International Journal of Mathematics andComputer Science, 7(2012), no. 1, 11–83

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MCS

Some infinite series involving the Riemannzeta function

Donal F. Connon

ElmhurstDundle Road

Matfield, Kent TN12 7HD, United Kingdom

email: dconnon@btopenworld.com

(Received May 16, 2010, Accepted July 12, 2012)

Abstract

This paper considers some infinite series involving the Riemannzeta function. Some examples are set out below

∞∑n=0

1n + 1

n∑k=0

(nk

)(−1)kxk

(k + y)s= s Φ(x, s + 1, y) − log x Φ(x, s, y)

∞∑n=0

(−1)nn∑

k=0

(nk

)2k

(k + 1)s=

12

(1 − 21−s

)ζ(s)

∞∑n=0

12n+1

n∑k=0

(nk

) [(−1)k

k + y+

(−1)k

k + 1 − y

]= π cosec(πy)

∞∑n=1

22n sin2n u

n3

(2nn

)−1

= 4u2 log(2 sin u)+2Cl3(2u)+4u Cl2(2u)−2ζ(3)

74ζ(3) − 1

2πG =

∞∑n=0

24n

(2n + 1)3[n!]4

[(2n)!]2

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

y

s − 1

∞∑n=0

1

n + 1

n∑k=0

(nk

)(−1)kyk

(k + 1)s−1= Lis(y) − log y

s − 1Lis−1(y) (1)

where Lis(y) is the polylogarithm function [39]

Lis(y) =∞∑

n=1

yn

ns

and, with y = 1, this becomes the formula originally discovered by Hasse [32]in 1930

1

s − 1

∞∑n=0

1

s + 1

n∑k=0

(nk

)(−1)k

(k + 1)s−1= Lis(1) = ζ(s) (2)

The gamma function is defined as Γ(s) =∫ ∞0

ts−1e−tdt,�s > 0.Using the substitution t = (k + y)u, we obtain

1

(k + y)s=

1

Γ(s)

∫ ∞

0

us−1e−(k+y)udu (3)

We now consider the finite sum set out below

Sn(x, y) =n∑

k=0

(nk

)xk

(k + y)s(4)

and combine (3) and (4) to obtain

Sn(x, y) =n∑

k=0

(nk

)xk

(k + y)s=

n∑k=0

(nk

)xk 1

Γ(s)

∫ ∞

0

us−1e−(k+y)udu

=1

Γ(s)

∫ ∞

0

us−1

{n∑

k=0

(nk

)xke−(k+y)u

}du

Some infinite series involving the Riemann zeta function 13

=1

Γ(s)

∫ ∞

0

us−1e−yu

{n∑

k=0

(nk

) [xe−u

]k

}du

Using the binomial theorem this becomes

Sn(x, y) =1

Γ(s)

∫ ∞

0

us−1e−yu(1 + xe−u

)ndu (5)

Making the summation, we see that

∞∑n=0

tnn∑

k=0

(nk

)xk

(k + y)s=

1

Γ(s)

∞∑n=0

tn∫ ∞

0

us−2e−yu(1 + xe−u

)ndu

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

∞∑n=0

[t(1 + xe−u

)]n=

1

1 − t (1 + xe−u)

and we then have

∞∑n=0

tnn∑

k=0

(nk

)xk

(k + y)s=

1

Γ(s)

∫ ∞

0

us−1e−yu

1 − t (1 + xe−u)du (6)

We now integrate (6) with respect to t

∞∑n=0

wn+1

n + 1

n∑k=0

(nk

)xk

(k + y)s=

1

Γ(s)

∫ w

0

dt

∫ ∞

0

us−1e−yu

1 − t(1 + xe−u)du

= − 1

Γ(s)

∫ ∞

0

us−1e−yu log [1 − w(1 + xe−u)]

1 + xe−udu

and obtain

14 Donal F. Connon

∞∑n=0

wn+1

n + 1

n∑k=0

(nk

)xk

(k + y)s= − 1

Γ(s)

∫ ∞

0

us−1e−(y−1)u log [1 − w(1 + xe−u)]

eu + xdu

(7)When w = 1 and x → −x we get

∞∑n=0

1

n + 1

n∑k=0

(nk

)(−1)kxk

(k + y)s= − 1

Γ(s)

∫ ∞

0

us−1e−(y−1)u log [xe−u]

eu − xdu (8)

=1

Γ(s)

∫ ∞

0

use−(y−1)u

eu − xdu − log x

Γ(s)

∫ ∞

0

us−1e−(y−1)u

eu − xdu

We see from (3) that

xk

(k + y)s=

xk

Γ(s)

∫ ∞

0

us−1e−(k+y)udu

and we have the summation

∞∑k=0

xk

(k + y)s=

1

Γ(s)

∫ ∞

0

us−1e−yu

∞∑k=0

xke−kudu

=1

Γ(s)

∫ ∞

0

us−1e−yu

1 − xe−udu

We therefore obtain the well-known formula [43, p. 121] for the Hurwitz-Lerch zeta function Φ(x, s, y)

Φ(x, s, y) =∞∑

k=0

xk

(k + y)s=

1

Γ(s)

∫ ∞

0

us−1e−(y−1)u

eu − xdu (9)

and with y = 1 we obtain

Φ(x, s, 1) =∞∑

k=0

xk

(k + 1)s=

1

x

∞∑k=1

xk

ks=

Lis(x)

x

Some infinite series involving the Riemann zeta function 15

giving us [46, p. 280]

Lis(x) =x

Γ(s)

∫ ∞

0

us−1

eu − xdu (10)

Reference to (8) then shows that

∞∑n=0

1

n + 1

n∑k=0

(nk

)(−1)kxk

(k + y)s= s Φ(x, s + 1, y) − log x Φ(x, s, y) (11)

With x = 1 we obtain Hasse’s formula (2)

∞∑n=0

1

n + 1

n∑k=0

(nk

)(−1)k

(k + y)s= s Φ(1, s + 1, y) = sζ(s + 1, y)

With y = 1 equation (11) becomes

∞∑n=0

1

n + 1

n∑k=0

(nk

)(−1)kxk

(k + 1)s= s Φ(x, s + 1, 1) − log x Φ(x, s, 1)

or equivalently

∞∑n=0

1

n + 1

n∑k=0

(nk

)(−1)kxk

(k + 1)s=

s

xLis+1(x) − 1

xlog xLis(x) (12)

which corresponds with (1). A closed form expression may be obtained forexample with x = 1/2 and s = 2.We see from (6) that

∞∑n=0

tnn∑

k=0

(nk

)xk

(k + y)s=

1

(1 − t)Γ(s)

∫ ∞

0

us−1e−yu

1 − e−utx/(1 − t)du

=1

(1 − t)Γ(s)

∫ ∞

0

us−1e−(y−1)u

eu − tx/(1 − t)du

16 Donal F. Connon

and hence referring to (9) we have

∞∑n=0

tnn∑

k=0

(nk

)xk

(k + y)s=

1

1 − tΦ

(tx

1 − t, s, y

)(13)

With t = 1/2 we obtain

∞∑n=0

1

2n+1

n∑k=0

(nk

)xk

(k + y)s= Φ (x, s, y) (14)

which is clearly an example of Euler’s transformation of series [33, p. 244].Letting w = tx

1−t, (13) may be represented by

∞∑n=0

(w

w + x

)n n∑k=0

(nk

)xk

(k + y)s=

w + x

xΦ (w, s, y) (15)

and with x = −1 this becomes

∞∑n=0

( −w

1 − w

)n n∑k=0

(nk

)(−1)k

(k + y)s= (1 − w)Φ (w, s, y) (16)

as previously reported by Guillera and Sondow [57].Letting w = 1 and x = −1 in (15) results in another identity due to Hasse

∞∑n=0

1

2n+1

n∑k=0

(nk

)(−1)k

(k + y)s= Φ (−1, s, y) = ζa (s, y) (17)

where ζa (s, y) is the alternating Hurwitz-Lerch zeta function

ζa (s, y) =∞∑

k=0

(−1)k

(k + y)s

With y = 1 we have

∞∑n=0

1

2n+1

n∑k=0

(nk

)(−1)k

(k + 1)s= ζa (s, 1) = ζa (s) (18)

Some infinite series involving the Riemann zeta function 17

Letting w = x = 1 in (15) results in

∞∑n=0

1

2n+1

n∑k=0

(nk

)1

(k + y)s= Φ (1, s, y) = ζ (s, y) (19)

Letting s = 1 and y = 1/2 in (17) results in

∞∑n=0

1

2n

n∑k=0

(nk

)(−1)k

2k + 1= ζa (1, 1/2)

We have from [15, p. 523]

ζa (1, 1/2) =π

2

and we therefore obtain

∞∑n=0

1

2n

n∑k=0

(nk

)(−1)k

2k + 1=

π

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 < y< 1

∞∑n=0

1

2n+1

n∑k=0

(nk

)[(−1)k

k + y+

(−1)k

k + 1 − y

]= π cosec(πy)

We now divide (7) by w and integrate to obtain

∞∑n=0

vn+1

(n + 1)2

n∑k=0

(nk

)xk

(k + y)s= − 1

Γ(s)

∫ v

0

log [1 − w(1 + xe−u)]

wdw

∫ ∞

0

us−2e−(y−1)u

eu + xdu

Since

18 Donal F. Connon

∫log(1 − ax)

xdx = −Li2(ax)

we have

∞∑n=0

vn+1

(n + 1)2

n∑k=0

(nk

)xk

(k + y)s=

1

Γ(s)

∫ ∞

0

us−1e−(y−1)uLi2 [v(1 + xe−u)]

eu + xdu

(20)Further operations of the same kind will result in

∞∑n=0

vn+1

(n + 1)r

n∑k=0

(nk

)xk

(k + y)s=

1

Γ(s)

∫ ∞

0

us−1e−(y−1)uLir [v(1 + xe−u)]

eu + xdu

(21)

With t = −1 in (6) we obtain

∞∑n=0

(−1)n

n∑k=0

(nk

)xk

(k + y)s=

1

Γ(s)

∫ ∞

0

us−1e−yu

2 + xe−udu

and letting x → −x we get

∞∑n=0

(−1)n

n∑k=0

(nk

)(−1)kxk

(k + y)s=

1

Γ(s)

∫ ∞

0

us−1e−yu

2 − xe−udu

With x = 2 we have

∞∑n=0

(−1)n

n∑k=0

(nk

)(−1)k2k

(k + y)s=

1

2Γ(s)

∫ ∞

0

us−1e−yu

1 − e−udu =

1

2Γ(s)

∫ ∞

0

us−1e−(y−1)u

eu − 1du

Hence we obtain using (9)

∞∑n=0

(−1)n

n∑k=0

(nk

)(−1)k2k

(k + y)s=

1

2Φ(1, s, y) (22)

Some infinite series involving the Riemann zeta function 19

and with s = n + 1 and y = 1 this becomes

∞∑n=0

(−1)n

n∑k=0

(nk

)(−1)k2k

(k + 1)n+1=

1

2ζ(n + 1) (23)

This identity, in the case where n is a positive integer, was recently reportedby Alzer and Koumandos [6] where it was derived in a very different manner.With x = −2 we have

∞∑n=0

(−1)n

n∑k=0

(nk

)2k

(k + y)s=

1

2Γ(s)

∫ ∞

0

us−2e−yu

eu + 1du

=1

2Φ(−1, s, y)

we see that with y = 1

=1

2Φ(−1, s, 1) =

1

2ζa(s)

Hence we have

∞∑n=0

(−1)n

n∑k=0

(nk

)2k

(k + 1)s=

1

2

(1 − 22−s

)ζ(s) (24)

With t = x = 12

in (6) we obtain

∞∑n=0

1

2n

n∑k=0

(nk

)1

2k(k + y)s=

2

Γ(s)

∫ ∞

0

us−2e−yu

eu − 1du

= 2Φ(1, s, y) = 2ζ(s, y)

and therefore we have with y = 1

ζ(s) =∞∑

n=0

1

2n+1

n∑k=0

(nk

)1

2k(k + 1)s(25)

With t = −1, x = 1/2 and y = 1 in (6) we have

20 Donal F. Connon

∞∑n=0

(−1)n

n∑k=0

(nk

)1

2k(k + 1)s=

1

2Γ(s)

∫ ∞

0

us−1

eu + 14

du = −2Lis(−1/4)

(26)

2 Some series involving the central binomial

numbers

From, for example, Knopp’s book [33, p. 271] we have the well-knownMaclaurin expansion

(sin−1 y

)2=

1

2

∞∑n=1

[n!]2

n2(2n)!(2y)2n,−1 ≤ y ≤ 1 (27)

and therefore upon letting x = sin−1 y we get

x2 =1

2

∞∑n=1

[n!]2

n2(2n)!22n sin2n x,−π/2 ≤ x ≤ π/2 (28)

which was known by Euler [34].Using Burmann’s theorem, it is an exercise in Whittaker & Watson [46, p.130] to prove

x2 = sin2 x + (2

3)1

2sin4 x + (

2.4

3.5)1

3sin6 x + ... =

∞∑n=1

An sin2n x

where

An =(2n − 2)!!

(2n − 1)!!

1

n.

Using the definitions of the double factorials

(2n)!! = 2.4...(2n) = 2nn!, (2n + 1)!! = 1.3.5...(2n + 1) =(2n + 1)!

2nn!

Some infinite series involving the Riemann zeta function 21

we see this is the same as the above identity.

We now multiply equation (28) by cotx and integrate to obtain

∫ t

0

x2 cot x dx =1

2

∞∑n=1

[n!]2

n2(2n)!22n

∫ t

0

sin2n−1 x cos x dx

which, for −π/2 ≤ x ≤ π/2, results in [12, p. 234]

∫ t

0

x2 cot x dx =1

4

∞∑n=1

22n sin2n t

n3

(2nn

)−1

(29)

where

(2nn

)= (2n)!

[n!]2is the central binomial coefficient.

This integral also appears in [12, p. 234].

Using integration by parts we have

∫ t

0

x2 cot x dx = x2 log sin x

∣∣∣∣t

0

− 2

∫ t

0

x log sin x dx

Since x2 log sin x = x2 log sin xx

+ x2 log x we see that limx→0

x2 log sin x = 0 and

thus

∫ t

0

x2 cot x dx = t2 log sin t − 2

∫ t

0

x log sin x dx

Hence we get for 0 ≤ t ≤ π/2

∫ t

0

x log sin x dx =1

2t2 log sin t − 1

8

∞∑n=1

22n sin2n t

n3

(2nn

)−1

(30)

According to Ayoub [8, p. 1084], Euler returned to the zeta function, forwhat appears to be the last time, in 1772 in a paper entitled “ExercitationesAnalyticae” [26]. Notwithstanding that by this time Euler had been blindfor six years, through what Ayoub describes as “a striking and elaboratescheme”, he was able to prove that

22 Donal F. Connon

∫ π2

0

x log sin x dx =7

16ζ(3) − π2

8log 2 (31)

A very elementary proof of (31) is given in equation (6.20a) in [22] where weused the basic identity

∫ b

a

p(x) cot x dx = 2∞∑

n=1

∫ b

a

p(x) sin 2nx dx (32)

which, as shown in [22], is valid for a wide class of suitably behaved func-tions. Specifically we require that p(x) is a twice continuously differentiablefunction. 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) isequivalent to the requirement that sinx has no zero in [a, b].For example, letting p(x) = x2 in (32) we have for −π < t < π

∫ t

0

x2 cot x dx = 2∞∑

n=1

∫ t

0

x2 sin 2nx dx

=1

2

∞∑n=1

cos 2nx

n3− x2

∞∑n=1

cos 2nx

n+ x

∞∑n=1

sin 2nx

n2

∣∣∣∣∣t

0

This gives us

∫ t

0

x2 cot x dx =1

2

∞∑n=1

cos 2nt

n3− t2

∞∑n=1

cos 2nt

n+ t

∞∑n=1

sin 2nt

n2− 1

2ζ(3) (33)

and with t = π/2 we obtain

∫ π2

0

x2 cot x dx =1

2

∞∑n=1

(−1)n

n3− π2

4

∞∑n=1

(−1)n

n− 1

2ζ(3)

Hence using the alternating zeta function

Some infinite series involving the Riemann zeta function 23

ζa(s) =∞∑

n=1

(−1)n+1

ns= (1 − 21−s)ζ(s)

we see that

∫ π2

0

x2 cot x dx = −7

8ζ(3) +

π2

4log 2 (34)

Another elementary evaluation of this integral has recently been provided byFujii and Suzuki [28].Using the substitutions y = sin x and z = tan x we may also note that

∫ t

0

xn cot x dx =

∫ sin t

0

(sin−1 y

)n

ydy

2

∫ t

0

xn

sin 2xdx =

∫ tan t

0

(tan−1 z)n

zdz

We also see from (29) and (33) that for −π/2 ≤ t ≤ π/2

1

4

∞∑n=1

22n sin2n t

n3

(2nn

)−1

=1

2

∞∑n=1

cos 2nt

n3−t2

∞∑n=1

cos 2nt

n+t

∞∑n=1

sin 2nt

n2−1

2ζ(3)

(35)and with t = π/2 we immediately see that

ζ(3) =2

7π2 log 2 − 2

7

∞∑n=1

22n

n3

(2nn

)−1

(36)

which is in agreement with Sherman’s compendium of formulae [42].We have from [12, p. 198]

B(n, 1/2) =22n

n

(2nn

)−1

(37)

where the beta function is defined by

24 Donal F. Connon

B(u, v) =

∫ 1

0

(1 − t)u−1 tv−1dt

This then gives us

22n

n

(2nn

)−1

=

∫ 1

0

(1 − t)n

(1 − t)√

tdt

and we have the summation involving the dilogarithm function

∞∑n=1

22n

n3

(2nn

)−1

=

∫ 1

0

Li2 (1 − t)

(1 − t)√

tdt

It may be noted that the Wolfram Mathematica Online Integrator evaluatesthis integral in terms of polylogarithms of order 2 and 3 and hence this willnot provide us with any new information regarding the nature of ζ(3). Itshould be easier to obtain a simpler version of the output by using integrationby parts, noting that∫

Li2 (1 − t)√t

dt = 2

∫Li2

(1 − u2

)du

and∫Li2

(1 − u2

)du = uLi2

(1 − u2

)−Li2 (1 − u)−Li2 (−u)−log u log(1+u)−2u log u−2u

We now attempt to use (37) in (29)

∫ t

0

x2 cot x dx =1

4

∞∑n=1

22n sin2n t

n3

(2nn

)−1

and this gives us

∞∑n=1

22n sin2n t

n3

(2nn

)−1

=

∫ 1

0

Li2[(1 − t) sin2 t

](1 − t)

√t

dt

Some infinite series involving the Riemann zeta function 25

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

Lin(z) =(−1)n−1

(n − 1)!

∫ 1

0

z logn−1 u

1 − zudu, Li2(z) = −

∫ 1

0

z log u

1 − zudu (38)

and thus

∫ 1

0

Li2[(1 − t) sin2 t

](1 − t)

√t

dt = −∫ 1

0

∫ 1

0

sin2 t log u√t[1 − u (1 − t) sin2 t

] dt du

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

∞∑n=1

x2n [n!]4

(2n)k [(2n)!]2=

(−1)k−1

(k − 3)!

∫ π2

0

1

sin y

∫ arcsin(x sin y/4)

0

φ logk−3[4 sin φ/x sin y]dφdy

Since sin(

π6

)= 1

2we see from (29) that

∞∑n=1

1

n3

(2nn

)−1

= −8

∫ π6

0

x log [2 sin x] dx (39)

which I first came across in van der Poorten’s 1979 paper “Some wonderfulformulae. . . an introduction to Polylogarithms” [45].In passing, we note from [12, p.122] that

∞∑n=1

(2t)2n

n3

(2nn

)−1

= 4

∫ t

0

(sin−1 x

)2

xdx = 2t2 4F3

[{1, 1, 1, 1},

{3

2, 2, 2

}, t2

]

in terms of the hypergeometric functions.

26 Donal F. Connon

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

∫ t

0

x3 cot x dx =1

2

∞∑n=1

[n!]2

n2(2n)!22n

∫ t

0

x sin2n−1 x cos x dx

Integration by parts yields∫ t

0

x sin2n−1 x cos x dx =t

2nsin2n t − 1

2n

∫ t

0

sin2n x dx

and therefore we have

∫ t

0

x3 cot x dx =1

4t

∞∑n=1

22n sin2n t

n3

(2nn

)−1

−1

4

∞∑n=1

22n

n3

(2nn

)−1 ∫ t

0

sin2n x dx

(40)With t = π/2 we obtain

∫ π2

0

x3 cot x dx =π

8

∞∑n=1

22n

n3

(2nn

)−1

− 1

4

∞∑n=1

22n

n3

(2nn

)−1 ∫ π2

0

sin2n x dx

and using [12, p. 195]

∫ π2

0

sin2n x dx =(2n)!

22n(n!)2

π

2=

1

22n

(2nn

)π

2(41)

we have

∫ π2

0

x3 cot x dx =π

8

∞∑n=1

22n

n3

(2nn

)−1

− π

8

∞∑n=1

1

n3

Using (36) we obtain

∫ π2

0

x3 cot x dx =π3

8log 2 − 9π

16ζ(3) (42)

As another example, letting p(x) = x3 in (32) we have for −π < t < π

Some infinite series involving the Riemann zeta function 27

∫ t

0

x3 cot x dx = 2∞∑

n=1

∫ t

0

x3 sin 2nx dx

=3

2x2

∞∑n=1

sin 2nx

n2− 3

4

∞∑n=1

sin 2nx

n4− x3

∞∑n=1

cos 2nx

n+

3

2x

∞∑n=1

cos 2nx

n3

∣∣∣∣∣t

0

=3

2t2

∞∑n=1

sin 2nt

n2− 3

4

∞∑n=1

sin 2nt

n4− t3

∞∑n=1

cos 2nt

n+

3

2t

∞∑n=1

cos 2nt

n3

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

0

x3 cot xdx =

1

4t

∞∑n=1

22n sin2n t

n3

(2nn

)−1

−1

4

∞∑n=1

22n

n3

(2nn

)−11

22n

[(2nn

)t +

n∑j=1

(−1)j

j

(2n

n − j

)sin 2jt

]

=1

4t

∞∑n=1

22n sin2n t

n3

(2nn

)−1

− t

4

∞∑n=1

1

n3−1

4

∞∑n=1

1

n3

(2nn

)−1 n∑j=1

(−1)j

j

(2n

n − j

)sin 2jt

This gives us

1

4t

∞∑n=1

22n sin2n t

n3

(2nn

)−1

− t

4ζ(3)−1

4

∞∑n=1

1

n3

(2nn

)−1 n∑j=1

(−1)j

j

(2n

n − j

)sin 2jt

=3

2t2

∞∑n=1

sin 2nt

n2− 3

4

∞∑n=1

sin 2nt

n4− t3

∞∑n=1

cos 2nt

n+

3

2t

∞∑n=1

cos 2nt

n3

and upon substituting (35) we have for −π/2 ≤ t ≤ π/2

28 Donal F. Connon

∞∑n=1

1

n3

(2nn

)−1 n∑j=1

(−1)j

j

(2n

n − j

)sin 2jt

= 3∞∑

n=1

sin 2nt

n4− 4t

∞∑n=1

cos 2nt

n3− 2t2

∞∑n=1

sin 2nt

n2− 3tζ(3)

(43)

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

∫ t

0

πx2 cot πx dx

= −[ζ ′(−2, t)+ζ ′(−2, 1−t)]+2t[ζ ′(−1, t)−ζ ′(−1, 1−t)]+t2 log(2 sin πt)−ζ(3)

2π2

(44)

where ζ ′(s, t) = ∂∂s

ζ(s, t) is a derivative of the Hurwitz zeta function.

Using (29) we see that for 0 ≤ t ≤ 1/2

∫ t

0

πx2 cot πx dx =1

π2

∫ πt

0

u2 cot u du =1

4π2

∞∑n=1

22n sin2n πt

n3

(2nn

)−1

(45)and hence we see from (44) and (45) that for 0 ≤ t ≤ 1/2

1

4π2

∞∑n=1

22n sin2n πt

n3

(2nn

)−1

= −[ζ ′(−2, t)+ζ ′(−2, 1−t)]+2t[ζ ′(−1, t)−ζ ′(−1, 1−t)]+t2 log(2 sin πt)−ζ(3)

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

1

4π2

∞∑n=1

22n sin2n πt

n3

(2nn

)−1

= −[ζ ′(−2, t)+ζ ′(−2, 1−t)]−2(1−t)[ζ ′(−1, t)−ζ ′(−1, 1−t)]

+(1 − t)2 log(2 sin πt) − ζ(3)

2π2

and subtraction of these two equations appeared to indicate that for

ζ ′(−1, t) − ζ ′(−1, 1 − t) =1

2(1 − 2t) log(2 sin πt)

We note from (55) below that

ζ ′(−1,

1

4

)− ζ ′

(−1,

3

4

)=

1

2πCl2(π/2) =

G

2π

and apparently this would lead to a closed-form expression for Catalan’sconstant. Unfortunately, the analysis is incorrect because the restrictionthat 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 functionsΓn(x) defined in [43, p. 24]. Adamchik [5] has shown that for �(x) > 0

ζ ′ (−n, x) − ζ ′ (−n) = (−1)n

n∑k=0

k!Qk,n(x) log Γk+1(x) (47)

where Qk,n(x) are polynomials defined by

Qk,n(x) =n∑

j=k

(1 − x)n−j

(nj

){jk

}

and

{jk

}are the Stirling subset numbers defined by

{jk

}= k

{n − 1

k

}+

{n − 1k − 1

},

{n0

}=

{1, n = 00, 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)

ζ ′ (−2, x) − ζ ′ (−2) = 2 log Γ3(x) + (3 − 2x) log G(x) − (1 − x)2 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)

ζ ′ (−2, 1 − x)−ζ ′ (−2) = 2 log Γ3(1−x)+(1+2x) log G(1−x)−x2 log Γ(1−x)

Hence we have

ζ ′ (−1, x) − ζ ′ (−1, 1 − x) = x log[Γ(x)Γ(1 − x)] + logG(1 − x)

G(1 + x)(49)

which we have previously derived in [19].Letting t = 1

2in (46) we obtain

1

4π2

∞∑n=1

22n

n3

(2nn

)−1

= −2ζ ′(−2,

1

2

)+

1

4log 2 − ζ(3)

2π2

and using (36) we obtain the known result

ζ ′(−2,

1

2

)=

3ζ(3)

16π2(50)

which is also derived, for example, in equation (4.3.168d) of [19].When t = 1

4in (46) we get

Some infinite series involving the Riemann zeta function 31

1

4π2

∞∑n=1

2n

n3

(2nn

)−1

= −[ζ ′

(−2,

1

4

)+ ζ ′

(−2,

3

4

)]+

1

2

[ζ ′

(−1,

1

4

)− ζ ′

(−1,

3

4

)]+

1

32log 2−ζ(3)

2π2

(51)

We now refer to Adamchik’s result [2]

ζ ′ (−2n − 1, t) − ζ ′ (−2n − 1, 1 − t) =(2n + 1)!

(2π)2n+1Cl2n+2(2πt) (52)

ζ ′ (−2n, t) + ζ ′ (−2n, 1 − t) = (−1)n (2n)!

(2π)2nCl2n+1(2πt) (53)

where Cln(t) is the Clausen function defined by

Cl2n+1(t) =∞∑

k=1

cos kt

k2n+1Cl2n(t) =

∞∑k=1

sin kt

k2n(54)

This is also derived in equation 4.3.167 of [19].We then see that (as noted by Adamchik [2])

ζ ′(−1,

1

4

)− ζ ′

(−1,

3

4

)=

1

2πCl2(π/2) =

G

2π(55)

where G is Catalan’s constant.Using the formula [18, equation 5.11] (where I have corrected a misprint)

Cl2n+1(π/2) =1 − 22n

24n+1ζ(2n + 1) (56)

this then gives us for t = 1/4

ζ ′(−2n,

1

4

)+ζ ′

(−2n,

3

4

)= (−1)n (2n)!

(2π)2nCl2n+1(π/2) = (−1)n 1 − 22n

24n+1

(2n)!

(2π)2nζ(2n+1)

32 Donal F. Connon

and in particular we have for n = 1

ζ ′(−2,

1

4

)+ ζ ′

(−2,

3

4

)=

3ζ(3)

64π2

We therefore have

∞∑n=1

2n

n3

(2nn

)−1

= −35ζ(3)

16+

π2

8log 2 + πG (57)

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

∫ π4

0

x2 cot xdx = −35

64ζ(3) +

π2

32log 2 +

πG

4(58)

With t = 16

in (46) we obtain

1

4π2

∞∑n=1

1

n3

(2nn

)−1

= −[ζ ′

(−2,

1

6

)− ζ ′

(−2,

5

6

)]+

1

3

[ζ ′

(−1,

1

6

)− ζ ′

(−1,

5

6

)]− ζ(3)

2π2

(59)

We see from (53) that

ζ ′(−2,

1

6

)+ ζ ′

(−2,

5

6

)= − 1

2π2Cl3

(π

3

)and from Lewin’s monograph [39, p. 198] we have

Cl2n+1

(π

3

)=

1

2(1 − 2−2n)(1 − 3−2n)ζ(2n + 1)

which gives us

Cl3

(π

3

)=

1

4ζ(3)

Some infinite series involving the Riemann zeta function 33

This results in

ζ ′(−2,

1

6

)+ ζ ′

(−2,

5

6

)= −ζ(3)

8π2(60)

We also have from (52)

ζ ′(−1,

1

6

)− ζ ′

(−1,

5

6

)=

1

2πCl2

(π

3

)and from (59) we obtain

1

4π2

∞∑n=1

1

n3

(2nn

)−1

= −[ζ ′

(−2,

1

6

)− ζ ′

(−2,

5

6

)]+

1

6πCl2

(π

3

)− ζ(3)

2π2

(61)Accordingly we have two simultaneous equations involving ζ ′ (−2, 1

6

)and

ζ ′ (−2, 56

)(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 theRiemann zeta function ζ(s, t) as reported in Titchmarsh’s treatise [43b, p.37]

ζ(s, t) = 2Γ(1 − s)

[sin

(πs

2

) ∞∑n=1

cos 2nπt

(2πn)1−s+ cos

(πs

2

) ∞∑n=1

sin 2nπt

(2πn)1−s

](62)

where �(s) < 0 and 0 < t ≤ 1. In 2000, Boudjelkha [52] showed that thisformula also applies in the region Re(s)< 1. It may be noted that when t = 1this reduces to Riemann’s functional equation for ζ(s). Letting s → 1− s wemay write this as

ζ(1 − s, t) = 2Γ(s)

[cos

(πs

2

) ∞∑n=1

cos 2nπt

(2πn)s+ sin

(πs

2

) ∞∑n=1

sin 2nπt

(2πn)s

](63)

Using (63) it is easily shown that

34 Donal F. Connon

2ζ ′(−1, t) − B2(t) (1 − γ − log(2π)) = −4∞∑

n=1

log n cos 2nπt

(2πn)2+

1

2πCl2(2πt)

(64)

ζ ′(−2, t)−B3(t)

[1

2− 1

3γ − 1

2log(2π)

]= −4

∞∑n=1

log n sin 2nπt

(2πn)3− 1

(2π)2Cl3(2πt)

(65)With t = 1/6 in (65) we obtain

ζ ′(−2,

1

6

)− 5

216

[1

2− 1

3γ − 1

2log(2π)

]= −4

∞∑n=1

log n sin(nπ/3)

(2πn)3− ζ(3)

16π2

and this indicates the complexities involved in determining a closed formexpression for

ζ ′(−2,

1

6

).

Differentiating (73) gives us

−∞∑

n=1

log n sin(nπ/3)

ns=

√3

{3−s − 1

2ζ ′(s) − 3−s log 3

2ζ(s)

}

+√

3

{6−s

[ζ ′

(s,

1

6

)+ ζ ′

(s,

1

3

)]− 6−s log 6

[ζ

(s,

1

6

)+ ζ

(s,

1

3

)]}

so that with s = 3 we have

−∞∑

n=1

log n sin(nπ/3)

n3=

√3

{3−3 − 1

2ζ ′(3) − 3−3 log 3

2ζ(3)

}

Some infinite series involving the Riemann zeta function 35

+√

3

{6−3

[ζ ′

(3,

1

6

)+ ζ ′

(3,

1

3

)]− 6−3 log 6

[ζ

(3,

1

6

)+ ζ

(3,

1

3

)]}

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

2ζ ′(−1,

1

4

)+

1

48[1 − γ − log(2π)] = −4

∞∑n=1

log n cos(nπ/2)

(2πn)2+

1

2πCl2

(π

2

)

2ζ ′(−1,

1

4

)+

1

48[1 − γ − log(2π)] = −4

∞∑n=1

(−1)n log 2n

(4πn)2+

G

2π

=log 2

4π2

∞∑n=1

(−1)n+1

n2+

1

4π2

∞∑n=1

(−1)n+1 log n

n2+

G

2π

=log 2

4π2ζa(2) − 1

4π2ζ ′a(2) +

G

2π

Since ζa(2) = 12ζ(2) and ζ ′

a(2) = 12ζ(2) log 2 + 1

2ζ ′(2) we have

2ζ ′(−1,

1

4

)+

1

48[1 − γ − log(2π)] = − 1

8π2ζ ′(2) +

G

2π

It is easily found from the functional equation for the Riemann zeta functionthat

ζ ′ (−1) =1

12(1 − γ − log(2π)) +

1

2π2ζ ′(2)

and hence we have (as originally determined by Adamchik [2])

ζ ′(−1,

1

4

)=

G

4π− 1

8ζ ′ (−1) (66)

Ghusayni [29] showed in 1998 that

36 Donal F. Connon

ζ(3) =π

2

∞∑n=1

sin (nπ/3)

n2− 3

4

∞∑n=1

1

n3

(2nn

)−1

(67)

Later in 2000, Ghusayni [30], having noted an earlier paper [31], reportedthat

∞∑n=1

sin(nπ/3)

n2=

√3

2

{1

12+

1

22− 1

42− 1

52+

1

72+

1

82− ...

}(68)

=

√3

2

{1

3ψ(1)

(1

3

)− 2

9π2

}and Ghusayni concluded that

ζ(3) = −√

3

18π3 +

3√

3

4π

∞∑n=1

1

(3n − 2)2− 3

4

∞∑n=1

1

n3

(2nn

)−1

(69)

It is shown in [30] that

1

12+

1

22− 1

42− 1

52+

1

72+

1

82− ... = − 2

27π2 − 2

∫ 1

0

log x

1 + x3dx

Mathematica evaluates this integral as

324

∫ 1

0

log x

1 + x3dx = −8π2 − 6PolyGamma

[1,

1

6

]+ 3PolyGamma

[1,

5

6

]

−3PolyGamma

[1,

1

3

]+ 6PolyGamma

[1,

2

3

]

The Mathworld website for the central binomial coefficient reports the fol-lowing formula which was experimentally obtained by Plouffe [41] (and infact determined analytically in 1985 by Zucker [49])

Some infinite series involving the Riemann zeta function 37

∞∑n=1

1

n3

(2nn

)−1

=1

18π√

3

[ψ(1)

(1

3

)− ψ(1)

(2

3

)]− 4

3ζ(3) (70)

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

ψ(1 − x) − ψ(x) = π cot πx

and differentiation gives us

ψ(k)(1 − x) + (−1)(k+1)ψ(k)(x) = (−1)kπdk

dxkcot πx

and therefore we see that

ψ(1)

(1

3

)+ ψ(1)

(2

3

)= π2/ sin2

(π

3

)=

4

3π2

We then obtain

∞∑n=1

1

n3

(2nn

)−1

=1

9π√

3 ψ(1)

(1

3

)− 4

3ζ(3) − 2

27π3√

3 (71)

We have

ψ(k)(p

q) = (−1)k+1k!ζ(k + 1,

p

q)

and therefore we see that

ψ′(1

3) = ζ(2,

1

3) = 9

∞∑n=0

1

(3n + 1)2

or, equivalently, changing the order of summation

ψ′(1

3) = 9

∞∑n=1

1

(3n − 2)2

Hence we see that (70) is equivalent to Ghusayni’s result (69)

38 Donal F. Connon

ζ(3) = −√

3

18π3 +

√3

12π ψ(1)

(1

3

)− 3

4

∞∑n=1

1

n3

(2nn

)−1

We now refer back to (35) which is valid for −1/2 ≤ t ≤ 1/2

1

4

∞∑n=1

22n sin2n πt

n3

(2nn

)−1

=1

2

∞∑n=1

cos 2nπt

n3−π2t2

∞∑n=1

cos 2nπt

n+πt

∞∑n=1

sin 2nπt

n2−1

2ζ(3)

so that with t = 1/6 we have

1

4

∞∑n=1

1

n3

(2nn

)−1

=1

2

∞∑n=1

cos(nπ/3)

n3− 1

36π2

∞∑n=1

cos(nπ/3)

n+

1

6π

∞∑n=1

sin(nπ/3)

n2−1

2ζ(3)

Lewin [39] and Srivastava and Tsumura [43, p. 293] reported for Re(s)> 1

∞∑n=1

cos(nπ/3)

ns=

1

2(61−s − 31−s − 21−s + 1)ζ(s) (72)

∞∑n=1

sin(nπ/3)

ns=

√3

{3−s − 1

2ζ(s) + 6−s

[ζ

(s,

1

6

)+ ζ

(s,

1

3

)]}(73)

Hence we have with s = 3 and s = 2 respectively

∞∑n=1

cos(n π/3)

n3=

1

3ζ(3) (74)

∞∑n=1

sin(n π/3)

n2=

√3

{3−2 − 1

2ζ(2) + 6−2

[ζ

(2,

1

6

)+ ζ

(2,

1

3

)]}(75)

We have the Fourier series [44, p. 148] for 0 < t < 1

Some infinite series involving the Riemann zeta function 39

∞∑n=1

cos 2nπt

n= − log(2 sin πt)

Using (75) we then obtain Ghusayni’s result

ζ(3) =π

2

√3

{3−2 − 1

2ζ(2) + 6−2

[ζ

(2,

1

6

)+ ζ

(2,

1

3

)]}−3

4

∞∑n=1

1

n3

(2nn

)−1

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

∞∑n=1

cos(2n π/3)

ns=

1

2(31−s − 1)ζ(s) (76)

∞∑n=1

sin(2n π/3)

ns=

√3

{3−s − 1

2ζ(s) + 3−sζ

(s,

1

3

)}(77)

∞∑n=1

cos(nπ/2)

ns= 2−s(21−s − 1)ζ(s) (78)

∞∑n=1

sin(n π/2)

ns= (2−s − 1)ζ(s) + 21−2sζ

(s,

1

4

)(79)

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)

∞∑n=1

22n sin2n πt

n3

(2nn

)−1

= 4π2t2 log(2 sin πt)+2Cl3(2πt)+4πt Cl2(2πt)−2ζ(3)

(80)

or equivalently

40 Donal F. Connon

∞∑n=1

22n sin2n u

n3

(2nn

)−1

= 4u2 log(2 sin u) + 2Cl3(2u) + 4u Cl2(2u) − 2ζ(3)

(81)Having recently seen a paper by Bradley et al. [14], I noted that this isequivalent to the identity previously discovered by Zucker [49] in 1985 (andI thank John Zucker for subsequently sending me a reprint of his originalpaper).With u = π/6 and u = π/4 we obtain respectively

∞∑n=1

1

n3

(2nn

)−1

= 2Cl3(π/3) +2π

3Cl2(π/3) − 2ζ(3) (82)

∞∑n=1

2n

n3

(2nn

)−1

=π2

8log 2 + 2Cl3(π/2) + πCl2(π/2) − 2ζ(3) (83)

Then using (56) we see that

Cl3(π/2) = − 3

32ζ(3)

and hence from (83) we obtain another derivation of (57)

∞∑n=1

2n

n3

(2nn

)−1

= −35ζ(3)

16+

π2

8log 2 + πG

We also have a number of formulae involving Cl2

(pπq

)in Browkin’s paper in

Lewin’s survey [59, p. 244], including for example

Cl2

(π

6

)+ Cl2

(5π

6

)=

4

3G =

4

3Cl2

(π

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

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

Cl2n+1(2π) = ζ(2n + 1)

We also have

Cl2(π/2) = G = −Cl2(3π/2)

1

2Cl2(2x) = Cl2(x) − Cl2(π − x)

which implies that

Cl2

(2π

3

)=

2

3Cl2

(π

3

)The Clausen function may be expressed in closed form in at least three othercases and from Lewin’s book [39, p. 198] we have

Cl2n+1

(π

2

)= −2−2n−1(1 − 2−2n)ζ(2n + 1)

Cl2n+1

(π

3

)=

1

2(1 − 2−2n)(1 − 3−2n)ζ(2n + 1)

Cl2n+1

(2π

3

)= −1

2(1 − 3−2n)ζ(2n + 1)

For example, we see from the definition that

Cl2n+1

(π

2

)= − 1

22n+1+

1

42n+1− 1

62n+1+ ...

= − 1

22n+1

[1

12n+1− 1

22n+1+ ...

]= − 1

22n+1ζa(2n + 1)

We then see Cl3(π/3) = ζ(3)/3 and from (82) that

42 Donal F. Connon

∞∑n=1

1

n3

(2nn

)−1

=2π

3Cl2(π/3) − 4

3ζ(3)

and comparing this with (71) we obtain

2√

3 Cl2(π/3) = ψ(1)

(1

3

)− 2

3π2 (84)

as previously derived by Fettis [55].More generally, we have [39a, p. 358]

ψ(1)

(p

q

)=

π2

2cosec2

(πp

q

)+ 2q

�(q−1)/2�∑m=1

sin

(2mπp

q

)Cl2

(2mπp

q

)(85)

and a particular case of this is (84). Another example is

Cl2(π/6) =1

24

[√3 ψ(1)

(1

3

)+ 16G − 2π2/

√3

](86)

Using PSLQ, Bailey et al. [50] discovered experimentally that

6Cl2 (α) − 6Cl2 (2α) + 2Cl2 (3α) = 7Cl2

(2π

7

)+ 7Cl2

(4π

7

)− 7Cl2

(6π

7

)(87)

where α = 2 tan−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

∞∑n=1

22n

n3

(2nn

)−1 ∫ x

0

sin2n u du (88)

=4

3x3 log 2 + 4

∫ x

0

u2 log sin u du + 2Cl4(2x) − 2x Cl3(2x) − 2x ζ(3)

Some infinite series involving the Riemann zeta function 43

In the above we used the fact that∫ x

0u sin 2ku du = sin 2kx

4k2 − x cos 2kx2k

and thisresults in ∫ x

0

u Cl2(2u) du =1

4Cl4(2x) − 1

2x Cl3(2x).

When x = π/2 we get

∞∑n=1

22n

n3

(2nn

)−1 ∫ π2

0

sin2n u du =∞∑

n=1

22n

n3

(2nn

)−1π

2

1

22n

(2nn

)=

π

2

∞∑n=1

1

n3=

π

2ζ(3)

=π3

6log 2 + 4

∫ π2

0

u2 log sin u du − π Cl3(π) − π ζ(3)

We see that

Cl3(π) =∞∑

k=1

(−1)k

k3= −ζa(3) = −3

4ζ(3)

which gives us

π

2ζ(3) =

π3

6log 2 + 4

∫ π2

0

u2 log sin u du − π

4ζ(3)

and we therefore obtain a new derivation of Euler’s integral

∫ π2

0

u2 log sin u du = −π3

24log 2 +

3π

16ζ(3) (89)

Wiener [47] has shown that

∫ x

0

sin2n u du =1

22n

[(2nn

)x +

n∑j=1

(−1)j

j

(2n

n − j

)sin 2jx

](90)

and we therefore obtain from (88)

44 Donal F. Connon

∞∑n=1

1

n3

(2nn

)−1 n∑j=1

(−1)j

j

(2n

n − j

)sin 2jx (91)

=4

3x3 log 2 + 4

∫ x

0

u2 log sin u du + 2Cl4(2x) − 2x Cl3(2x) − 3x ζ(3)

Using integration by parts we have∫ x

0

u3 cot u du = u3 log sin u

∣∣∣∣x

0

− 3

∫ x

0

u2 log sin u du

= x3 log sin x − 3

∫ x

0

u2 log sin u du

Using (32) we have

∫ x

0

u3 cot u du = 2∞∑

n=1

∫ x

0

u3 sin 2nu du

=3

4u2

∞∑n=1

sin 2nu

n2− 3

8

∞∑n=1

sin 2nu

n4− 1

2u3

∞∑n=1

cos 2nu

n+

3

4u

∞∑n=1

cos 2nu

n3

∣∣∣∣∣x

0

=3

4x2

∞∑n=1

sin 2nx

n2− 3

8

∞∑n=1

sin 2nx

n4− 1

2x3

∞∑n=1

cos 2nx

n+

3

4x

∞∑n=1

cos 2nx

n3

=3

4x2Cl2(2x) − 3

8Cl4(2x) − 1

2x3Cl1(2x) +

3

4x Cl3(2x)

Therefore we may write (91) as

∞∑n=1

1

n3

(2nn

)−1 n∑j=1

(−1)j

j

(2n

n − j

)sin 2jx (92)

Some infinite series involving the Riemann zeta function 45

= 13x3 log sin x + 4

3x3 log 2 + 2Cl4(2x) − 2x Cl3(2x) − 3x ζ(3)

−14x2Cl2(2x) + 1

8Cl4(2x) + 1

3x3Cl1(2x) − 1

4x Cl3(2x)

=1

3x3 log sin x+

4

3x3 log 2+

17

8Cl4(2x)−9

4x Cl3(2x)−1

4x2Cl2(2x)+

1

3x3Cl1(2x)−3x ζ(3)

Differentiating (28) gives us

x =1

2

∞∑n=1

22n sin2n−1 x cos x

n

(2nn

)−1

(93)

and with x → πx this becomes

πx =1

2

∞∑n=1

22n sin2n−1 πx cos πx

n

(2nn

)−1

(94)

We now multiply this across by cot πx and integrate to obtain

∫ t

0

πx cot πx dx =1

2

∞∑n=1

22n

n

(2nn

)−1 ∫ t

0

sin2n−2 πx cos2 πx dx

=1

2

∞∑n=1

22n

n

(2nn

)−1 ∫ t

0

(sin2n−2 πx − sin2n πx

)dx

The Wolfram Mathematica Online Integrator evaluates the above integral interms of the hypergeometric function 3F2

(32, 3

2,−n; 5

2; cos2 πt

).

Alternatively, using integration by parts, we get∫ t

0

sin2n−2 πx cos2 πx dx =

∫ t

0

sin2n−2 πx cos πx cos πx dx

= cos πxsin2n−1 πx

(2n − 1)π

∣∣∣∣t

0

+1

(2n − 1)

∫ t

0

sin2n πx dx

46 Donal F. Connon

and with t = 1/2 we have

∫ 12

0

sin2n−2 πx cos2 πx dx =1

2n − 1

∫ 12

0

sin2n πx dx.

Since

∫ 12

0

sin2n πx dx =1

π

∫ π2

0

sin2n x dx

we have

∫ π2

0

sin2n x dx =(2n − 1)!!

(2n)!!

π

2=

1

22n

(2nn

)π

2

Therefore we see that

∫ 12

0

sin2n−2 πx cos2 πx dx =1

2n − 1

1

22n+1

(2nn

)Hence we obtain

∫ 12

0

πx cot πx dx =1

4

∞∑n=1

1

n(2n − 1)

Clearly

∞∑n=1

1

n(2n − 1)=

∞∑n=1

(1

n − 12

− 1

n

)

Since the digamma function may be represented by [43, p. 14]

ψ(a) = −γ − 1

a−

∞∑n=1

(1

n + a− 1

n

)

we see that

∞∑n=1

(1

n − 12

− 1

n

)= −ψ

(−1

2

)− γ + 2

Some infinite series involving the Riemann zeta function 47

We know that [43, p. 22]

ψ

(−1

2

)= 2 − γ − 2 log 2

and we therefore deduce that

∞∑n=1

(1

n − 12

− 1

n

)= 2 log 2

Hence we have the well-known integral

∫ 12

0

x cot πx dx =log2

2π

We now divide (93) by sin x and integrate to obtain

∫ t

0

x

sin xdx =

1

2

∞∑n=1

22n sin2n−1 t

n(2n − 1)

(2nn

)−1

Alternatively, with n → n + 1 in the summation this may be written as

∫ t

0

x

sin xdx =

∞∑n=0

22n sin2n+1 t

(2n + 1)2

(2nn

)−1

(95)

We recall from [22] that

∫ b

a

p(x)

sin xdx = 2

∞∑n=0

∫ b

a

p(x) sin(2n + 1)x dx (96)

where p(x) is a twice continuously differentiable function. It should be notedthat 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 thenp(a) must also be zero. Condition (i) is equivalent to the requirement thatsin x has no zero in [a, b].

Letting p(x) = x in (96) we get

48 Donal F. Connon

∫ t

0

x

sin xdx = −2t

∞∑n=0

cos(2n + 1)t

2n + 1+ 2

∞∑n=0

sin(2n + 1)t

(2n + 1)2(97)

and hence we obtain

∞∑n=0

22n sin2n+1 t

(2n + 1)2

(2nn

)−1

= −t∞∑

n=0

cos(2n + 1)t

2n + 1+

∞∑n=0

sin(2n + 1)t

(2n + 1)2(98)

In particular, we have from (97)

∫ π2

0

x

sin xdx = 2

∞∑n=0

(−1)n

(2n + 1)2= 2G (99)

and we then determine that

G =1

2

∞∑n=0

22n

(2n + 1)2

(2nn

)−1

(100)

in agreement with Sherman’s formula [42].

Using integration by parts Bradley [53] showed that

∫ z

0

log tan u du − z log tan z = −∫ z

0

u sec2 u

tan udu

= −1

4

∫ 2z

0

u du

tan(u/2) cos2(u/2)

= −1

2

∫ 2z

0

u

sin udu

= −∫ sin(2z)

0

2x sin−1 x√1 − x2

dx

x2

Differentiating (27) gives us

Some infinite series involving the Riemann zeta function 49

2x sin−1 x√1−x2 =

∑∞n=1

(2x)2n

n

(2nn

)−1

for |x| < 1 and integrating term by term

results in

∫ z

0

log tan u du − z log tan z = −1

4

∞∑n=0

(2 sin 2z)2n+1

(2n + 1)2

(2nn

)−1

for 0 ≤ z ≤ π/4.Therefore we obtain (95) again

∫ z

0

u

sin udu =

∞∑n=0

22n sin2n+1 z

(2n + 1)2

(2nn

)−1

We note from (102) below that

∫ t

0

x

sin xdx = t log tan

(t

2

)+ 2

∞∑n=0

sin(2n + 1)t

(2n + 1)2

and hence we have

t log tan t +∞∑

n=0

sin 2(2n + 1)t

(2n + 1)2=

∞∑n=0

22n sin2n+1 t

(2n + 1)2

(2nn

)−1

(101)

Reference to [44, p. 149] shows that for 0 < t < π (this may also be easilyderived using the methods outlined in [22])

∞∑n=0

cos(2n + 1)t

2n + 1= −1

2log tan

(t

2

)

and for 0 ≤ t ≤ π

∞∑n=0

sin(2n + 1)t

(2n + 1)2= −1

2

∫ t

0

log tan(x

2

)Referring to (97) we have

50 Donal F. Connon

∫ t

0

x

sin xdx = t log tan

(t

2

)+ 2

∞∑n=0

sin(2n + 1)t

(2n + 1)2(102)

and noting the Clausen function

Cl2(t) = −∫ t

0

log[2 sin(u

2)]du =

∞∑n=1

sin nt

n2

we see that

Cl2(π − t) =∞∑

n=1

sin n(π − t)

n2=

∞∑n=1

(−1)n sin nt

n2

Cl2(t) + Cl2(π − t) = 2∞∑

n=0

sin(2n + 1)t

(2n + 1)2

Hence, as shown by Lewin [39, p. 255] we have

∫ t

0

x

sin xdx = t log tan

(t

2

)+ Cl2(t) + Cl2(π − t) (103)

More generally, we have

∞∑n=0

sin(2n + 1)t

(2n + 1)2− t

∞∑n=0

cos(2n + 1)t

2n + 1=

∞∑n=0

22n sin2n+1 t

(2n + 1)2

(2nn

)−1

(104)

= t log tan

(t

2

)+ Cl2(t) + Cl2(π − t)

See also the recent paper by Cho et al. [16] who consider the integrals∫ t

0xm

sin xdx and, in particular,

∫ π3

0

x

sin xdx = −π

6log 3 −

√3

9π2 +

√3

18ζ

(2,

1

6

)

Some infinite series involving the Riemann zeta function 51

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

∫ t

0

sin2n+1 x dx =

∫ t

0

sin2n x sin x dx =

∫ t

0

(1 − cos2 x)n sin x dx

=

∫ t

0

n∑k=0

(nk

)(−1)k cos2k x sin x dx

and we therefore obtain

∫ t

0

sin2n+1 x dx =n∑

k=0

(nk

)(−1)k

2k + 1−

n∑k=0

(nk

)(−1)k cos2k+1 t

2k + 1

With t = π/2 we get

∫ π2

0

sin2n+1 x dx =n∑

k=0

(nk

)(−1)k

2k + 1

as compared with the frequently quoted form of the Wallis integral formula

∫ π2

0

sin2n+1 x dx =(2n)!!

(2n + 1)!!

We therefore have the following identity which is reported in [33, p. 270]

n∑k=0

(nk

)(−1)k 1

2k + 1=

(2n)!!

(2n + 1)!!=

[2nn!]2

(2n + 1)!(105)

so that

∫ π2

0

sin2n+1 x dx =[2nn!]2

(2n + 1)!(106)

52 Donal F. Connon

We recall (5)

n∑k=0

(nk

)xk

(k + y)s=

1

Γ(s)

∫ ∞

0

us−1e−yu(1 + xe−u

)ndu

and note that

n∑k=0

(nk

)(−1)k 1

2k + 1=

1

2

n∑k=0

(nk

)(−1)k 1

k + 12

=1

2

∫ ∞

0

e−u/2(1 − xe−u

)ndu

Integrating (104) gives us

2∞∑

n=0

1

(2n + 1)3− 2

∞∑n=0

cos(2n + 1)t

(2n + 1)3− t

∞∑n=0

sin(2n + 1)t

(2n + 1)2

=∞∑

n=0

22n

(2n + 1)2

(2nn

)−1 ∫ t

0

sin2n+1 x dx

and with t = π/2 we have

7

4ζ(3) +

π

2

∞∑n=0

(−1)n+1

(2n + 1)2=

∞∑n=0

22n

(2n + 1)2

(2nn

)−1 ∫ π2

0

sin2n+1 x dx

=∞∑

n=0

22n

(2n + 1)2

(2nn

)−1[2nn!]2

(2n + 1)!

Hence we obtain

7

4ζ(3) − 1

2πG =

∞∑n=0

24n

(2n + 1)3

[n!]4

[(2n)!]2(107)

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

∫ π2

0

x2 sin x dx =1

2

∞∑n=1

[n!]2

n2(2n)!22n

∫ π2

0

sin2n+1 x dx

and therefore we see that

π − 2 =1

2

∞∑n=1

24n [n!]4

n2(2n + 1) [(2n)!]2(108)

Let us now divide (28) by sin x and integrate to obtain

∫ π2

0

x2

sin xdx =

1

2

∞∑n=1

[n!]2

n2(2n)!22n

∫ π2

0

sin2n−1 x dx

In equation 6.29 in [22] we have previously shown that

∫ π2

0

x2

sin xdx = 2π G − 7

2ζ(3)

(the evaluation of this integral is also contained in [16]) and using the integralin [12, p. 113]Using

∫ π2

0

sin2n−1 xdx =n−1∑k=0

(n − 1

k

)(−1)k

2k + 1=

(2n − 2)!!

(2n − 1)!!

we obtain

8πG − 14ζ(3) =∞∑

n=1

24n [n!]4

n3 [(2n)!]2(109)

which was previously derived by Batir [9].Letting p(x) = x cos x in (96) gives us

∫ t

0

x cos x

sin xdx =

∫ t

0

x cot x dx = 2∞∑

n=0

∫ t

0

x cos x sin(2n + 1)x dx

54 Donal F. Connon

and we have using integration by parts for n ≥ 1

8

∫ t

0

x cos x sin(2n+1)x dx =sin(2nt)

n2−2t

cos(2nt)

n+

sin[2(n + 1)t]

(n + 1)2−2t

cos[2(n + 1)t]

n + 1

This gives us

2

∫ π2

0

x cos x sin(2n + 1)x dx =π

4

[(−1)n

n + 1− (−1)n

n

]

and thus

2∞∑

n=0

∫ π2

0

x cos x sin(2n + 1)x dx =π

2log 2

This, of course, concurs with the well-known Euler integral

∫ π2

0

x cot x dx =π

2log 2

Dividing (28) by sin2 x and integrating gives us

∫ t

0

x2

sin2 xdx =

1

2

∞∑n=1

[n!]2

n2(2n)!22n

∫ t

0

sin2n−2 x dx

and with t = π/2 we have after dealing separately with the case for n = 1

∫ π2

0

x2

sin2 xdx =

π

2+

1

2

∞∑n=2

[n!]2

n2(2n)!22n π

22n−1

(2n − 2n − 1

)

=π

2+ π

∞∑n=2

1

(2n − 1)(2n − 2)

=π

2+ π

∞∑n=2

(1

2n − 1− 1

2n − 2

)

Some infinite series involving the Riemann zeta function 55

=π

2+ π

∞∑n=1

(1

2n + 1− 1

2n

)

=π

2+

π

2

∞∑n=1

(1

n + 12

− 1

n

)

Since the digamma function may be represented by [43, p. 14]

ψ(a) = −γ − 1

a−

∞∑n=1

(1

n + a− 1

n

)

we see that

∞∑n=1

(1

n + 12

− 1

n

)= −ψ

(1

2

)− γ − 2

We know that [43, p. 20]

ψ

(1

2

)= −γ − 2 log 2

and we therefore deduce that

∫ π2

0

x2

sin2 xdx = π log 2

Using integration by parts we see that

∫ π2

0

x2

sin2 xdx = 2

∫ π2

0

x cot x dx

and thus we have

∫ π2

0

x cot x dx = −∫ π

2

0

log sin x dx

Euler [25] was the first person to show that

56 Donal F. Connon

∫ π2

0

log sin x dx = −π

2log 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

1

2

∫ x

0

un cot(u

2

)du = cos

(nπ

2

)n!ζ(n+1)−

n∑j=0

(−1)j(j+1)/2 Γ(n + 1)

Γ(n + 1 − j)xn−jClj+1(x)

(110)and also see the paper by Srivastava et al [61]. Ramanujan’s formula wasemployed by Muzaffar [40]. With x = π we have

2n

∫ π2

0

xn cot x dx = cos(nπ

2

)n!ζ(n+1)−

n∑j=0

(−1)j(j+1)/2 Γ(n + 1)

Γ(n + 1 − j)πn−jClj+1(π)

(111)Provided a �= b we readily determine that∫

sin ax sin bx dx =1

2

[sin(a − b)x

a − b− sin(a + b)x

a + b

]+ c

and hence we have

∫ t

0

sin px sin(2n + 1)x dx =1

2

[sin(p − 2n − 1)t

p − 2n − 1− sin(p + 2n + 1)t

p + 2n + 1

]

and more specifically

∫ π2

0

sin px sin(2n + 1)x dx =(−1)n+1 cos(pπ/2)

2

[1

p − 2n − 1+

1

p + 2n + 1

]

Therefore with p(x) = sin px in (96) we obtain

Some infinite series involving the Riemann zeta function 57

∫ π2

0

sin px

sin xdx = 2

∞∑n=0

∫ π2

0

sin px sin(2n + 1)x dx

so that

∫ π2

0

sin px

sin xdx = 2p cos(pπ/2)

∞∑n=0

(−1)n

(2n + 1)2 − p2(112)

and, using L’Hopital’s rule, we see that in the limit as p → 1

∫ π2

0

dx = limp→1

−pπ sin(pπ/2) + 2 cos(pπ/2)

−2p=

π

2

With p = 2 we obtain

∞∑n=0

(−1)n+1

(2n + 1)2 − 42=

1

2(113)

We may write (112) as

∞∑n=0

(−1)n

(2n + 1)2 − p2=

1

2p cos(pπ/2)

∫ π2

0

sin px

sin xdx

and, using L’Hoital’s rule, we obtain as p → 0

∞∑n=0

(−1)n

(2n + 1)2= lim

p→0

[1

−pπ sin(pπ/2) + 2 cos(pπ/2)

∫ π2

0

x cos px

sin xdx

]

Hence we obtain (99) again

2∞∑

n=0

(−1)n

(2n + 1)2=

∫ π2

0

x

sin xdx

We note from Fettis [56] that

58 Donal F. Connon

∫ π2

0

sin px

sin xdx =

π

2− 1

2cos

(pπ

2

) [ψ

(3 + p

4

)− ψ

(1 + p

4

)](114)

and

∫ π2

0

1 − cos px

sin xdx = ψ

(1 + p

2

)−ψ

(1

2

)−1

2sin

(pπ

2

) [ψ

(3 + p

4

)− ψ

(1 + p

4

)](115)

Comparing (112) with (114) gives us

2p cos(pπ/2)∞∑

n=0

(−1)n

(2n + 1)2 − p2=

π

2−1

2cos

(pπ

2

) [ψ

(3 + p

4

)− ψ

(1 + p

4

)](116)

Differentiating (114) results in

∫ π2

0

x cos px

sin xdx

= −1

8cos

(pπ

2

) [ψ′

(3 + p

4

)− ψ′

(1 + p

4

)]+

π

4sin

(pπ

2

) [ψ

(3 + p

4

)− ψ

(1 + p

4

)]

and with p = 0 we obtain

∫ π2

0

x

sin xdx = −1

8

[ψ′

(3

4

)− ψ′

(1

4

)]

Using (99) we see that

2G = −1

8

[ψ′

(3

4

)− ψ′

(1

4

)](117)

which concurs with Kolbig [58].

With p = 1 we have the well known integral

Some infinite series involving the Riemann zeta function 59

∫ π2

0

x cot x dx =π

4

[ψ (1) − ψ

(1

2

)]=

π

2log 2

With p = 2 we obtain

∫ π2

0

[ x

sin x− 2x sin x

]dx =

1

8

[ψ′

(1 +

1

4

)− ψ′

(3

4

)]and since ψ′ (1 + 1

4

)= ψ′ (1

4

) − 16 we obtain (117) again.Differentiating (115) results in

∫ π2

0

x sin px

sin xdx =

1

2ψ′

(1 + p

2

)− 1

8sin

(pπ

2

) [ψ′

(3 + p

4

)− ψ′

(1 + p

4

)]

−π

4cos

(pπ

2

) [ψ

(3 + p

4

)− ψ

(1 + p

4

)]and with p = 0 we obtain

1

2ψ′

(1

2

)=

π

4

[ψ

(3

4

)− ψ

(1

4

)]so that

ψ′(

1

2

)=

π2

2(118)

in accordance with Kolbig’s paper [58].With p = 1 we obtain

π2

8=

1

2ψ′ (1) − 1

8

[ψ′ (1) − ψ′

(1

2

)]and, since ψ′ (1) = ζ(2), this concurs with (118).We have provided p �= 2n + 1

∫ t

0

cos px sin(2n+1)x dx =1

2

[cos(p − 2n − 1)t

p − 2n − 1− cos(p + 2n + 1)t

p + 2n + 1

]− 2n + 1

p2 − (2n + 1)2

60 Donal F. Connon

and thus

∫ π2

0

cos px sin(2n+1)x dx =(−1)n+1

2

[1

p − 2n − 1+

1

p + 2n + 1

]− 2n + 1

p2 − (2n + 1)2

=(−1)n+1p − (2n + 1)

p2 − (2n + 1)2

Therefore letting p(x) = 1 − cos px in (96) gives us

∫ π2

0

1 − cos px

sin xdx = 2

∞∑n=0

[1

2n + 1+

(−1)np + (2n + 1)

p2 − (2n + 1)2

](119)

and comparing this with (115) results in

2∞∑

n=0

[1

2n + 1+

(−1)np + (2n + 1)

p2 − (2n + 1)2

]

= ψ

(1 + p

2

)− ψ

(1

2

)− 1

2sin

(pπ

2

)[ψ

(3 + p

4

)− ψ

(1 + p

4

)]

Multiplying across by cos(

pπ2

)gives us

2 cos(pπ

2

) ∞∑n=0

[1

2n + 1+

(−1)np + (2n + 1)

p2 − (2n + 1)2

]

= cos(pπ

2

) [ψ

(1 + p

2

)− ψ

(1

2

)]−1

2cos

(pπ

2

)sin

(pπ

2

)[ψ

(3 + p

4

)− ψ

(1 + p

4

)]

We have

2 cos(pπ

2

) ∞∑n=0

[1

2n + 1+

(−1)np + (2n + 1)

p2 − (2n + 1)2

]

Some infinite series involving the Riemann zeta function 61

=π

2−1

2cos

(pπ

2

) [ψ

(3 + p

4

)− ψ

(1 + p

4

)]+2 cos

(pπ

2

) ∞∑n=0

[1

2n + 1+

2n + 1

p2 − (2n + 1)2

]

=π

2−1

2cos

(pπ

2

) [ψ

(3 + p

4

)− ψ

(1 + p

4

)]+2p2 cos

(pπ

2

) ∞∑n=0

[1

2n + 1

1

p2 − (2n + 1)2

]

and therefore we see that

2p2

∞∑n=0

1

2n + 1

1

p2 − (2n + 1)2(120)

= ψ

(1 + p

2

)−ψ

(1

2

)+

1

2

[1 − sin

(pπ

2

)] [ψ

(3 + p

4

)− ψ

(1 + p

4

)]−π

2/ cos

(pπ

2

)This is reminiscent of the expression in Prudnikov et al. [41a, 5.1.25-13]

2p2

∞∑n=1

1

n

1

p2 − n2= ψ(1 + p) + ψ(1 − p) + 2γ (121)

and separating the even and odd terms gives us

∞∑n=1

1

n

1

p2 − n2=

∞∑n=1

1

2n

1

p2 − 4n2+

∞∑n=0

1

2n + 1

1

p2 − (2n + 1)2

Letting p → p/2 in (121) we see that

ψ(1 +

p

2

)+ ψ

(1 − p

2

)+ 2γ = 2p2

∞∑n=1

1

n

1

p2 − 4n2

and hence we obtain

ψ(1 + p) + ψ(1 − p) + 2γ − 1

2

[ψ

(1 +

p

2

)+ ψ

(1 − p

2

)+ 2γ

]

62 Donal F. Connon

= ψ

(1 + p

2

)−ψ

(1

2

)+

1

2

[1 − sin

(pπ

2

)] [ψ

(3 + p

4

)− ψ

(1 + p

4

)]−π

2/ cos

(pπ

2

)

which may be written as

ψ(1 + p) + ψ(1 − p) − 1

2

[ψ

(1 +

p

2

)+ ψ

(1 − p

2

)](122)

= ψ

(1 + p

2

)+2 log 2+

1

2

[1 − sin

(pπ

2

)] [ψ

(3 + p

4

)− ψ

(1 + p

4

)]−π

2/ cos

(pπ

2

)

For example, with p = 1/2 we obtain

ψ

(1 +

1

2

)+ ψ

(1

2

)− 1

2

[ψ

(1 +

1

4

)+ ψ

(3

4

)]

= ψ

(3

4

)+ 2 log 2 +

1

2

[1 − 1

2

√2

] [ψ

(7

8

)− ψ

(3

8

)]− π

√2

2

and this may be verified by substituting the specific values for the digammafunction in [43, p. 20] (albeit one of the relevant values is reported incorrectlyin [43, p. 20]).Ramanujan [11, Part I, p. 263] reported the following Maclaurin series (inthe slightly modified form employed by Borwein and Chamberland [51])

(sin−1 y

)4=

3

2

∞∑n=1

H(2)n−1

n2

(2nn

)−1

(2y)2n,−1 ≤ y ≤ 1 (123)

where H(m)n is the generalized harmonic number

H(m)n =

n∑k=1

1

km

Therefore upon letting x = sin−1 y we get

Some infinite series involving the Riemann zeta function 63

x4 =3

2

∞∑n=1

H(2)n−1

n2

(2nn

)−1

22n sin2n x,−π/2 ≤ x ≤ π/2 (124)

Integration results in

π5

160=

3

2

∞∑n=1

H(2)n−1

n2

(2nn

)−1

22n

∫ π2

0

sin2n x dx

and hence we obtain the Euler sum

π5

160=

3π

4

∞∑n=1

H(2)n−1

n2

or equivalently

π4

120=

∞∑n=1

H(2)n−1

n2=

∞∑n=1

H(2)n

n2− ζ(4)

Since ζ(4) = π4

90this may be written in its more familiar form as (see for

example [19])

7

4ζ(4) =

∞∑n=1

H(2)n

n2

Alternatively, we now multiply equation (124) by cotx and integrate to ob-tain

∫ t

0

x4 cot x dx =3

2

∞∑n=1

H(2)n−1

n2

(2nn

)−1

22n

∫ t

0

sin2n−1 x cos x dx

which results in

∫ t

0

x4 cot x dx =3

4

∞∑n=1

22nH(2)n−1 sin2n t

n3

(2nn

)−1

(125)

64 Donal F. Connon

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

4t(t2 − 6) cos t + (t4 − 12t2 + 24) sin t =3

2

∞∑n=1

22nH(2)n−1

(2n + 1)n2

(2nn

)−1

sin2n+1 t

and with t = π/2 we obtain

π4

16− 3π2 + 24 =

3

2

∞∑n=1

22nH(2)n−1

(2n + 1)n2

(2nn

)−1

(126)

We showed in [19] that for 0 ≤ t < 1

∫ t

0

πx cot πx dx = ζ ′(−1, t) − ζ ′(−1, 1 − t) + t log(2 sin πt) (127)

and we now refer to the well-known formula [12, p. 130]

πx cot πx = −2∞∑

n=0

ζ(2n)x2n, (|x| < 1) (128)

(where ζ(0) = −1/2). This gives us

−2∞∑

n=0

ζ(2n)

2n + 1t2n+1 = ζ ′(−1, t) − ζ ′(−1, 1 − t) + t log(2 sin πt) (129)

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

0

[xψ(1 − x) − xψ(x)]dx = ζ ′(−1, t) − ζ ′(−1, 1 − t) + t log(2 sin πt)

Some infinite series involving the Riemann zeta function 65

Integration by parts results in

∫ t

0

[xψ(1 − x) − xψ(x)]dx = −t log[Γ(t)Γ(1 − t)] +

∫ t

0

log[Γ(x)Γ(1 − x)] dx

where upon using Euler’s reflection formula for the gamma function [43, p.3]

Γ(t)Γ(1 − t) =π

sin πtthis becomes

= t log sin πt −∫ t

0

log sin πx dx

Therefore, as noted in equation (4.3.158a) in [19], we have∫ t

0

log[2 sin πx] dx = −[ζ ′(−1, t) − ζ ′(−1, 1 − t)] (130)

which could of course have been obtained more directly by using integrationby 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

∫ 12

0

log sin πx dx = −1

2log 2

Let us now differentiate (129) to obtain

−2∞∑

n=0

ζ(2n)t2n =d

dt[ζ ′(−1, t) − ζ ′(−1, 1 − t)] + log(2 sin πt) + π cot πt

Since the Hurwitz zeta function is analytic in the whole complex plane exceptfor s �= 1, its partial derivatives commute in the region where the function isanalytic: we therefore have

66 Donal F. Connon

∂

∂t

∂

∂sζ(s, t) =

∂

∂s

∂

∂tζ(s, t) = − ∂

∂s[sζ(s + 1, t)]

and we then see that

∂

∂t

∂

∂sζ(s, t) = −ζ(s + 1, t) − s

∂

∂sζ(s + 1, t) (131)

Hence with s = −1 we see that

d

dtζ ′(−1, t) = ζ ′(0, t) − ζ(0, t)

d

dtζ ′(−1, 1 − t) = −ζ ′(0, 1 − t) + ζ(0, 1 − t)

and we then obtain

−2∞∑

n=0

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

Using Lerch’s identity [10] for t > 0

ζ ′(0, t) = log Γ(t) − 1

2log(2π)

and the well-known result [7, p. 264] involving the Bernoulli polynomialsBn(t)

ζ(1 − n, t) = −Bn(t)n

for n ≥ 1this becomes

−2∞∑

n=0

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

Using Euler’s reflection formula for the gamma function

Γ(t)Γ(1 − t) =π

sin πt

Some infinite series involving the Riemann zeta function 67

we return to where we started from

−2∞∑

n=0

ζ(2n) t2n = πt cot πt

Dividing this by t, dealing separately with the n = 0 term, and integratingresults in

∞∑n=1

ζ(2n)t2n

n= log t − log sin πt + c

and in the limit as t → 0 we see that the integration constant is c = log π.We thus obtain the known result [43, p. 161]

∞∑n=1

ζ(2n)t2n

n= log(πt) − log sin πt

We now multiply this by t and integrate to obtain

∞∑n=0

ζ(2n)

n(2n + 2)u2n+2 = u log(πu) − u −

∫ u

0

t log sin πt dt

and, in the case u = 1/2, we obtain using (31)

ζ(3) =2π2

7

[log π − 1

2−

∞∑n=1

ζ(2n)

n(n + 1)22n

]

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 formof Euler’s infinite product identity for the sine function

log sin x = log x +∞∑

n=1

log

[1 − x2

n2π2

]

Upon integrating (131) with respect to t we see that

−s

∫ v

0

ζ ′(s + 1, t)dt =

∫ v

0

∂

∂t

∂

∂sζ(s, t)dt +

∫ v

0

ζ(s + 1, t) dt

68 Donal F. Connon

We therefore get

−s

∫ v

0

ζ ′(s + 1, t)dt = ζ ′(s, v) − ζ ′(s, 0) +

∫ v

0

ζ(s + 1, t) dt

and with s = −n we have

n

∫ v

0

ζ ′(1 − n, t)dt = ζ ′(−n, v) − ζ ′(−n, 0) +

∫ v

0

ζ(1 − n, t) dt

Then using the well-known formulaζ(1 − n, v) = −Bn(v)

nfor n ≥ 1

we obtain

n

∫ v

0

ζ ′(1 − n, t) dt =Bn+1 − Bn+1(v)

n(n + 1)+ ζ ′(−n, v) − ζ ′(−n, 0) (132)

This identity was originally derived by Adamchik [1] in a different mannerin 1998. With n = 2 we obtain∫ v

0

ζ ′(−1, t) dt = − 1

12B3(v) +

1

2ζ ′(−2, v) +

ζ(3)

8π2(133)

since ζ ′(−n, 0) = ζ ′(−n) and ζ ′(−2) = − ζ(3)4π2 .

With v = 1 we note that ∫ 1

0

ζ ′(1 − n, t) dt = 0 (134)

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 thetriple sine function. The multiple sine functions are defined for r = 2, 3, · · ·by

Some infinite series involving the Riemann zeta function 69

Sr(x) = exp

(xr−1

r − 1

) ∞∏n=−∞,n�=0

[Pr

(x

n

)]nr−1

(135)

= exp

(xr−1

r − 1

) ∞∏n=1

[Pr

(x

n

)Pr

(−x

n

)]nr−1

where

Pr(u) = (1 − u) exp

(u

1+

u2

2+ · · · + ur

r

)For example, the triple sine function is defined by

S3(x) = ex2

2

∞∏n=1

[(1 − x2

n2

)n2

ex2

](136)

and we then have

log S3(x) =1

2x2 +

∞∑n=1

[n2 log

(1 − x2

n2

)+ x2

]

where upon differentiation results in

S ′3(x)

S3(x)= x +

∞∑n=1

[2n2x

x2 − n2+ 2x

]

= x + x2

∞∑n=1

2x

x2 − n2

= x2

(1

x+

∞∑n=1

2x

x2 − n2

)

We have the well-known decomposition formula [12, p.131]

π cot πx =1

x+

∞∑n=1

2x

x2 − n2(137)

70 Donal F. Connon

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

log S3(x) =

∫ x

0

πt2 cot πt dt (138)

The double sine function is defined by

S2(x) = ex

∞∏n=1

[(1 − x/n

1 + x/n

)n

e2x

](139)

It is easily seen that

log S2(x) = x +∞∑

n=1

[n

{log

(1 − x

n

)− log

(1 − x

n

)}+ 2x

](140)

and, in the same manner as before, we easily find that

log S2(x) =

∫ x

0

πt cot πt dt (141)

The function S1(x) is defined by

S1(x) = 2πx∞∏

n=1

[(1 − x2

n2

)]= 2 sin πx (142)

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]

G(1 + x) = (2π)x2 exp[−1

2(γx2 + x2 + x)]

∞∏k=1

[(1 +x

k)k exp(

x2

2k− x)]

(143)

We showed in [23] that the Barnes double gamma function could be repre-sented by

Some infinite series involving the Riemann zeta function 71

log G(1+x) =1

2x log(2π)−1

2x(1+x)−1

2γx2 =

∞∑n=1

[1

2nx2 − x + n log

(1 +

x

n

)](144)

and letting x → −x gives us

log G(1−x) = −1

2x log(2π)+

1

2x(1−x)−1

2γx2 =

∞∑n=1

[1

2nx2 + x + n log

(1 − x

n

)]

Subtraction results in

log G(1 − x) − log G(1 + x)

= x − x log(2π) +∞∑

n=1

[2x + n log

(1 − x

n

)− n log

(1 +

x

n

)]

and, using (140), we therefore see that

log S2(x) = log G(1 − x) − log G(1 + x) + x log(2π) (145)

We have therefore rediscovered the well-known formula originally found byKinkelin in 1860 [43, p. 30]

logG(1 + x)

G(1 − x)= x log(2π) −

∫ x

0

πt cot πt dt (146)

which was generalized in 1992 by Freund and Zabrodin [27] who reported themore general identity for

Γn(u)[Γn(−u)](−1)n+1

= exp[−π

∫ u

0

xn−1 cot πxdx]

for n ≥ 2.

(147)

72 Donal F. Connon

where we have followed the Vigneras notation [43, p. 39]

Γn(x) = [Gn(x)](−1)n−1

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

log Sn(x) =

∫ x

0

πtn−1 cot πt dt (148)

Reference should also be made to Kurokawa’s paper [36].It is an exercise in Bromwich’s book [15, p. 526] to show that

∞∑n=1

[n log

(2n + 1

2n − 1

)− 1

]=

1

2(1 − log 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]

log Γ(x + 1) =∞∑

n=1

[x log

(1 +

1

n

)− log

(1 +

x

n

)]

and this may be written as

log Γ(x + 1) =∞∑

n=1

[x log

(1 +

1

n

)+

∞∑k=1

(−1)kxk

knk

]

=∞∑

n=1

[x

[log

(1 +

1

n

)− 1

n

]+

∞∑k=2

(−1)kxk

knk

]

=∞∑

n=1

[log

(1 +

1

n

)− 1

n

]x +

∞∑n=1

∞∑k=2

(−1)kxk

knk

=∞∑

n=1

[log

(1 +

1

n

)− 1

n

]x +

∞∑k=2

(−1)kxk

k

∞∑n=1

1

nk

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

log Γ(x + 1) = −γx +∞∑

k=2

(−1)kζ(k)

kxk (149)

Now referring back to (144)

log G(1+x) =1

2x log(2π)−1

2x(1+x)−1

2γx2+

∞∑n=1

[1

2nx2 − x + n log

(1 +

x

n

)]

and applying the same procedure as with log Γ(x + 1) above this becomes

=1

2x log(2π) − 1

2x(1 + x) − 1

2γx2 +

∞∑n=1

[1

2nx2 − x −

∞∑k=1

(−1)kxk

knk−1

]

=1

2x log(2π) − 1

2x(1 + x) − 1

2γx2 −

∞∑n=1

∞∑k=3

(−1)kxk

knk−1

=1

2x log(2π) − 1

2x(1 + x) − 1

2γx2 −

∞∑n=1

∞∑m=2

(−1)m+1xm+1

(m + 1)nm

=1

2x log(2π) − 1

2x(1 + x) − 1

2γx2 −

∞∑m=2

(−1)m+1xm+1

m + 1

∞∑n=1

1

nm

=1

2x log(2π) − 1

2x(1 + x) − 1

2γx2 −

∞∑m=2

(−1)m+1ζ(m)xm+1

m + 1

Hence, as reported in [43, p. 211], we obtain

log G(1+x) =1

2x log(2π)− 1

2x(1+x)− 1

2γx2 −

∞∑k=2

(−1)k+1ζ(k)xk+1

k + 1(150)

74 Donal F. Connon

We now recall the Gosper/Vardi functional equation [5] for the double gammafunction (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

2x log(2π)−1

2x(1+x)−1

2γx2−

∞∑k=2

(−1)k+1ζ(k)

k + 1xk+1+γx2+

∞∑k=2

(−1)k+1ζ(k)

kxk+1

= ζ ′(−1) − ζ ′(−1, 1 + x)

This may be written as

ζ ′(−1)−ζ ′(−1, 1+x) =1

2x log(2π)− 1

2x(1+x)+

1

2γx2+

∞∑k=2

(−1)k+1ζ(k)

k(k + 1)xk+1

(152)which is contained in a slightly disguised form in [43, p. 222].Differentiation results in

− d

dxζ ′(−1, 1 + x) =

1

2log(2π) − 1

2− x + γx +

∞∑k=2

(−1)k+1ζ(k)

kxk

and as shown previously we have

d

dxζ ′(−1, 1 + x) = ζ ′(0, 1 + x) − ζ(0, 1 + x)

= log Γ(1 + x) − 1

2log(2π) +

1

2+ x

Therefore we simply recover (149)

log Γ(x + 1) = −γx +∞∑

k=2

(−1)kζ(k)

kxk

Some infinite series involving the Riemann zeta function 75

We have

log S3(x) =1

2x2 +

∞∑n=1

[n2 log

(1 − x2

n2

)+ x2

]

=1

2x2 +

∞∑n=1

[x2 −

∞∑k=1

x2k

kn2k−2

]

=1

2x2 −

∞∑n=1

∞∑k=2

x2k

kn2k−2=

1

2x2 −

∞∑k=2

x2k

k

∞∑n=1

1

n2k−2

=1

2x2 −

∞∑k=2

ζ(2k − 2)

kx2k

This gives us

log S3(x) =1

2x2 −

∞∑k=1

ζ(2k)

k + 1x2k+2 = −

∞∑k=0

ζ(2k)

k + 1x2k+2 (153)

The following identity for |x|<1 appears in the book by Srivastava and Choi[43, p. 216]

∞∑k=1

ζ(2k)

k + 1x2k+2 =

1

2[1−log(2π)]x2+x log

G(1 + x)

G(1 − x)−

∫ x

0

log G(1+t) dt−∫ −x

0

log G(1+t) dt

We note that

∫ −x

0

log G(1 + t) dt = −∫ x

0

log G(1 − t) dt

and integration by parts shows that

x logG(1 + x)

G(1 − x)−

∫ x

0

log G(1+t)dt−∫ −x

0

log G(1+t)dt =

∫ x

0

td

dtlog

G(1 + t)

G(1 − t)dt

76 Donal F. Connon

Using Kinkelin’s formula (146) we see that

d

dtlog

G(1 + t)

G(1 − t)= log(2π) − πt cot πt

and hence we have∫ x

0

td

dtlog

G(1 + t)

G(1 − t)dt =

1

2log(2π)x2 −

∫ x

0

πt2 cot πt dt

This results in

∞∑k=1

ζ(2k)

k + 1x2k+2 =

1

2x2 −

∫ x

0

πt2 cot πt dt (154)

We have the following well-known identity

t cot t =∞∑

n=0

(−1)n 22nB2n

(2n)!t2n, (|t| < π) (155)

Combining this with Euler’s formula

ζ(2n) = (−1)n+1 22n−1π2nB2n

(2n)!, (n ≥ 1) (156)

and, letting t → πt, we obtain

πt cot πt = −2∞∑

n=0

ζ(2n) t2n, (|t| < 1) (157)

Since the first term of the series (155) is equal to B0 = 1, to be consistentwith (157), we define ζ(0) = −1

2(which in fact also coincides with the value

determined by the analytic continuation of ζ(s) at s = 0.) We may nowmultiply (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

0

log G(1+t)dt =

(1

4− 2 log A

)x+

1

4log(2π)x2− 1

6x3−(1−x) log G(1+x)

+ log G(x) − 2 log Γ3(1 + x) + 2 log Γ3(x)

and hence we may express the triple sine function in terms of the double andtriple gamma functions. This type of representation naturally arises fromthe combination of the facts that

log Sn(x) =

∫ x

0

πtn−1 cot πt dt

− cot πt = ψ(t) − ψ(−t) +1

t

and the knowledge that integrals of the form∫ x

0tn−1ψ(t)dt result in multiple

gamma functions [43, p. 208].The triple gamma function is defined by [17]

Γ3(1 + x) = exp(c1x + c2x2 + c3x

3)F (x)

where

F (x) =∞∏

k=1

{(1 +

x

k

)− 12k(k+1)

exp

(1

2(k + 1)x − 1

4

(1 +

1

k

)x2 +

1

6

x3

k+

1

6

x3

k2

)}

=∞∏

k=1

{(1 +

x

k

)− 12k(k+1)

exp

([1 +

1

k

] {1

2kx − 1

4x2 +

1

6

x3

k

})}

and

c1 =3

8− 1

4log(2π) − log A, c2 =

1

4

[γ + log(2π) +

1

2

]

c3 = −1

6

[γ + ζ(2) +

3

2

]

78 Donal F. Connon

References

[1] V. S. Adamchik, Polygamma Functions of Negative Order, J. Comp.and Applied Math., 100, (1998), 191-199.

[2] V. S. Adamchik, A Class of Logarithmic Integrals, Proceedings of the1997 International Symposium on Symbolic and Algebraic Computation,ACM, Academic Press, 2001, 1-8.http://www-2.cs.cmu.edu/˜adamchik/articles/logs.htm

[3] V. S. Adamchik, Certain Series Associated with Catalan’s Constant,Journal for Analysis and its Applications (ZAA), 21, 3 (2002), 817-826.http://www-2.cs.cmu.edu/˜adamchik/articles/csum.html

[4] V. S. Adamchik, Contributions to the Theory of the Barnes Function,Computer Physics Communications, 2003.http://www-2.cs.cmu.edu/˜adamchik/articles/barnes1.pdf

[5] V. S. Adamchik, The multiple gamma function and its application tocomputation of series, The Ramanujan Journal, 9, (2005), 271-288.http://www-2.cs.cmu.edu/˜adamchik/Barnes/

[6] H. Alzer, S. Koumandos, Series representations for γ and other mathe-matical constants, Analysis Mathematica, 34, (2008), 1-8.

[7] T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, Heidelberg and Berlin, 1976.

[8] R. Ayoub, Euler and the Zeta Function, Amer. Math. Monthly, 81,(1974), 1067-1086.

[9] N. Batir, Integral representations of some series involving∑∞k=1

(2kk

)−1

k−n and some related series, Appl. Math. and

Comput., 147, (2004), 645-667.

[10] B. C. Berndt, The Gamma Function and the Hurwitz Zeta Function,Amer. Math. Monthly, 92, (1985), 126-130.

Some infinite series involving the Riemann zeta function 79

[11] B. C. Berndt, Ramanujan’s Notebooks, Parts I-III, Springer-Verlag,1985-1991.

[12] G. Boros, V. H. Moll, Irresistible Integrals: Symbolics, Analysis andExperiments in the Evaluation of Integrals, Cambridge University Press,2004.

[13] D. M. Bradley, Representations of Catalan’s constant, (an unpublishedcatalogue of formulae for the alternating sum of the reciprocals of theodd positive squares), 1998.http://germain.umemat.maine.edu/faculty/bradley/papers/pub.html

[14] D. M. Bradley, A. E. Ozluk, C. Snyder, On a class number formula forreal quadratic number fields, Bulletin of the Australian MathematicalSociety, 65, no. 2, (2002), 259-270.

[15] T. J. I’a Bromwich, Introduction to the theory of infinite series, 2ndedition, Macmillan & Co Ltd, 1965.

[16] Y. J. Cho, M. Jung, J. Choi, H. M. Srivastava, Closed-form evaluationsof definite integrals and associated infinite series involving the Riemannzeta function, Int. J. Comput. Math., 83, nos. 5-6, (2006), 461-472.

[17] J. Choi, Y. J. Cho, H. M. Srivastava, Series involving the Zeta functionand multiple Gamma functions, Appl. Math. Comput., 159, (2004),509-537.

[18] J. Choi, H. M. Srivastava, V. S. Adamchik, Multiple Gamma and Re-lated Functions, The Ramanujan Journal, 2003.

[19] D. F. Connon, Some series and integrals involving the Riemann zetafunction, binomial coefficients and the harmonic numbers, Volume II(a),2007.

[20] D. F. Connon, Some series and integrals involving the Riemann zetafunction, binomial coefficients and the harmonic numbers, Volume II(b),2007.

80 Donal F. Connon

[21] D. F. Connon, Some series and integrals involving the Riemann zetafunction, binomial coefficients and the harmonic numbers, Volume III,2007.

[22] D. F. Connon, Some series and integrals involving the Riemann zetafunction, binomial coefficients and the harmonic numbers, Volume V,2007.

[23] D. F. Connon, Some series and integrals involving the Riemann zetafunction, binomial coefficients and the harmonic numbers, Volume VI,2007.

[24] O. Espinosa, V. H. Moll, On some integrals involving the Hurwitz zetafunction: Part 2, The Ramanujan Journal, 6, (2002), 449-468.http://www.math.tulane.edu/˜vhm/papers html/hurwitz2.pdf

[25] L. Euler, De summis serierum numeros Bernoullianos involventium, Novicommentarii acadamiae scienrarum Petropolitanae, 14, (1769), 129-167.

[26] L. Euler, Exercitiatones analyticae,Novi commentarii acadamiae scien-rarum Petropolitanae, 17, (1772), 173-204.

[27] P. G. O. Freund, A. V. Zabrodin, A Hierarchical Array of IntegrableModels, J. Math. Phys., 34, (1993), 5832-5842.

[28] K. Fujii, T. Suzuki, Introduction of an elementary method to expressζ(2n + 1) in terms of ζ(2k) with k ≥ 1, 2008.

[29] B. Ghusayni, Some Representations of ζ(3), Missouri Journal of Math-ematical Sciences, 10, (1998), 169-175.

[30] B. Ghusayni,. Euler-type formula using Maple, Palma Research Journal,7, (2001), 175-180.

[31] F. Haring, G. T. Nelson, G. Bach, ζ(n), ψ(n) and an Infinite Series,Amer. Math. Monthly, 81, (1974), 180-181.

Some infinite series involving the Riemann zeta function 81

[32] H. Hasse, Ein Summierungsverfahren fr Die Riemannsche ζ- Reithe,Math. Z., 32, (1930), 458-464.http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?ht=VIEW&did=D23956&p=462

[33] K. Knopp, Theory and Application of Infinite Series, Second EnglishEd., Dover Publications Inc, New York, 1990.

[34] S. Koyama,N. Kurokawa, Certain Series Related to the Triple Sine Func-tion, 2003.http://131.113.30.71:8080/dspace/bitstream/2302/703/1/03002.pdf

[35] S. Koyama, N. Kurokawa, Euler’s integrals and multiple sine functions,Proc. Amer. Math. Soc., 133, (2004), 1257-1265.

[36] N. Kurokawa, Multiple sine functions and Selbergzeta functions, Proc. Japan Acad., 67, (1991), 61-64.http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pja/1195512182

[37] N. Kurokawa, M. Wakayama, Extremal values of double and tripletrigonometric functions, Kyushu J. Math., 58, (2004), 141-166.http://www.jstage.jst.go.jp/article/kyushujm/58/1/58 141/ article

[38] D. H. Lehmer, Interesting series involving the central binomial coeffi-cient, Amer. Math. Monthly, 92, (1985), 449-457.

[39] L. Lewin, Polylogarithms and Associated Functions, Elsevier (North-Holland), New York, London and Amsterdam, 1981.

[40] H. Muzaffar, Some interesting series arising from the power series expan-sion of

(sin−1 x

)q, Int. J. of Math. and Math. Sci., 14, (2005), 2329-2336.

[41] S. Plouffe, The art of inspired guessing, 1998.

[42] T. Sherman, Summation of Glaisher- and Apery-like series, 2000.http://math.arizona.edu/˜rta/001/sherman.travis/series.pdf

[43] H. M. Srivastava, J. Choi, Series Associated with the Zeta and RelatedFunctions, Kluwer Academic Publishers, Dordrecht, Netherlands, 2001.

82 Donal F. Connon

[44] G. P. Tolstov, Fourier Series, (Translated from the Russian by R. A.Silverman), Dover Publications Inc, New York, 1976.

[45] A. van der Poorten, Some wonderful formulae. . . an introduction to Poly-logarithms, Queen’s Papers in Pure Appl. Math., 54, (1979), 269-286.http://www.ega-math.narod.ru/Apery2.htm#ref10txt

[46] E. T. Whittaker, G. N. Watson, A Course of Modern Analysis: AnIntroduction to the General Theory of Infinite Processes and of AnalyticFunctions; With an Account of the Principal Transcendental Functions.Fourth Ed., Cambridge University Press, Cambridge, London and NewYork, 1963.

[47] J. Wiener, Integrals of cos2n x and sin2n x, The College MathematicsJournal, 31, (2000), 60-61.

[48] Zhang Nan-Yue, K. S. Williams, Values of the Riemann zeta functionand integrals involving log

(2 sinh θ

2

)and log

(2 sin θ

2

), Pacific J. Math.,

168, (1995), 271-289.http://projecteuclid.org/Dienst/UI/1.0/Summarize/euclid.pjm/1102620561

[49] I. J. Zucker, On the series∑∞

k=1

(2kk

)−1

k−n and related sums, J.

Number Theory, 20, (1985), 92-102.

[50] D. H. Bailey, J.M. Borwein, V. Kapoor, E. Weisstein, Ten problemsin experimental mathematics, Amer. Math. Monthly, June-July 2006,481-509.

[51] J. M. Borwein, M. Chamberland, Integer powers of arcsin, InternationalJournal of Mathematics and Mathematical Sciences, Volume 2007,Article ID 19381, 10 pages, doi:10.1155/2007/19381.http://www.math.helsinki.fi/EMIS/journals/HOA/IJMMS/Volume2007/19381.pdf

[52] M. T. Boudjelkha, A proof that extends Hurwitz formula into the criticalstrip, Applied Mathematics Letters, 14, (2001) 309-403.

Some infinite series involving the Riemann zeta function 83

[53] D. M. Bradley, A class of series acceleration formulae for Catalan’s con-stant, The Ramanujan Journal, 3, no. 2, (1999), 159-173.

[54] M. W. Coffey, Alternative evaluation of a ln tan integral arising in quan-tum field theory, 2008.

[55] H. E. Fettis, W. A. Newcomb, On Clausen’s Integral, Amer. Math.Monthly, 92, (1985), 436-437.

[56] H. E. Fettis, More trigonometric integrals, Math. of Comp., 43, (1984),557-564.

[57] J. Guillera, J. Sondow, Double integrals and infinite products for someclassical constants via analytic continuations of Lerch’s transcendent,The Ramanujan Journal, 16, no. 3, (2008), 247-270.

[58] K.S. Kolbig, The polygamma function ψ(k)(x) for x = 14

and x = 34,

Journal of Computational and Applied Mathematics, 75, (1996), 43-46.

[59] L. Lewin (Editor), Structural Properties of Polylogarithms, (Mathe-matical Surveys and Monographs, 37), American Mathematical Society,Providence, Rhode Island, USA, 1991.

[60] A. P. Prudnikov, Yu. A. Brychkov, O. I. Marichev, Integrals and series,2: Special Functions. Gordon and Breach, New York, 1986.

[61] H. M. Srivastava, M.L. Glasser, V. S. Adamchik, Some definite integralsassociated with the Riemann zeta function, Z. Anal. Anwendungen, 129,(2000), 77-84.

[62] E. C. Titchmarsh, The Zeta-Function of Riemann, Oxford University(Clarendon) Press, Oxford, London and New York, 1951, Second Ed.(Revised by D. R. Heath- Brown), 1986.