 [n!] π 2= 2 2 (108) 2 n=1 n (2n +1)[(2n)!] Let us now divide (28) by sin x and integrate to obtain

π π 2 ∞ 2 2 x 1 [n!] 2n 2 2n−1 dx = 2 2 sin xdx 0 sin x 2 n=1 n (2n)! 0 In equation 6.29 in [22] we have previously shown that

π 2 2 x 7 dx =2πG− ζ(3) 0 sin x 2 (the evaluation of this integral is also contained in [16]) and using the integral in [12, p. 113] Using

π n−1 k 2 2n−1 n − 1 (−1) (2n − 2)!! sin xdx = = k − 0 k=0 2k +1 (2n 1)!! we obtain

∞ 4n 4 − 2 [n!] 8πG 14ζ(3) = 3 2 (109) n=1 n [(2n)!] which was previously derived by Batir [9]. Letting p(x)=x cos x in (96) gives us

∞ t x cos x t t dx = x cot xdx=2 x cos x sin(2n +1)xdx 0 sin x 0 n=0 0 54 Donal F. Connon and we have using integration by parts for n ≥ 1

t sin(2nt)− cos(2nt) sin[2(n +1)t]− cos[2(n +1)t] 8 x cos x sin(2n+1)xdx= 2 2t + 2 2t 0 n n (n +1) n +1

This gives us π n n 2 π (−1) (−1) 2 x cos x sin(2n +1)xdx= − 0 4 n +1 n and thus

∞ π 2 π 2 x cos x sin(2n +1)xdx= log 2 n=0 0 2 This, of course, concurs with the well-known Euler integral

π 2 π x cot xdx= log 2 0 2 Dividing (28) by sin2 x and integrating gives us

t 2 ∞ 2 t x 1 [n!] 2n 2n−2 2 dx = 2 2 sin xdx 0 sin x 2 n=1 n (2n)! 0 and with t = π/2 we have after dealing separately with the case for n =1

π 2 ∞ 2 2 x π 1 [n!] 2n π 2n − 2 dx = + 2 2 2 2n−1 n − 1 0 sin x 2 2 n=2 n (2n)! 2

∞ π 1 = + π − − 2 n=2 (2n 1)(2n 2) ∞ π 1 − 1 = + π − − 2 n=2 2n 1 2n 2 Some inﬁnite series involving the Riemann zeta function 55

∞ π 1 1 = + π − 2 n=1 2n +1 2n ∞ π π 1 1 = + 1 − 2 2 n=1 n + 2 n Since the digamma function may be represented by [43, p. 14]

∞ 1 1 1 ψ(a)=−γ − − − a n=1 n + a n we see that

∞ 1 1 1 1 − = − ψ − γ − 2 n=1 n + 2 n 2 We know that [43, p. 20] 1 ψ = −γ − 2log2 2 and we therefore deduce that

π 2 2 x 2 dx = π log 2 0 sin x Using integration by parts we see that

π 2 π 2 x 2 2 dx =2 x cot xdx 0 sin x 0 and thus we have

π π 2 2 x cot xdx= − log sin xdx 0 0 Euler [25] was the ﬁrst person to show that 56 Donal F. Connon

π 2 π log sin xdx= − log 2 0 2 and we have therefore come full circle! It may also be noted that Ramanujan [11, Part I, p. 261] showed that for |x|< 2π

x n 1 n u nπ − − j(j+1)/2 Γ(n +1) n−j u cot du =cos n!ζ(n+1) ( 1) − x Clj+1(x) 2 0 2 2 j=0 Γ(n +1 j) (110) and also see the paper by Srivastava et al [61]. Ramanujan’s formula was employed by Muzaﬀar [40]. With x = π we have

π n n 2 n nπ − − j(j+1)/2 Γ(n +1) n−j 2 x cot xdx=cos n!ζ(n+1) ( 1) − π Clj+1(π) 0 2 j=0 Γ(n +1 j) (111) Provided a = b we readily determine that 1 sin(a − b)x sin(a + b)x sin ax sin bx dx = − + c 2 a − b a + b and hence we have

t 1 sin(p − 2n − 1)t sin(p +2n +1)t sin px sin(2n +1)xdx= − 0 2 p − 2n − 1 p +2n +1 and more speciﬁcally

π n+1 2 (−1) cos(pπ/2) 1 1 sin px sin(2n +1)xdx= + 0 2 p − 2n − 1 p +2n +1

Therefore with p(x) = sin px in (96) we obtain Some inﬁnite series involving the Riemann zeta function 57

π ∞ π 2 sin px 2 dx =2 sin px sin(2n +1)xdx 0 sin x n=0 0 so that

π ∞ n 2 sin px (−1) dx p pπ/ =2 cos( 2) 2 − 2 (112) 0 sin x n=0 (2n +1) p and, using L’H¨opital’s rule, we see that in the limit as p → 1

π 2 −pπ sin(pπ/2) + 2 cos(pπ/2) π dx = lim = 0 p→1 −2p 2 With p =2weobtain

∞ (−1)n+1 1 2 − 2 = (113) n=0 (2n +1) 4 2 We may write (112) as

π ∞ n (−1) 1 2 sin px = dx 2 − 2 n=0 (2n +1) p 2p cos(pπ/2) 0 sin x and, using L’H¨oital’s rule, we obtain as p → 0

π ∞ n (−1) 1 2 x cos px = lim dx 2 p→0 − n=0 (2n +1) pπ sin(pπ/2) + 2 cos(pπ/2) 0 sin x

Hence we obtain (99) again

π ∞ n (−1) 2 x 2 2 = dx n=0 (2n +1) 0 sin x We note from Fettis [56] that 58 Donal F. Connon

π 2 sin px π 1 pπ 3+p 1+p dx = − cos ψ − ψ (114) 0 sin x 2 2 2 4 4 and

π 2 1 − cos px 1+p 1 1 pπ 3+p 1+p dx = ψ −ψ − sin ψ − ψ 0 sin x 2 2 2 2 4 4 (115) Comparing (112) with (114) gives us

∞ (−1)n π 1 pπ 3+p 1+p p pπ/ − ψ − ψ 2 cos( 2) 2 − 2 = cos n=0 (2n +1) p 2 2 2 4 4 (116) Diﬀerentiating (114) results in

π 2 x cos px dx 0 sin x 1 pπ 3+p 1+p π pπ 3+p 1+p = − cos ψ − ψ + sin ψ − ψ 8 2 4 4 4 2 4 4 and with p =0weobtain

π 2 x 1 3 1 dx = − ψ − ψ 0 sin x 8 4 4 Using (99) we see that 1 3 1 2G = − ψ − ψ (117) 8 4 4 which concurs with K¨olbig [58]. With p = 1 we have the well known integral Some inﬁnite series involving the Riemann zeta function 59

π 2 π 1 π x cot xdx= ψ (1) − ψ = log 2 0 4 2 2 With p =2weobtain

π 2 x 1 1 3 − 2x sin x dx = ψ 1+ − ψ sin x 8 4 4 0 1 1 − and since ψ 1+ 4 = ψ 4 16 we obtain (117) again. Diﬀerentiating (115) results in

π 2 x sin px 1 1+p 1 pπ 3+p 1+p dx = ψ − sin ψ − ψ 0 sin x 2 2 8 2 4 4 π pπ 3+p 1+p − cos ψ − ψ 4 2 4 4 and with p =0weobtain 1 1 π 3 1 ψ = ψ − ψ 2 2 4 4 4 so that 1 π2 ψ = (118) 2 2 in accordance with K¨olbig’s paper [58]. With p =1weobtain π2 1 1 1 = ψ (1) − ψ (1) − ψ 8 2 8 2 and, since ψ (1) = ζ(2), this concurs with (118). We have provided p =2 n +1 t − − 1 cos(p 2n 1)t − cos(p +2n +1)t − 2n +1 cos px sin(2n+1)xdx= 2 2 0 2 p − 2n − 1 p +2n +1 p − (2n +1) 60 Donal F. Connon and thus

π n+1 2 − ( 1) 1 1 − 2n +1 cos px sin(2n+1)xdx= + 2 2 0 2 p − 2n − 1 p +2n +1 p − (2n +1)

(−1)n+1p − (2n +1) = p2 − (2n +1)2 Therefore letting p(x)=1− cos px in (96) gives us

π ∞ n 2 1 − cos px 1 (−1) p +(2n +1) dx =2 + (119) 2 − 2 0 sin x n=0 2n +1 p (2n +1) and comparing this with (115) results in

∞ 1 (−1)np +(2n +1) 2 + 2 − 2 n=0 2n +1 p (2n +1) 1+p 1 1 pπ 3+p 1+p = ψ − ψ − sin ψ − ψ 2 2 2 2 4 4 pπ Multiplying across by cos 2 gives us ∞ pπ 1 (−1)np +(2n +1) 2cos + 2 − 2 2 n=0 2n +1 p (2n +1)

pπ 1+p 1 1 pπ pπ 3+p 1+p =cos ψ − ψ − cos sin ψ − ψ 2 2 2 2 2 2 4 4

We have

∞ pπ 1 (−1)np +(2n +1) 2cos + 2 − 2 2 n=0 2n +1 p (2n +1) Some inﬁnite series involving the Riemann zeta function 61

∞ π 1 pπ 3+p 1+p pπ 1 2n +1 − ψ − ψ = cos +2 cos + 2 − 2 2 2 2 4 4 2 n=0 2n +1 p (2n +1)

∞ π 1 pπ 3+p 1+p pπ 1 1 − ψ − ψ p2 = cos +2 cos 2 − 2 2 2 2 4 4 2 n=0 2n +1p (2n +1) and therefore we see that

∞ 1 1 2p2 2 − 2 (120) n=0 2n +1p (2n +1) 1+p 1 1 pπ 3+p 1+p π pπ = ψ −ψ + 1 − sin ψ − ψ − / cos 2 2 2 2 4 4 2 2 This is reminiscent of the expression in Prudnikov et al. [41a, 5.1.25-13]

∞ 1 1 p2 ψ p ψ − p γ 2 2 − 2 = (1 + )+ (1 )+2 (121) n=1 n p n and separating the even and odd terms gives us

∞ ∞ ∞ 1 1 1 1 1 1 2 − 2 = 2 − 2 + 2 − 2 n=1 n p n n=1 2n p 4n n=0 2n +1p (2n +1) Letting p → p/2 in (121) we see that ∞ p p 1 1 ψ ψ − γ p2 1+ + 1 +2 =2 2 − 2 2 2 n=1 n p 4n and hence we obtain

1 p p ψ(1 + p)+ψ(1 − p)+2γ − ψ 1+ + ψ 1 − +2γ 2 2 2 62 Donal F. Connon

1+p 1 1 pπ 3+p 1+p π pπ = ψ −ψ + 1 − sin ψ − ψ − / cos 2 2 2 2 4 4 2 2 which may be written as 1 p p ψ(1 + p)+ψ(1 − p) − ψ 1+ + ψ 1 − (122) 2 2 2

1+p 1 pπ 3+p 1+p π pπ = ψ +2 log 2+ 1 − sin ψ − ψ − / cos 2 2 2 4 4 2 2

For example, with p =1/2weobtain 1 1 1 1 3 ψ 1+ + ψ − ψ 1+ + ψ 2 2 2 4 4

√ 3 1 1√ 7 3 π 2 = ψ +2log2+ 1 − 2 ψ − ψ − 4 2 2 8 8 2 and this may be verified by substituting the specific values for the digamma function in [43, p. 20] (albeit one of the relevant values is reported incorrectly in [43, p. 20]). Ramanujan [11, Part I, p. 263] reported the following Maclaurin series (in the slightly modified form employed by Borwein and Chamberland [51]) ∞ (2) −1 −1 4 3 Hn−1 2n 2n sin y = (2y) , −1 ≤ y ≤ 1 (123) 2 n 2 n=1 n (m) where Hn is the generalized harmonic number n (m) 1 Hn = m k=1 k Therefore upon letting x =sin−1 y we get Some infinite series involving the Riemann zeta function 63

∞ (2) −1 4 3 Hn−1 2n 2n 2n x = 2 sin x, −π/2 ≤ x ≤ π/2 (124) 2 n 2 n=1 n Integration results in

π 5 ∞ (2) −1 π 3 Hn−1 2n 2n 2 2n = 2 sin xdx 2 n 160 2 n=1 n 0 and hence we obtain the Euler sum

5 ∞ (2) π 3π Hn−1 = 2 160 4 n=1 n or equivalently

4 ∞ (2) ∞ (2) π Hn−1 Hn − = 2 = 2 ζ(4) 120 n=1 n n=1 n π4 Since ζ(4) = 90 this may be written in its more familiar form as (see for example [19])

∞ (2) 7 Hn ζ(4) = 2 4 n=1 n Alternatively, we now multiply equation (124) by cot x and integrate to obtain

t ∞ (2) −1 t 4 3 Hn−1 2n 2n 2n−1 x cot xdx= 2 sin x cos xdx 2 n 0 2 n=1 n 0 which results in t ∞ 2n (2) 2n −1 4 3 2 Hn−1 sin t 2n x cot xdx= (125) 3 n 0 4 n=1 n 64 Donal F. Connon

This integral may also be evaluated in terms of, for example, the Clausen functions by using (31). We multiply (124) by cos x and integrate to obtain for −π/2 ≤ x ≤ π/2

∞ 2n (2) −1 2 4 2 3 2 Hn−1 2n 2n+1 4t(t − 6) cos t +(t − 12t + 24) sin t = sin t 2 n 2 n=1 (2n +1)n and with t = π/2weobtain 4 ∞ 2n (2) −1 π 2 3 2 Hn−1 2n − 3π +24= (126) 2 n 16 2 n=1 (2n +1)n We showed in [19] that for 0 ≤ t<1

t πx cot πx dx = ζ(−1,t) − ζ(−1, 1 − t)+t log(2 sin πt) (127) 0 and we now refer to the well-known formula [12, p. 130] ∞ πx cot πx = −2 ζ(2n)x2n, (|x| < 1) (128) n=0 (where ζ(0) = −1/2). This gives us

∞ ζ(2n) −2 t2n+1 = ζ(−1,t) − ζ(−1, 1 − t)+t log(2 sin πt) (129) n=0 2n +1 We also have [43, p. 12] in terms of the digamma function

πx cot πx = xψ(1 − x) − xψ(x) and therefore we have

t [xψ(1 − x) − xψ(x)]dx = ζ(−1,t) − ζ(−1, 1 − t)+t log(2 sin πt) 0 Some inﬁnite series involving the Riemann zeta function 65

Integration by parts results in

t t [xψ(1 − x) − xψ(x)]dx = −t log[Γ(t)Γ(1 − t)] + log[Γ(x)Γ(1 − x)] dx 0 0 where upon using Euler’s reﬂection formula for the gamma function [43, p. 3] π Γ(t)Γ(1 − t)= sin πt this becomes

t = t log sin πt − log sin πx dx 0 Therefore, as noted in equation (4.3.158a) in [19], we have

t log[2 sin πx] dx = −[ζ(−1,t) − ζ(−1, 1 − t)] (130) 0 which could of course have been obtained more directly by using integration by parts on equation (127). This incidentally shows us that

ζ(−1, 0) = ζ(−1, 1) = ζ(−1) Letting t =1/2 in (130) immediately gives us Euler’s integral

1 2 1 log sin πx dx = − log 2 0 2 Let us now diﬀerentiate (129) to obtain

∞ d −2 ζ(2n)t2n = [ζ(−1,t) − ζ(−1, 1 − t)] + log(2 sin πt)+π cot πt n=0 dt Since the Hurwitz zeta function is analytic in the whole complex plane except for s = 1, its partial derivatives commute in the region where the function is analytic: we therefore have 66 Donal F. Connon

∂ ∂ ∂ ∂ ∂ ζ(s, t)= ζ(s, t)=− [sζ(s +1,t)] ∂t ∂s ∂s ∂t ∂s andwethenseethat ∂ ∂ ∂ ζ(s, t)=−ζ(s +1,t) − s ζ(s +1,t) (131) ∂t ∂s ∂s Hence with s = −1weseethat d ζ(−1,t)=ζ(0,t) − ζ(0,t) dt d ζ(−1, 1 − t)=−ζ(0, 1 − t)+ζ(0, 1 − t) dt andwethenobtain

∞ −2 ζ(2n)t2n =[ζ(0,t)+ζ(0, 1−t)]−[ζ(0,t)−ζ(0, 1−t)]+log(2 sin πt)+π cot πt n=0

Using Lerch’s identity [10] for t>0 1 ζ(0,t)=logΓ(t) − log(2π) 2 and the well-known result [7, p. 264] involving the Bernoulli polynomials Bn(t) − − Bn(t) ≥ ζ(1 n, t)= n for n 1 this becomes

∞ −2 ζ(2n)t2n =log[Γ(t)Γ(1 − t)] − log(2π) + log(2 sin πt)+π cot πt n=0

Using Euler’s reﬂection formula for the gamma function π Γ(t)Γ(1 − t)= sin πt Some inﬁnite series involving the Riemann zeta function 67 we return to where we started from ∞ −2 ζ(2n) t2n = πt cot πt n=0 Dividing this by t, dealing separately with the n = 0 term, and integrating results in

∞ t2n ζ(2n) =logt − log sin πt + c n=1 n and in the limit as t → 0 we see that the integration constant is c =logπ. We thus obtain the known result [43, p. 161]

∞ t2n ζ(2n) =log(πt) − log sin πt n=1 n We now multiply this by t and integrate to obtain

∞ ζ(2n) u u2n+2 = u log(πu) − u − t log sin πt dt n=0 n(2n +2) 0 and, in the case u =1/2, we obtain using (31) 2 ∞ 2π − 1 − ζ(2n) ζ(3) = log π 2n 7 2 n=1 n(n +1)2 This known result was also recently derived by Fujii and Suzuki [28] where, in what appears to be a new approach, they employed the logarithmic form of Euler’s inﬁnite product identity for the sine function ∞ 2 − x log sin x =logx + log 1 2 2 n=1 n π Upon integrating (131) with respect to t we see that

v v ∂ ∂ v −s ζ(s +1,t)dt = ζ(s, t)dt + ζ(s +1,t) dt 0 0 ∂t ∂s 0 68 Donal F. Connon

We therefore get

v v −s ζ(s +1,t)dt = ζ(s, v) − ζ(s, 0) + ζ(s +1,t) dt 0 0 and with s = −n we have

v v n ζ(1 − n, t)dt = ζ(−n, v) − ζ(−n, 0) + ζ(1 − n, t) dt 0 0 Then using the well-known formula − − Bn(v) ≥ ζ(1 n, v)= n for n 1 we obtain

v Bn+1 − Bn+1(v) n ζ(1 − n, t) dt = + ζ(−n, v) − ζ(−n, 0) (132) 0 n(n +1) This identity was originally derived by Adamchik [1] in a diﬀerent manner in 1998. With n =2weobtain

v − − 1 1 − ζ(3) ζ ( 1,t) dt = B3(v)+ ζ ( 2,v)+ 2 (133) 0 12 2 8π − − − − ζ(3) since ζ ( n, 0) = ζ ( n)andζ ( 2) = 4π2 . With v = 1 we note that

1 ζ(1 − n, t) dt = 0 (134) 0 which may also be obtained by integrating (63).

3 A brief survey of multiple sine functions

The identity (33) was also found by Koyama and Kurokawa [34] using the triple sine function. The multiple sine functions are deﬁned for r =2, 3, ··· by Some inﬁnite series involving the Riemann zeta function 69

∞ r− xr−1 x n 1 Sr(x)=exp − Pr (135) r 1 n=−∞,n=0 n ∞ r−1 nr−1 x x −x =exp − Pr Pr r 1 n=1 n n where u u2 ur Pr(u)=(1− u)exp + + ···+ 1 2 r For example, the triple sine function is deﬁned by ∞ 2 n2 x2 x x2 2 − S3(x)=e 1 2 e (136) n=1 n and we then have ∞ 2 1 2 2 − x 2 log S3(x)= x + n log 1 2 + x 2 n=1 n where upon diﬀerentiation results in

∞ S (x) 2n2x 3 = x + +2x 2 − 2 S3(x) n=1 x n ∞ 2x = x + x2 2 − 2 n=1 x n ∞ 1 2x = x2 + 2 − 2 x n=1 x n We have the well-known decomposition formula [12, p.131]

∞ 1 2x π πx cot = + 2 − 2 (137) x n=1 x n 70 Donal F. Connon

and, since S3(0) = 1, we then see that

x 2 log S3(x)= πt cot πt dt (138) 0 The double sine function is deﬁned by ∞ n x 1 − x/n 2x S2(x)=e e (139) n=1 1+x/n It is easily seen that

∞ x x log S2(x)=x + n log 1 − − log 1 − +2x (140) n=1 n n and, in the same manner as before, we easily ﬁnd that

x log S2(x)= πt cot πt dt (141) 0

The function S1(x)isdefinedby ∞ 2 − x S1(x)=2πx 1 2 = 2 sin πx (142) n=1 n and it is well known that logarithmic differentiation of this results in (137) above. The Barnes double gamma function is defined, inter alia, by [43, p. 25]

∞ 2 x 1 2 2 x k x G(1 + x)=(2π) 2 exp[− (γx + x + x)] [(1 + ) exp( − x)] 2 k=1 k 2k

(143) We showed in [23] that the Barnes double gamma function could be represented by Some inﬁnite series involving the Riemann zeta function 71

∞ 1 1 1 1 x log G(1+x)= x log(2π)− x(1+x)− γx2 = x2 − x + n log 1+ 2 2 2 n=1 2n n (144) and letting x →−x gives us

∞ 1 1 1 1 x log G(1−x)=− x log(2π)+ x(1−x)− γx2 = x2 + x + n log 1 − 2 2 2 n=1 2n n

Subtraction results in

log G(1 − x) − log G(1 + x)

∞ x x = x − x log(2π)+ 2x + n log 1 − − n log 1+ n=1 n n and, using (140), we therefore see that

log S2(x)=logG(1 − x) − log G(1 + x)+x log(2π) (145) We have therefore rediscovered the well-known formula originally found by Kinkelin in 1860 [43, p. 30]

G(1 + x) x log = x log(2π) − πt cot πt dt (146) G(1 − x) 0 which was generalized in 1992 by Freund and Zabrodin [27] who reported the more general identity for

u (−1)n+1 n−1 Γn(u)[Γn(−u)] =exp[−π x cot πxdx] 0 for n ≥ 2.

(147) 72 Donal F. Connon where we have followed the Vign´eras notation [43, p. 39]

(−1)n−1 Γn(x)=[Gn(x)] and Γ1(x)=G1(x)=Γ(x), Γ2(x)=1/G2(x)=1/G(x) Koyama and Kurokawa [35] have shown that

x n−1 log Sn(x)= πt cot πt dt (148) 0 Reference should also be made to Kurokawa’s paper [36]. It is an exercise in Bromwich’s book [15, p. 526] to show that ∞ 2n +1 − 1 − n log − 1 = (1 log 2) n=1 2n 1 2 and it may be noted that this agrees with (140) when x =1/2. The gamma function may be deﬁned by [43, p. 2]

∞ 1 x log Γ(x +1)= x log 1+ − log 1+ n=1 n n and this may be written as ∞ ∞ 1 (−1)kxk log Γ(x +1)= x log 1+ + k n=1 n k=1 kn ∞ ∞ − k k 1 − 1 ( 1) x = x log 1+ + k n=1 n n k=2 kn ∞ ∞ ∞ − k k 1 − 1 ( 1) x = log 1+ x + k n=1 n n n=1 k=2 kn ∞ ∞ − k k ∞ 1 − 1 ( 1) x 1 = log 1+ x + k n=1 n n k=2 k n=1 n Hence we have the well-known Maclaurin series [43, p. 160] for log Γ(x +1) Some inﬁnite series involving the Riemann zeta function 73

∞ (−1)kζ(k) log Γ(x +1)=−γx + xk (149) k=2 k Now referring back to (144)

∞ 1 1 1 1 x log G(1+x)= x log(2π)− x(1+x)− γx2+ x2 − x + n log 1+ 2 2 2 n=1 2n n and applying the same procedure as with log Γ(x + 1) above this becomes

∞ ∞ − k k 1 − 1 − 1 2 1 2 − − ( 1) x = x log(2π) x(1 + x) γx + x x k−1 2 2 2 n=1 2n k=1 kn

∞ ∞ − k k 1 − 1 − 1 2 − ( 1) x = x log(2π) x(1 + x) γx k−1 2 2 2 n=1 k=3 kn ∞ ∞ − m+1 m+1 1 − 1 − 1 2 − ( 1) x = x log(2π) x(1 + x) γx m 2 2 2 n=1 m=2 (m +1)n

∞ − m+1 m+1 ∞ 1 − 1 − 1 2 − ( 1) x 1 = x log(2π) x(1 + x) γx m 2 2 2 m=2 m +1 n=1 n

∞ 1 1 1 (−1)m+1ζ(m)xm+1 = x log(2π) − x(1 + x) − γx2 − 2 2 2 m=2 m +1 Hence, as reported in [43, p. 211], we obtain

∞ 1 1 1 (−1)k+1ζ(k)xk+1 log G(1 + x)= x log(2π) − x(1 + x) − γx2 − (150) 2 2 2 k=2 k +1 74 Donal F. Connon

We now recall the Gosper/Vardi functional equation [5] for the double gamma function (a further derivation of this is contained in [19])

log G(1 + x) − x log Γ(1 + x)=ζ(−1) − ζ(−1, 1+x) (151) and using (149) and (151) we obtain

∞ ∞ 1 1 1 (−1)k+1ζ(k) (−1)k+1ζ(k) x log(2π)− x(1+x)− γx2− xk+1+γx2+ xk+1 2 2 2 k=2 k +1 k=2 k

= ζ(−1) − ζ(−1, 1+x) This may be written as

∞ 1 1 1 (−1)k+1ζ(k) ζ(−1)−ζ(−1, 1+x)= x log(2π)− x(1+x)+ γx2 + xk+1 2 2 2 k=2 k(k +1) (152) which is contained in a slightly disguised form in [43, p. 222]. Diﬀerentiation results in

∞ d 1 1 (−1)k+1ζ(k) − ζ(−1, 1+x)= log(2π) − − x + γx + xk dx 2 2 k=2 k and as shown previously we have d ζ(−1, 1+x)=ζ(0, 1+x) − ζ(0, 1+x) dx 1 1 =logΓ(1+x) − log(2π)+ + x 2 2 Therefore we simply recover (149)

∞ (−1)kζ(k) log Γ(x +1)=−γx + xk k=2 k Some inﬁnite series involving the Riemann zeta function 75

We have ∞ 2 1 2 2 − x 2 log S3(x)= x + n log 1 2 + x 2 n=1 n ∞ ∞ 2k 1 2 2 − x = x + x 2k−2 2 n=1 k=1 kn

∞ ∞ 2k ∞ 2k ∞ 1 2 − x 1 2 − x 1 = x 2k−2 = x 2k−2 2 n=1 k=2 kn 2 k=2 k n=1 n ∞ 1 ζ(2k − 2) = x2 − x2k 2 k=2 k This gives us ∞ ∞ 1 2 ζ(2k) 2k+2 ζ(2k) 2k+2 log S3(x)= x − x = − x (153) 2 k=1 k +1 k=0 k +1 The following identity for |x|<1 appears in the book by Srivastava and Choi [43, p. 216]

∞ x −x ζ(2k) 2k+2 1 − 2 G(1 + x)− − x = [1 log(2π)]x +x log − log G(1+t) dt log G(1+t) dt k=1 k +1 2 G(1 x) 0 0

We note that

−x x log G(1 + t) dt = − log G(1 − t) dt 0 0 and integration by parts shows that

G(1 + x) x −x x d G(1 + t) x log − log G(1+t)dt− log G(1+t)dt = t log dt G(1 − x) 0 0 0 dt G(1 − t) 76 Donal F. Connon

Using Kinkelin’s formula (146) we see that

d G(1 + t) log =log(2π) − πt cot πt dt G(1 − t) and hence we have

x d G(1 + t) 1 x t log dt = log(2π)x2 − πt2 cot πt dt 0 dt G(1 − t) 2 0 This results in

∞ ζ(2k) 1 x x2k+2 = x2 − πt2 cot πt dt (154) k=1 k +1 2 0 We have the following well-known identity

∞ 2n 2 B2n t cot t = (−1)n t2n, (|t| <π) (155) n=0 (2n)! Combining this with Euler’s formula

2n−1 2n 2 π B2n ζ(2n)=(−1)n+1 , (n ≥ 1) (156) (2n)! and, letting t → πt,weobtain ∞ πt cot πt = −2 ζ(2n) t2n, (|t| < 1) (157) n=0

Since the first term of the series (155) is equal to B0 =1, to be consistent − 1 with (157), we define ζ(0) = 2 (which in fact also coincides with the value determined by the analytic continuation of ζ(s)ats =0.)Wemaynow multiply (157) by t and integrate this to obtain another derivation of (154). We note from [43, p. 207] that Some infinite series involving the Riemann zeta function 77

x 1 1 1 log G(1+t)dt = − 2logA x+ log(2π)x2 − x3 −(1−x)logG(1+x) 0 4 4 6

+logG(x) − 2logΓ3(1 + x)+2logΓ3(x) and hence we may express the triple sine function in terms of the double and triple gamma functions. This type of representation naturally arises from the combination of the facts that

x n−1 log Sn(x)= πt cot πt dt 0 1 − cot πt = ψ(t) − ψ(−t)+ t x n−1 and the knowledge that integrals of the form 0 t ψ(t)dt result in multiple gamma functions [43, p. 208]. The triple gamma function is deﬁned by [17]

2 3 Γ3(1 + x)=exp(c1x + c2x + c3x )F (x) where

∞ − 1 k(k+1) 3 3 x 2 1 − 1 1 2 1 x 1 x F (x)= 1+ exp (k +1)x 1+ x + + 2 k=1 k 2 4 k 6 k 6 k

∞ − 1 k(k+1) 3 x 2 1 1 1 1 x = 1+ exp 1+ kx − x2 + k=1 k k 2 4 6 k and 3 1 1 1 c1 = − log(2π) − log A, c2 = γ +log(2π)+ 8 4 4 2 1 3 c3 = − γ + ζ(2) + 6 2 78 Donal F. Connon

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