International Journal of Mathematics and
Computer Science, 7(2012), no. 1, 11–83
M
CS
Some infinite series involving the Riemann
zeta function
Donal F. Connon
Elmhurst
Dundle 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 Riemann
zeta function. Some examples are set out below
∞
n=0
n
1 n+1
k=0
∞
n=0
∞
n=0
n
k
(−1)k xk
= s Φ(x, s + 1, y) − log x Φ(x, s, y)
(k + y)s
n 2k
1
n
(−1)
=
1 − 21−s ζ(s)
s
k
(k + 1)
2
1
2n+1
n
k=0
n (−1)k
(−1)k
n
+
= π cosec(πy)
k
k+y
k+1−y
k=0
−1
∞
22n sin2n u 2n
= 4u2 log(2 sin u)+2Cl3 (2u)+4u Cl2 (2u)−2ζ(3)
n
n3
n=1
∞
7
1
24n
[n!]4
ζ(3) − πG =
4
2
(2n + 1)3 [(2n)!]2
n=0
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.
Donal F. Connon
12
1
Some Hasse-type series
It was shown in [21] that
y 1 s − 1 n=0 n + 1 k=0
∞
n
n
k
(−1)k y k
log y
= Lis (y) −
Lis−1 (y)
s−1
(k + 1)
s−1
(1)
where Lis (y) is the polylogarithm function [39]
Lis (y) =
∞
yn
n=1
ns
and, with y = 1, this becomes the formula originally discovered by Hasse [32]
in 1930
1 1 s − 1 n=0 s + 1 k=0
∞
n
(−1)k
= Lis (1) = ζ(s)
(k + 1)s−1
∞
The gamma function is defined as Γ(s) = 0 ts−1 e−t dt, s > 0.
Using the substitution t = (k + y)u, we obtain
∞
1
1
=
us−1 e−(k+y)u du
(k + y)s
Γ(s) 0
n
k
(2)
(3)
We now consider the finite sum set out below
n xk
n
Sn (x, y) =
k (k + y)s
k=0
(4)
and combine (3) and (4) to obtain
∞
n n xk
n
n
k 1
Sn (x, y) =
=
us−1 e−(k+y)u du
x
k (k + y)s
k
Γ(s) 0
k=0
k=0
1
=
Γ(s)
0
∞
n n
us−1
xk e−(k+y)u du
k
k=0
Some infinite series involving the Riemann zeta function
=
1
Γ(s)
13
n n k
du
us−1 e−yu
xe−u
k
∞
0
k=0
Using the binomial theorem this becomes
1
Sn (x, y) =
Γ(s)
∞
n
us−1 e−yu 1 + xe−u du
(5)
0
Making the summation, we see that
n ∞
xk
1 n ∞ s−2 −yu n
−u n
1
+
xe
t
=
t
u
e
du
k (k + y)s
Γ(s) n=0
0
n=0
k=0
∞
n
The geometric series gives us for |t (1 + xe−u )|< 1
∞
n
t 1 + xe−u
=
n=0
1
1 − t (1 + xe−u )
and we then have
∞
n xk
1
us−1 e−yu
n
t
=
du
k (k + y)s
Γ(s) 0 1 − t (1 + xe−u )
n=0
k=0
∞
n
We now integrate (6) with respect to t
w ∞
∞
n wn+1 n
us−1 e−yu
1
xk
=
dt
du
n + 1 k=0 k (k + y)s
Γ(s) 0
1 − t(1 + xe−u )
0
n=0
1
=−
Γ(s)
and obtain
0
∞
us−1 e−yu log [1 − w(1 + xe−u )]
du
1 + xe−u
(6)
Donal F. Connon
14
∞ s−1 −(y−1)u
∞
n xk
1
wn+1 n
u e
log [1 − w(1 + xe−u )]
=
−
du
n + 1 k=0 k (k + y)s
Γ(s) 0
eu + x
n=0
(7)
When w = 1 and x → −x we get
∞
n=0
1 n + 1 k=0
n
1
=
Γ(s)
n
k
(−1)k xk
1
=−
s
(k + y)
Γ(s)
∞
0
∞
us−1 e−(y−1)u log [xe−u ]
du (8)
eu − x
0
us e−(y−1)u
log x
du −
u
e −x
Γ(s)
∞
0
us−1 e−(y−1)u
du
eu − x
We see from (3) that
xk
xk
=
(k + y)s
Γ(s)
∞
us−1 e−(k+y)u du
0
and we have the summation
∞
k=0
∞
∞
1
xk
s−1 −yu
=
u
e
xk e−ku du
s
(k + y)
Γ(s) 0
k=0
∞ s−1 −yu
1
u e
=
du
Γ(s) 0 1 − xe−u
We therefore obtain the well-known formula [43, p. 121] for the HurwitzLerch zeta function Φ(x, s, y)
Φ(x, s, y) =
∞
k=0
1
xk
=
s
(k + y)
Γ(s)
∞
0
us−1 e−(y−1)u
du
eu − x
and with y = 1 we obtain
Φ(x, s, 1) =
∞
k=0
xk
1 xk
Lis (x)
=
=
s
s
(k + 1)
x k=1 k
x
∞
(9)
Some infinite series involving the Riemann zeta function
15
giving us [46, p. 280]
x
Lis (x) =
Γ(s)
∞
0
us−1
du
eu − x
(10)
Reference to (8) then shows that
∞
n=0
1 n + 1 k=0
n
n
k
(−1)k xk
= s Φ(x, s + 1, y) − log x Φ(x, s, y)
(k + y)s
(11)
With x = 1 we obtain Hasse’s formula (2)
∞
n=0
1 n + 1 k=0
n
n
k
(−1)k
= s Φ(1, s + 1, y) = sζ(s + 1, y)
(k + y)s
With y = 1 equation (11) becomes
∞
n=0
1 n + 1 k=0
n
n
k
(−1)k xk
= s Φ(x, s + 1, 1) − log x Φ(x, s, 1)
(k + 1)s
or equivalently
∞
n=0
1 n + 1 k=0
n
n
k
(−1)k xk
s
1
= Lis+1 (x) − log x Lis (x)
s
(k + 1)
x
x
(12)
which corresponds with (1). A closed form expression may be obtained for
example with x = 1/2 and s = 2.
We see from (6) that
∞
n us−1 e−yu
1
xk
n
t
=
du
k (k + y)s
(1 − t)Γ(s) 0 1 − e−u tx/(1 − t)
n=0
k=0
∞
n
1
=
(1 − t)Γ(s)
0
∞
us−1 e−(y−1)u
du
eu − tx/(1 − t)
Donal F. Connon
16
and hence referring to (9) we have
n xk
1
tx
n
t
=
Φ
, s, y
k (k + y)s
1−t
1−t
n=0
k=0
∞
n
(13)
With t = 1/2 we obtain
∞
n xk
1 n
= Φ (x, s, y)
2n+1 k=0 k (k + y)s
n=0
(14)
which is clearly an example of Euler’s transformation of series [33, p. 244].
tx
, (13) may be represented by
Letting w = 1−t
n n ∞ w
xk
w+x
n
Φ (w, s, y)
=
s
k (k + y)
w+x
x
n=0
k=0
(15)
and with x = −1 this becomes
n n ∞ −w
(−1)k
n
= (1 − w)Φ (w, s, y)
k (k + y)s
1−w
n=0
k=0
(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 1 n
(−1)k
= Φ (−1, s, y) = ζa (s, y)
2n+1 k=0 k (k + y)s
n=0
(17)
where ζa (s, y) is the alternating Hurwitz-Lerch zeta function
∞
(−1)k
ζa (s, y) =
(k + y)s
k=0
With y = 1 we have
∞
n 1 n
(−1)k
= ζa (s, 1) = ζa (s)
2n+1 k=0 k (k + 1)s
n=0
(18)
Some infinite series involving the Riemann zeta function
17
Letting w = x = 1 in (15) results in
∞
n 1
1 n
= Φ (1, s, y) = ζ (s, y)
n+1
k (k + y)s
2
n=0
k=0
(19)
Letting s = 1 and y = 1/2 in (17) results in
∞
n 1 n
(−1)k
= ζa (1, 1/2)
n
k 2k + 1
2
n=0
k=0
We have from [15, p. 523]
ζa (1, 1/2) =
π
2
and we therefore obtain
∞
n (−1)k
π
1 n
=
n
2 k=0 k 2k + 1
2
n=0
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 (−1)k
(−1)k
1 n
+
= π cosec(πy)
n+1
k
2
k
+
y
k
+
1
−
y
n=0
k=0
We now divide (7) by w and integrate to obtain
∞
n=0
v
∞ s−2 −(y−1)u
n v n+1 n
log [1 − w(1 + xe−u )]
u e
1
xk
=−
dw
du
2
s
(n + 1) k=0 k (k + y)
Γ(s) 0
w
eu + x
0
Since
Donal F. Connon
18
log(1 − ax)
dx = −Li2 (ax)
x
we have
∞ s−1 −(y−1)u
n v n+1 n
u e
Li2 [v(1 + xe−u )]
1
xk
=
du
(n + 1)2 k=0 k (k + y)s
Γ(s) 0
eu + x
n=0
(20)
Further operations of the same kind will result in
∞
∞ s−1 −(y−1)u
n xk
1
v n+1 n
u e
Lir [v(1 + xe−u )]
=
du
(n + 1)r k=0 k (k + y)s
Γ(s) 0
eu + x
n=0
(21)
With t = −1 in (6) we obtain
∞
∞ s−1 −yu
n xk
1
u e
n
(−1)
=
du
s
k (k + y)
Γ(s) 0 2 + xe−u
n=0
k=0
∞
n
and letting x → −x we get
∞ s−1 −yu
n ∞
u e
1
n (−1)k xk
n
(−1)
=
du
k (k + y)s
Γ(s) 0 2 − xe−u
n=0
k=0
With x = 2 we have
∞ s−1 −yu
∞ s−1 −(y−1)u
n u e
u e
1
1
n (−1)k 2k
du
(−1)
=
du =
−u
k (k + y)s
2Γ(s) 0 1 − e
2Γ(s) 0
eu − 1
n=0
k=0
∞
n
Hence we obtain using (9)
n 1
n (−1)k 2k
(−1)
= Φ(1, s, y)
s
k (k + y)
2
n=0
k=0
∞
n
(22)
Some infinite series involving the Riemann zeta function
19
and with s = n + 1 and y = 1 this becomes
∞
n 1
(−1)k 2k
n
n
(−1)
= ζ(n + 1)
k (k + 1)n+1
2
n=0
k=0
(23)
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 = −2 we have
∞ s−2 −yu
n 2k
1
u e
n
(−1)
=
du
k (k + y)s
2Γ(s) 0
eu + 1
n=0
k=0
∞
n
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 2k
1
n
(−1)
=
1 − 22−s ζ(s)
s
k (k + 1)
2
n=0
k=0
∞
With t = x =
1
2
n
(24)
in (6) we obtain
∞ s−2 −yu
∞
n 1 n
u e
2
1
=
du
n
k
s
u−1
k 2 (k + y)
2
Γ(s)
e
0
n=0
k=0
= 2Φ(1, s, y) = 2ζ(s, y)
and therefore we have with y = 1
∞
n 1
1 n
ζ(s) =
n+1
k
k 2 (k + 1)s
2
n=0
k=0
With t = −1, x = 1/2 and y = 1 in (6) we have
(25)
Donal F. Connon
20
∞ s−1
n ∞
u
1
1
n
n
(−1)
=
du = −2Lis (−1/4)
k
s
u+ 1
k 2 (k + 1)
2Γ(s)
e
0
4
n=0
k=0
(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
sin
−1
y
2
1 [n!]2
=
(2y)2n , −1 ≤ y ≤ 1
2 n=1 n2 (2n)!
∞
(27)
and therefore upon letting x = sin−1 y we get
x2 =
1 [n!]2 2n 2n
2 sin x, −π/2 ≤ x ≤ π/2
2 n=1 n2 (2n)!
∞
(28)
which was known by Euler [34].
Using Bürmann’s theorem, it is an exercise in Whittaker & Watson [46, p.
130] to prove
2 1
2.4 1
x = sin x + ( ) sin4 x + ( ) sin6 x + ... =
An sin2n x
3 2
3.5 3
n=1
2
∞
2
where
(2n − 2)!! 1
.
(2n − 1)!! n
Using the definitions of the double factorials
An =
(2n)!! = 2.4...(2n) = 2n n!, (2n + 1)!! = 1.3.5...(2n + 1) =
(2n + 1)!
2n n!
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 cot x and integrate to obtain
0
t
1 [n!]2 2n
x cot x dx =
2
2 n=1 n2 (2n)!
∞
2
t
sin2n−1 x cos x dx
0
which, for −π/2 ≤ x ≤ π/2, results in [12, p. 234]
t
0
1 22n sin2n t
x cot x dx =
4 n=1
n3
∞
2
2n
n
−1
(29)
2n
where
= (2n)!
is the central binomial coefficient.
[n!]2
n
This integral also appears in [12, p. 234].
Using integration by parts we have
t
0
t
t
x cot x dx = x log sin x − 2
x log sin x dx
2
2
0
0
Since x2 log sin x = x2 log sinx x + x2 log x we see that lim x2 log sin x = 0 and
x→0
thus
0
t
2
2
x cot x dx = t log sin t − 2
t
x log sin x dx
0
Hence we get for 0 ≤ t ≤ π/2
0
t
1
1 22n sin2n t
x log sin x dx = t2 log sin t −
2
8 n=1
n3
∞
2n
n
−1
(30)
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
Donal F. Connon
22
π
2
x log sin x dx =
0
π2
7
ζ(3) −
log 2
16
8
(31)
A very elementary proof of (31) is given in equation (6.20a) in [22] where we
used the basic identity
b
p(x) cot x dx = 2
a
∞ n=1
b
p(x) sin 2nx dx
(32)
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 < π
t
2
x cot x dx = 2
0
∞ n=1
t
x2 sin 2nx dx
0
t
∞
∞
∞
1 cos 2nx
cos
2nx
sin
2nx
2
−
x
=
+
x
3
2
2 n=1 n
n
n
n=1
n=1
0
This gives us
0
t
∞
∞
∞
cos 2nt
sin 2nt 1
1 cos 2nt
2
+t
x cot x dx =
−t
− ζ(3) (33)
3
2 n=1 n
n
n2
2
n=1
n=1
2
and with t = π/2 we obtain
0
π
2
x2 cot x dx =
∞
∞
1 (−1)n π 2 (−1)n 1
−
− ζ(3)
2 n=1 n3
4 n=1 n
2
Hence using the alternating zeta function
Some infinite series involving the Riemann zeta function
ζa (s) =
∞
(−1)n+1
n=1
ns
23
= (1 − 21−s )ζ(s)
we see that
π
2
0
π2
7
log 2
x2 cot x dx = − ζ(3) +
8
4
(34)
Another elementary evaluation of this integral has recently been provided by
Fujii and Suzuki [28].
Using the substitutions y = sin x and z = tan x we may also note that
sin t −1 n
t
sin y
dy
xn cot x dx =
y
0
0
t
tan t
n
xn
(tan−1 z)
2
dx =
dz
z
0 sin 2x
0
We also see from (29) and (33) that for −π/2 ≤ t ≤ π/2
1 22n sin2n t
4 n=1
n3
−1
∞
∞
∞
1 cos 2nt 2 cos 2nt sin 2nt 1
−t
− ζ(3)
+t
2
2 n=1 n3
n
n
2
n=1
n=1
(35)
and with t = π/2 we immediately see that
∞
2n
n
=
2
2 22n
ζ(3) = π 2 log 2 −
7
7 n=1 n3
∞
2n
n
−1
(36)
which is in agreement with Sherman’s compendium of formulae [42].
We have from [12, p. 198]
22n
B(n, 1/2) =
n
where the beta function is defined by
2n
n
−1
(37)
Donal F. Connon
24
1
B(u, v) =
0
(1 − t)u−1 tv−1 dt
This then gives us
22n
n
2n
n
−1
=
0
1
(1 − t)n
√ dt
(1 − t) t
and we have the summation involving the dilogarithm function
−1 1
∞
22n 2n
Li2 (1 − t)
√ dt
=
3
n
n
t
(1
−
t)
0
n=1
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)
√
dt = 2 Li2 1 − u2 du
t
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)
0
t
1 22n sin2n t
x cot x dx =
4 n=1
n3
2
∞
2n
n
−1
and this gives us
−1 1
∞
Li2 (1 − t) sin2 t
22n sin2n t 2n
√
dt
=
n
n3
(1 − t) t
0
n=1
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
Lin (z) =
(n − 1)!
0
1
z logn−1 u
du, Li2 (z) = −
1 − zu
0
1
z log u
du
1 − zu
(38)
and thus
0
1
1
1
Li2 (1 − t) sin2 t
sin2 t log u
√
√ dt du
dt = −
t 1 − u (1 − t) sin2 t
(1 − t) t
0
0
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
arcsin(x sin y/4)
(−1)k−1 2 1
x2n [n!]4
=
φ logk−3 [4 sin φ/x sin y]dφdy
k [(2n)!]2
(k
−
3)!
sin
y
(2n)
0
0
n=1
Since sin π6 = 12 we see from (29) that
∞
−1
π
∞
6
1
2n
=
−8
x log [2 sin x] dx
n
n3
0
n=1
(39)
which I first came across in van der Poorten’s 1979 paper “Some wonderful
formulae. . . an introduction to Polylogarithms” [45].
In passing, we note from [12, p.122] that
−1
t −1 2
∞
sin x
3
(2t)2n 2n
2
=4
dx = 2t 4 F3 {1, 1, 1, 1},
, 2, 2 , t2
3
n
n
x
2
0
n=1
in terms of the hypergeometric functions.
Donal F. Connon
26
We now multiply (28) by x cot x and integrate
0
t
1 [n!]2 2n
x cot x dx =
2
2 n=1 n2 (2n)!
∞
3
t
x sin2n−1 x cos x dx
0
Integration by parts yields
t
t
t
1
2n−1
2n
x sin
x cos x dx =
sin2n x dx
sin t −
2n
2n 0
0
and therefore we have
−1
−1 t
∞
∞
1 22n sin2n t 2n
1 22n 2n
x cot x dx = t
−
sin2n x dx
3
3
n
n
4
n
4
n
0
0
n=1
n=1
(40)
With t = π/2 we obtain
t
0
3
π
2
−1
−1 π
∞
∞
2
π 22n 2n
1 22n 2n
x cot x dx =
−
sin2n x dx
n
n
8 n=1 n3
4 n=1 n3
0
3
and using [12, p. 195]
π
2
0
(2n)! π
1
sin x dx = 2n
= 2n
2
2 (n!) 2
2
2n
2n
n
π
2
(41)
we have
0
π
2
−1
∞
∞
π 22n 2n
π 1
x cot x dx =
−
n
8 n=1 n3
8 n=1 n3
3
Using (36) we obtain
0
π
2
x3 cot x dx =
π3
9π
log 2 −
ζ(3)
8
16
As another example, letting p(x) = x3 in (32) we have for −π < t < π
(42)
Some infinite series involving the Riemann zeta function
t
3
x cot x dx = 2
0
∞ n=1
t
27
x3 sin 2nx dx
0
t
∞
∞
∞
∞
3 2 sin 2nx 3 sin 2nx
cos
2nx
cos
2nx
3
3
= x
−
−
x
+
x
3
2 n=1 n2
4 n=1 n4
n
2
n
n=1
n=1
0
∞
∞
∞
∞
cos 2nt 3 cos 2nt
3 sin 2nt 3 sin 2nt
3
= t2
−
−
t
+
t
2 n=1 n2
4 n=1 n4
n
2 n=1 n3
n=1
Using (40) together with (90) we see that
t
x3 cot xdx =
∞
2n
2n
1
2 sin t
t
4 n=1
n3
2n
n
−1
0
−1
∞
n
2n
j
1
2
1
(−1)
2n
2n
2n
−
t+
sin 2jt
3
2n
n
n
n
−j
4 n=1 n
2
j
j=1
−1
−1 n
∞
∞
∞
t 1 1 1
(−1)j
1 22n sin2n t 2n
2n
2n
−
−
sin 2jt
= t
n
n
n−j
4 n=1
n3
4 n=1 n3 4 n=1 n3
j
j=1
This gives us
−1
−1 n
∞
∞
1 22n sin2n t 2n
(−1)j
t
1 1
2n
2n
− ζ(3)−
t
sin 2jt
n
n
n−j
4 n=1
n3
4
4 n=1 n3
j
j=1
∞
∞
∞
∞
cos 2nt 3 cos 2nt
3 2 sin 2nt 3 sin 2nt
3
+ t
−
−t
= t
2 n=1 n2
4 n=1 n4
n
2 n=1 n3
n=1
and upon substituting (35) we have for −π/2 ≤ t ≤ π/2
Donal F. Connon
28
−1 n
∞
1
(−1)j
2n
2n
sin 2jt
3
n−j
n
n
j
n=1
j=1
=3
∞
sin 2nt
n=1
n4
− 4t
∞
cos 2nt
n=1
n3
2
− 2t
∞
sin 2nt
n=1
n2
− 3tζ(3)
(43)
It was shown in equation (4.3.168a) of [19] that for 0 ≤ t< 1
t
πx2 cot πx dx
0
= −[ζ (−2, t)+ζ (−2, 1−t)]+2t[ζ (−1, t)−ζ (−1, 1−t)]+t2 log(2 sin πt)−
ζ(3)
2π 2
(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
−1
∞
1 22n sin2n πt 2n
u cot u du = 2
n
4π n=1
n3
0
0
(45)
and hence we see from (44) and (45) that for 0 ≤ t ≤ 1/2
t
1
πx cot πx dx = 2
π
2
πt
2
−1
∞
1 22n sin2n πt 2n
n
4π 2 n=1
n3
= −[ζ (−2, t)+ζ (−2, 1−t)]+2t[ζ (−1, t)−ζ (−1, 1−t)]+t2 log(2 sin πt)−
ζ(3)
2π 2
(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
∞
1 22n sin2n πt 2n
= −[ζ (−2, t)+ζ (−2, 1−t)]−2(1−t)[ζ (−1, t)−ζ (−1, 1−t)]
n
4π 2 n=1
n3
ζ(3)
2π 2
and subtraction of these two equations appeared to indicate that for
+(1 − t)2 log(2 sin πt) −
1
ζ (−1, t) − ζ (−1, 1 − t) = (1 − 2t) log(2 sin πt)
2
We note from (55) below that
1
3
G
1
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
ζ (−n, x) − ζ (−n) = (−1)
n
k!Qk,n (x) log Γk+1 (x)
k=0
where Qk,n (x) are polynomials defined by
n
n
j
n−j
Qk,n (x) =
(1 − x)
j
k
and
j
k
j=k
are the Stirling subset numbers defined by
j
k
=k
n−1
k
Particular cases of (47) are
+
n−1
k−1
n
1, n = 0
,
=
0
0, n = 0
(47)
Donal F. Connon
30
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)] + log
G(1 − x)
G(1 + x)
(49)
which we have previously derived in [19].
Letting t = 12 in (46) we obtain
−1
∞
1
1 22n 2n
1
ζ(3)
= −2ζ −2,
+ log 2 −
2
3
n
4π n=1 n
2
4
2π 2
and using (36) we obtain the known result
1
3ζ(3)
ζ −2,
=
2
16π 2
which is also derived, for example, in equation (4.3.168d) of [19].
When t = 14 in (46) we get
(50)
Some infinite series involving the Riemann zeta function
31
−1
∞
1 2n 2n
n
4π 2 n=1 n3
1 1
ζ(3)
1
3
1
3
+ ζ −2,
+ ζ −1,
− ζ −1,
+ log 2− 2
= − ζ −2,
4
4
2
4
4
32
2π
(51)
We now refer to Adamchik’s result [2]
ζ (−2n − 1, t) − ζ (−2n − 1, 1 − t) =
ζ (−2n, t) + ζ (−2n, 1 − t) = (−1)n
(2n + 1)!
Cl2n+2 (2πt)
(2π)2n+1
(52)
(2n)!
Cl2n+1 (2πt)
(2π)2n
(53)
where Cln (t) is the Clausen function defined by
Cl2n+1 (t) =
∞
cos kt
k=1
k 2n+1
Cl2n (t) =
∞
sin kt
k=1
k 2n
This is also derived in equation 4.3.167 of [19].
We then see that (as noted by Adamchik [2])
1
1
3
G
− ζ −1,
=
Cl2 (π/2) =
ζ −1,
4
4
2π
2π
(54)
(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
ζ(2n + 1)
24n+1
(56)
this then gives us for t = 1/4
ζ
2n
(2n)!
(2n)!
1
3
n1 − 2
−2n,
Cl
(π/2)
=
(−1)
ζ(2n+1)
+ζ −2n,
= (−1)n
2n+1
2n
4n+1
4
4
(2π)
2
(2π)2n
Donal F. Connon
32
and in particular we have for n = 1
3ζ(3)
1
3
ζ −2,
+ ζ −2,
=
4
4
64π 2
We therefore have
−1
∞
2n 2n
35ζ(3) π 2
+
log 2 + πG
=
−
3
n
n
16
8
n=1
(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
0
With t =
1
6
π
4
x2 cot xdx = −
35
π2
πG
ζ(3) +
log 2 +
64
32
4
(58)
in (46) we obtain
−1
∞
1 1
2n
n
4π 2 n=1 n3
1 ζ(3)
1
5
1
5
− ζ −2,
+
ζ −1,
− ζ −1,
−
= − ζ −2,
6
6
3
6
6
2π 2
(59)
We see from (53) that
π 1
5
1
ζ −2,
+ ζ −2,
= − 2 Cl3
6
6
2π
3
and from Lewin’s monograph [39, p. 198] we have
Cl2n+1
π 3
1
= (1 − 2−2n )(1 − 3−2n )ζ(2n + 1)
2
which gives us
Cl3
π 3
1
= ζ(3)
4
Some infinite series involving the Riemann zeta function
33
This results in
ζ
1
−2,
6
ζ(3)
5
+ ζ −2,
=− 2
6
8π
(60)
We also have from (52)
π 1
5
1
ζ −1,
− ζ −1,
= Cl2
6
6
2π
3
and from (59) we obtain
−1
∞
π ζ(3)
1 1
1
5
1
2n
−2,
−2,
=
−
ζ
−
ζ
+
Cl
−
2
n
4π 2 n=1 n3
6
6
6π
3
2π 2
(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 πs cos 2nπt
sin 2nπt
+ cos
ζ(s, t) = 2Γ(1 − s) sin
1−s
2 n=1 (2πn)
2 n=1 (2πn)1−s
(62)
where (s) < 0 and 0 < t ≤ 1. In 2000, Boudjelkha [52] showed that this
formula also applies in the region Re(s)< 1. It may be noted that when t = 1
this reduces to Riemann’s functional equation for ζ(s). Letting s → 1 − s we
may write this as
ζ(1 − s, t) = 2Γ(s) cos
∞
πs cos 2nπt
2
n=1
Using (63) it is easily shown that
(2πn)s
+ sin
∞
πs sin 2nπt
2
n=1
(2πn)s
(63)
Donal F. Connon
34
2ζ (−1, t) − B2 (t) (1 − γ − log(2π)) = −4
∞
log n cos 2nπt
n=1
(2πn)2
+
1
Cl2 (2πt)
2π
(64)
∞
1 1
1
1
log n sin 2nπt
−
Cl3 (2πt)
− γ − log(2π) = −4
ζ (−2, t)−B3 (t)
2 3
2
(2πn)3
(2π)2
n=1
(65)
With t = 1/6 in (65) we obtain
ζ
1
−2,
6
∞
ζ(3)
5 1 1
1
log n sin(nπ/3)
−
−
− γ − log(2π) = −4
3
216 2 3
2
(2πn)
16π 2
n=1
and this indicates the complexities involved in determining a closed form
expression for
1
.
ζ −2,
6
Differentiating (73) gives us
−
∞
log n sin(n π/3)
n=1
ns
=
√
3
3−s − 1 3−s log 3
ζ (s) −
ζ(s)
2
2
√
1
1
1
1
−s
−s
ζ s,
+ 3 6
+ ζ s,
− 6 log 6 ζ s,
+ ζ s,
6
3
6
3
so that with s = 3 we have
−
∞
log n sin(n π/3)
n=1
n3
√
= 3
3−3 − 1 3−3 log 3
ζ (3) −
ζ(3)
2
2
Some infinite series involving the Riemann zeta function
35
√
1
1
1
1
−3
−3
ζ 3,
+ 3 6
+ ζ 3,
− 6 log 6 ζ 3,
+ ζ 3,
6
3
6
3
With t = 1/4 in (64) we obtain
∞
π 1
1
1
log n cos(nπ/2)
2ζ −1,
+
+ [1 − γ − log(2π)] = −4
Cl2
4
48
(2πn)2
2π
2
n=1
2ζ
1
−1,
4
∞
G
1
log 2n
(−1)n
+
+ [1 − γ − log(2π)] = −4
2
48
(4πn)
2π
n=1
∞
∞
log 2 (−1)n+1
1 G
log n
=
+ 2
(−1)n+1 2 +
2
2
4π n=1
n
4π n=1
n
2π
=
log 2
1
G
ζa (2) − 2 ζa (2) +
2
4π
4π
2π
Since ζa (2) = 12 ζ(2) and ζa (2) = 12 ζ(2) log 2 + 12 ζ (2) we have
1
G
1
1
2ζ −1,
+ [1 − γ − log(2π)] = − 2 ζ (2) +
4
48
8π
2π
It is easily found from the functional equation for the Riemann zeta function
that
1
1
(1 − γ − log(2π)) + 2 ζ (2)
12
2π
and hence we have (as originally determined by Adamchik [2])
1
1
G
ζ −1,
=
− ζ (−1)
4
4π 8
ζ (−1) =
Ghusayni [29] showed in 1998 that
(66)
Donal F. Connon
36
−1
∞
∞
π sin (nπ/3) 3 1
2n
ζ(3) =
−
n
2 n=1
n2
4 n=1 n3
(67)
Later in 2000, Ghusayni [30], having noted an earlier paper [31], reported
that
√ ∞
sin(n π/3)
3 1
1
1
1
1
1
=
+
−
−
+
+
− ...
(68)
n2
2
12 22 42 52 72 82
n=1
√ 2 2
3 1 (1) 1
ψ
− π
=
2
3
3
9
and Ghusayni concluded that
√
√
−1
∞
∞
3 3 3 3 1
3 1
2n
ζ(3) = −
−
π +
π
2
n
18
4
(3n
−
2)
4 n=1 n3
n=1
(69)
It is shown in [30] that
1
1
1
1
1
1
2
+ 2 − 2 − 2 + 2 + 2 − ... = − π 2 − 2
2
1
2
4
5
7
8
27
1
0
log x
dx
1 + x3
Mathematica evaluates this integral as
1
324
0
1
5
log x
2
dx = −8π − 6PolyGamma 1, + 3PolyGamma 1,
3
1+x
6
6
2
1
−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
−1
∞
1
2
4
1
1 √
2n
(1)
(1)
−ψ
− ζ(3)
= π 3 ψ
3
n
n
18
3
3
3
n=1
37
(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)k π
dk
cot πx
dxk
and therefore we see that
4
1
2
(1)
(1)
2
2 π
ψ
+ψ
= π / sin
= π2
3
3
3
3
We then obtain
−1
∞
1 √ (1) 1
1
4
2 3√
2n
=
3
ψ
π
−
ζ(3)
−
π 3
n
n3
9
3
3
27
n=1
We have
p
p
ψ (k) ( ) = (−1)k+1 k!ζ(k + 1, )
q
q
and therefore we see that
∞
1
1
1
ψ ( ) = ζ(2, ) = 9
3
3
(3n + 1)2
n=0
or, equivalently, changing the order of summation
∞
1
1
ψ( )=9
3
(3n − 2)2
n=1
Hence we see that (70) is equivalent to Ghusayni’s result (69)
(71)
Donal F. Connon
38
√
−1
∞
3 3
3
1
3 1
2n
(1)
ζ(3) = −
π +
πψ
−
n
18
12
3
4 n=1 n3
√
We now refer back to (35) which is valid for −1/2 ≤ t ≤ 1/2
1 22n sin2n πt
4 n=1
n3
∞
2n
n
−1
∞
∞
∞
sin 2nπt 1
1 cos 2nπt 2 2 cos 2nπt
=
−π t
− ζ(3)
+πt
3
2
2 n=1
n
n
n
2
n=1
n=1
so that with t = 1/6 we have
1 1
4 n=1 n3
∞
2n
n
−1
∞
∞
∞
1 cos(nπ/3) 1 2 cos(nπ/3) 1 sin(nπ/3) 1
+ π
=
− π
− ζ(3)
2 n=1
n3
36 n=1
n
6 n=1
n2
2
Lewin [39] and Srivastava and Tsumura [43, p. 293] reported for Re(s)> 1
∞
cos(n π/3)
ns
n=1
∞
sin(n π/3)
n=1
ns
=
√
3
1
= (61−s − 31−s − 21−s + 1)ζ(s)
2
1
3−s − 1
1
−s
ζ s,
ζ(s) + 6
+ ζ s,
2
6
3
(72)
(73)
Hence we have with s = 3 and s = 2 respectively
∞
cos(n π/3)
n=1
∞
sin(n π/3)
n=1
n2
=
√
3
n3
1
= ζ(3)
3
1
3−2 − 1
1
−2
ζ 2,
ζ(2) + 6
+ ζ 2,
2
6
3
We have the Fourier series [44, p. 148] for 0 < t < 1
(74)
(75)
Some infinite series involving the Riemann zeta function
∞
cos 2nπt
= − log(2 sin πt)
n
n=1
39
Using (75) we then obtain Ghusayni’s result
π√
ζ(3) =
3
2
−1
∞
1
3−2 − 1
1
3 1
2n
−2
ζ 2,
ζ(2) + 6
+ ζ 2,
−
n
2
6
3
4 n=1 n3
Srivastava and Tsumura [43, p. 293] have also reported for Re(s)> 1
∞
cos(2n π/3)
ns
n=1
∞
sin(2n π/3)
n=1
ns
=
√
3
∞
cos(n π/2)
n=1
∞
sin(n π/2)
n=1
ns
ns
1
= (31−s − 1)ζ(s)
2
3−s − 1
1
−s
ζ(s) + 3 ζ s,
2
3
(77)
= 2−s (21−s − 1)ζ(s)
(78)
−s
= (2
(76)
1−2s
− 1)ζ(s) + 2
1
ζ s,
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)
−1
∞
22n sin2n πt 2n
= 4π 2 t2 log(2 sin πt)+2Cl3 (2πt)+4πt Cl2 (2πt)−2ζ(3)
3
n
n
n=1
(80)
or equivalently
Donal F. Connon
40
−1
∞
22n sin2n u 2n
= 4u2 log(2 sin u) + 2Cl3 (2u) + 4u Cl2 (2u) − 2ζ(3)
3
n
n
n=1
(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/6 and u = π/4 we obtain respectively
−1
∞
1
2π
2n
= 2Cl3 (π/3) +
Cl2 (π/3) − 2ζ(3)
3
n
n
3
n=1
−1
∞
2n 2n
π2
log 2 + 2Cl3 (π/2) + πCl2 (π/2) − 2ζ(3)
=
n
n3
8
n=1
(82)
(83)
Then using (56) we see that
3
ζ(3)
32
and hence from (83) we obtain another derivation of (57)
Cl3 (π/2) = −
−1
∞
35ζ(3) π 2
2n 2n
=−
+
log 2 + πG
n
n3
16
8
n=1
in Browkin’s paper in
We also have a number of formulae involving Cl2 pπ
q
Lewin’s survey [59, p. 244], including for example
π π 4
4
5π
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
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
Cl2 (2x) = Cl2 (x) − Cl2 (π − x)
2
which implies that
Cl2
2π
3
π 2
= Cl2
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 = −2−2n−1 (1 − 2−2n )ζ(2n + 1)
Cl2n+1
2
π 1
= (1 − 2−2n )(1 − 3−2n )ζ(2n + 1)
3
2
2π
1
Cl2n+1
= − (1 − 3−2n )ζ(2n + 1)
3
2
Cl2n+1
For example, we see from the definition that
Cl2n+1
=−
1
π 2
=−
1
22n+1 12n+1
−
1
22n+1
1
22n+1
+
1
42n+1
−
1
62n+1
+ ...
1
+ ... = − 2n+1 ζa (2n + 1)
2
We then see Cl3 (π/3) = ζ(3)/3 and from (82) that
Donal F. Connon
42
−1
∞
1
2π
4
2n
=
Cl2 (π/3) − ζ(3)
3
n
n
3
3
n=1
and comparing this with (71) we obtain
√
2 3 Cl2 (π/3) = ψ (1)
1
2
− π2
3
3
(84)
as previously derived by Fettis [55].
More generally, we have [39a, p. 358]
ψ
(1)
(q−1)/2
p
πp
2mπp
2mπp
π2
2
sin
= cosec
+ 2q
Cl2
q
2
q
q
q
m=1
(85)
and a particular case of this is (84). Another example is
√
1 √ (1) 1
2
Cl2 (π/6) =
3ψ
+ 16G − 2π / 3
24
3
(86)
Using PSLQ, Bailey et al. [50] discovered experimentally that
4π
7
6π
+ 7Cl2
− 7Cl2
6Cl2 (α) − 6Cl2 (2α) + 2Cl2 (3α) = 7Cl2
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
2π
7
−1 x
∞
22n 2n
sin2n u du
3
n
n
0
n=1
4
= x3 log 2 + 4
3
x
0
u2 log sin u du + 2Cl4 (2x) − 2x Cl3 (2x) − 2x ζ(3)
(88)
Some infinite series involving the Riemann zeta function
43
x
− x cos2k2kx and this
In the above we used the fact that 0 u sin 2ku du = sin4k2kx
2
results in
x
1
1
u Cl2 (2u) du = Cl4 (2x) − x Cl3 (2x).
4
2
0
When x = π/2 we get
−1 π
−1
∞
∞
∞
2
π 1
π
22n 2n
22n 2n
π 1
2n
2n
=
sin u du =
= ζ(3)
3
3
2n
3
n
n
n
n
n
22
2 n=1 n
2
0
n=1
n=1
π3
=
log 2 + 4
6
π
2
0
u2 log sin u du − π Cl3 (π) − π ζ(3)
We see that
Cl3 (π) =
∞
(−1)k
k=1
k3
3
= −ζa (3) = − ζ(3)
4
which gives us
π
π3
ζ(3) =
log 2 + 4
2
6
0
π
2
u2 log sin u du −
π
ζ(3)
4
and we therefore obtain a new derivation of Euler’s integral
0
π
2
u2 log sin u du = −
π3
3π
log 2 +
ζ(3)
24
16
(89)
Wiener [47] has shown that
0
x
1
sin2n u du = 2n
2
2n
n
and we therefore obtain from (88)
x+
n
(−1)j
j=1
j
2n
n−j
sin 2jx
(90)
Donal F. Connon
44
−1 n
∞
1
(−1)j
2n
2n
sin 2jx
n−j
n
n3
j
n=1
j=1
4
= x3 log 2 + 4
3
x
0
(91)
u2 log sin u du + 2Cl4 (2x) − 2x Cl3 (2x) − 3x ζ(3)
Using integration by parts we have
x
x
3
3
u cot u du = u log sin u − 3
0
= x3 log sin x − 3
x
u2 log sin u du
0
0
x
u2 log sin u du
0
Using (32) we have
x
3
u cot u du = 2
0
∞ n=1
x
u3 sin 2nu du
0
x
∞
∞
∞
∞
3 2 sin 2nu 3 sin 2nu 1 3 cos 2nu 3 cos 2nu −
− u
+ u
= u
4 n=1 n2
8 n=1 n4
2 n=1
n
4 n=1 n3 0
∞
∞
∞
∞
3 2 sin 2nx 3 sin 2nx 1 3 cos 2nx 3 cos 2nx
= x
−
− x
+ x
4 n=1 n2
8 n=1 n4
2 n=1
n
4 n=1 n3
3
3
1
3
= x2 Cl2 (2x) − Cl4 (2x) − x3 Cl1 (2x) + x Cl3 (2x)
4
8
2
4
Therefore we may write (91) as
−1 n
∞
1
(−1)j
2n
2n
sin 2jx
n
n−j
n3
j
n=1
j=1
(92)
Some infinite series involving the Riemann zeta function
45
= 13 x3 log sin x + 43 x3 log 2 + 2Cl4 (2x) − 2x Cl3 (2x) − 3x ζ(3)
− 14 x2 Cl2 (2x) + 18 Cl4 (2x) + 13 x3 Cl1 (2x) − 14 x Cl3 (2x)
1
4
17
9
1
1
= x3 log sin x+ x3 log 2+ Cl4 (2x)− x Cl3 (2x)− x2 Cl2 (2x)+ x3 Cl1 (2x)−3x ζ(3)
3
3
8
4
4
3
Differentiating (28) gives us
1 22n sin2n−1 x cos x
x=
2 n=1
n
∞
2n
n
−1
(93)
and with x → πx this becomes
1 22n sin2n−1 πx cos πx
πx =
2 n=1
n
∞
2n
n
−1
(94)
We now multiply this across by cot πx and integrate to obtain
0
t
1 22n
πx cot πx dx =
2 n=1 n
∞
1 22n
=
2 n=1 n
∞
2n
n
−1 2n
n
t
0
−1 t
sin2n−2 πx cos2 πx dx
0
sin2n−2 πx − sin2n πx dx
The Wolfram Mathematica Online Integrator
integral in
evaluates the above
terms of the hypergeometric function 3 F2 32 , 32 , −n; 52 ; cos2 πt .
Alternatively, using integration by parts, we get
t
t
2n−2
2
sin
πx cos πx dx =
sin2n−2 πx cos πx cos πx dx
0
0
t
t
1
sin2n−1 πx +
sin2n πx dx
= cos πx
(2n − 1)π 0
(2n − 1) 0
Donal F. Connon
46
and with t = 1/2 we have
1
2
sin
1
πx cos πx dx =
2n − 1
2n−2
2
0
1
2
sin2n πx dx.
0
Since
1
2
1
sin πx dx =
π
2n
0
π
2
sin2n x dx
0
we have
0
π
2
(2n − 1)!! π
1
sin x dx =
= 2n
(2n)!! 2
2
2n
2n
n
π
2
Therefore we see that
1
2
sin
2n−2
0
1
1
πx cos πx dx =
2n − 1 22n+1
2
2n
n
Hence we obtain
1
2
0
1
1
πx cot πx dx =
4 n=1 n(2n − 1)
∞
Clearly
∞
n=1
1
=
n(2n − 1) n=1
∞
1
n−
1
2
1
−
n
Since the digamma function may be represented by [43, p. 14]
1 ψ(a) = −γ − −
a n=1
∞
1
1
−
n+a n
we see that
∞ n=1
1
n−
1
2
1
−
n
1
= −ψ −
2
−γ+2
Some infinite series involving the Riemann zeta function
47
We know that [43, p. 22]
1
ψ −
2
= 2 − γ − 2 log 2
and we therefore deduce that
∞ n=1
1
n−
1
2
1
−
n
= 2 log 2
Hence we have the well-known integral
1
2
x cot πx dx =
0
log2
2π
We now divide (93) by sin x and integrate to obtain
t
0
1 22n sin2n−1 t
x
dx =
sin x
2 n=1 n(2n − 1)
∞
2n
n
−1
Alternatively, with n → n + 1 in the summation this may be written as
0
t
22n sin2n+1 t
x
dx =
sin x
(2n + 1)2
n=0
∞
2n
n
−1
(95)
We recall from [22] that
a
b
∞ b
p(x)
p(x) sin(2n + 1)x dx
dx = 2
sin x
n=0 a
(96)
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
Donal F. Connon
48
0
t
∞
∞
x
cos(2n + 1)t
sin(2n + 1)t
dx = −2t
+2
sin x
2n + 1
(2n + 1)2
n=0
n=0
(97)
and hence we obtain
−1
∞
∞
∞
22n sin2n+1 t 2n
cos(2n + 1)t sin(2n + 1)t
= −t
(98)
+
n
(2n + 1)2
2n + 1
(2n + 1)2
n=0
n=0
n=0
In particular, we have from (97)
0
π
2
∞
x
(−1)n
= 2G
dx = 2
sin x
(2n + 1)2
n=0
(99)
and we then determine that
22n
1
G=
2 n=0 (2n + 1)2
∞
−1
2n
n
(100)
in agreement with Sherman’s formula [42].
Using integration by parts Bradley [53] showed that
0
z
log tan u du − z log tan z = −
=−
1
4
2z
0
=−
Differentiating (27) gives us
0
sin(2z)
0
0
u sec2 u
du
tan u
u du
tan(u/2) cos2 (u/2)
1
=−
2
z
2z
u
du
sin u
2x sin−1 x dx
√
1 − x2 x 2
Some infinite series involving the Riemann zeta function
2x√sin−1 x
1−x2
=
∞
(2x)2n
n=1
n
2n
n
49
−1
for |x| < 1 and integrating term by term
results in
0
z
1 (2 sin 2z)2n+1
log tan u du − z log tan z = −
4 n=0 (2n + 1)2
∞
2n
n
−1
for 0 ≤ z ≤ π/4.
Therefore we obtain (95) again
z
0
22n sin2n+1 z
u
du =
sin u
(2n + 1)2
n=0
∞
2n
n
−1
We note from (102) below that
t
0
x
dx = t log tan
sin x
∞
t
sin(2n + 1)t
+2
2
(2n + 1)2
n=0
and hence we have
t log tan t +
∞
sin 2(2n + 1)t
n=0
(2n + 1)2
−1
∞
22n sin2n+1 t 2n
=
n
(2n + 1)2
n=0
(101)
Reference to [44, p. 149] shows that for 0 < t < π (this may also be easily
derived using the methods outlined in [22])
∞
cos(2n + 1)t
2n + 1
n=0
1
= − log tan
2
t
2
and for 0 ≤ t ≤ π
∞
sin(2n + 1)t
n=0
(2n + 1)2
Referring to (97) we have
1
=−
2
t
log tan
0
x
2
Donal F. Connon
50
t
0
x
dx = t log tan
sin x
∞
sin(2n + 1)t
t
+2
2
(2n + 1)2
n=0
(102)
and noting the Clausen function
Cl2 (t) = −
t
0
sin nt
u
log[2 sin( )]du =
2
n2
n=1
∞
we see that
Cl2 (π − t) =
∞
sin n(π − t)
n=1
n2
=
∞
n=1
Cl2 (t) + Cl2 (π − t) = 2
(−1)n
sin nt
n2
∞
sin(2n + 1)t
n=0
(2n + 1)2
Hence, as shown by Lewin [39, p. 255] we have
t
x
dx = t log tan
sin x
0
t
+ Cl2 (t) + Cl2 (π − t)
2
(103)
More generally, we have
∞
sin(2n + 1)t
n=0
(2n + 1)2
−t
∞
cos(2n + 1)t
n=0
2n + 1
−1
∞
22n sin2n+1 t 2n
=
n
(2n + 1)2
n=0
(104)
t
= t log tan
+ Cl2 (t) + Cl2 (π − t)
2
See
t xmalso the recent paper by Cho et al. [16] who consider the integrals
dx and, in particular,
0 sin x
0
π
3
√ √
x
3 2
3
π
1
dx = − log 3 −
π +
ζ 2,
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
sin
2n+1
t
x dx =
2n
sin x sin x dx =
0
0
0
t
(1 − cos2 x)n sin x dx
t
n n
=
(−1)k cos2k x sin x dx
k
0
k=0
and we therefore obtain
t
sin
0
2n+1
n n (−1)k
n
n (−1)k cos2k+1 t
x dx =
−
k 2k + 1
k
2k + 1
k=0
k=0
With t = π/2 we get
π
2
sin
2n+1
0
n (−1)k
n
x dx =
k 2k + 1
k=0
as compared with the frequently quoted form of the Wallis integral formula
π
2
sin2n+1 x dx =
0
(2n)!!
(2n + 1)!!
We therefore have the following identity which is reported in [33, p. 270]
n (2n)!!
[2n n!]2
1
n
(−1)k
=
=
k
2k + 1
(2n + 1)!!
(2n + 1)!
k=0
(105)
so that
π
2
sin
0
2n+1
[2n n!]2
x dx =
(2n + 1)!
(106)
Donal F. Connon
52
We recall (5)
∞
n xk
1
n
s−1 −yu
−u n
1
+
xe
=
u
e
du
k (k + y)s
Γ(s) 0
k=0
and note that
n n 1 n
1
1
n
k
(−1)
(−1)k
=
k
2k + 1
2 k=0 k
k+
k=0
1
2
1
=
2
∞
0
n
e−u/2 1 − xe−u du
Integrating (104) gives us
2
∞
n=0
∞
∞
1
cos(2n + 1)t
sin(2n + 1)t
−2
−t
3
3
(2n + 1)
(2n + 1)
(2n + 1)2
n=0
n=0
=
∞
n=0
22n
(2n + 1)2
2n
n
−1 t
sin2n+1 x dx
0
and with t = π/2 we have
−1 π
∞
∞
2
π (−1)n+1
22n
7
2n
=
sin2n+1 x dx
ζ(3) +
2
2
n
4
2 n=0 (2n + 1)
(2n
+
1)
0
n=0
=
∞
n=0
22n
(2n + 1)2
2n
n
−1
[2n n!]2
(2n + 1)!
Hence we obtain
7
1
24n
[n!]4
ζ(3) − πG =
4
2
(2n + 1)3 [(2n)!]2
n=0
∞
which was also previously obtained by Batir [9].
We now multiply (28) by sin x and integrate to obtain
(107)
Some infinite series involving the Riemann zeta function
π
2
0
1 [n!]2 2n
2
x sin x dx =
2 n=1 n2 (2n)!
∞
2
π
2
53
sin2n+1 x dx
0
and therefore we see that
24n [n!]4
1
π−2=
2 n=1 n2 (2n + 1) [(2n)!]2
∞
(108)
Let us now divide (28) by sin x and integrate to obtain
π
2
0
x2
1 [n!]2 2n
dx =
2
sin x
2 n=1 n2 (2n)!
∞
π
2
sin2n−1 x dx
0
In equation 6.29 in [22] we have previously shown that
0
π
2
7
x2
dx = 2π G − ζ(3)
sin x
2
(the evaluation of this integral is also contained in [16]) and using the integral
in [12, p. 113]
Using
π
2
sin
2n−1
0
n−1 (−1)k
(2n − 2)!!
n−1
xdx =
=
k
2k + 1
(2n − 1)!!
k=0
we obtain
8πG − 14ζ(3) =
∞
24n [n!]4
n3 [(2n)!]2
n=1
which was previously derived by Batir [9].
Letting p(x) = x cos x in (96) gives us
0
t
x cos x
dx =
sin x
t
x cot x dx = 2
0
∞ n=0
0
t
x cos x sin(2n + 1)x dx
(109)
Donal F. Connon
54
and we have using integration by parts for n ≥ 1
t
8
x cos x sin(2n+1)x dx =
0
sin(2nt)
cos(2nt) sin[2(n + 1)t]
cos[2(n + 1)t]
−2t
−2t
+
n2
n
(n + 1)2
n+1
This gives us
π
2
2
0
π (−1)n (−1)n
−
x cos x sin(2n + 1)x dx =
4 n+1
n
and thus
2
∞ n=0
π
2
x cos x sin(2n + 1)x dx =
0
π
log 2
2
This, of course, concurs with the well-known Euler integral
π
2
x cot x dx =
0
π
log 2
2
2
Dividing (28) by sin x and integrating gives us
0
t
x2
1 [n!]2 2n
2
dx =
2 n=1 n2 (2n)!
sin2 x
∞
t
sin2n−2 x dx
0
and with t = π/2 we have after dealing separately with the case for n = 1
0
π
2
π 1 [n!]2 2n π
x2
dx
=
+
2 2n−1
2 2 n=2 n2 (2n)!
2
sin2 x
∞
∞
1
π
= +π
2
(2n − 1)(2n − 2)
n=2
∞ π
1
1
= +π
−
2
2n − 1 2n − 2
n=2
2n − 2
n−1
Some infinite series involving the Riemann zeta function
∞ π
1
1
= +π
−
2
2n + 1 2n
n=1
∞ 1
π π
= +
2
2 n=1 n +
1
2
1
−
n
Since the digamma function may be represented by [43, p. 14]
1 ψ(a) = −γ − −
a n=1
∞
1
1
−
n+a n
we see that
∞ n=1
1
n+
1
2
1
−
n
1
= −ψ
−γ−2
2
We know that [43, p. 20]
1
ψ
= −γ − 2 log 2
2
and we therefore deduce that
0
π
2
x2
dx = π log 2
sin2 x
Using integration by parts we see that
0
π
2
x2
dx = 2
sin2 x
π
2
x cot x dx
0
and thus we have
0
π
2
x cot x dx = −
π
2
log sin x dx
0
Euler [25] was the first person to show that
55
Donal F. Connon
56
π
2
0
π
log sin x dx = − log 2
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
un cot
0
u
2
du = cos
nπ 2
n
Γ(n + 1) n−j
n!ζ(n+1)− (−1)j(j+1)/2
x Clj+1 (x)
Γ(n
+
1
−
j)
j=0
(110)
and also see the paper by Srivastava et al [61]. Ramanujan’s formula was
employed by Muzaffar [40]. With x = π we have
n
π
2
n
x cot x dx = cos
2
0
nπ 2
n
Γ(n + 1)
n!ζ(n+1)− (−1)j(j+1)/2
π n−j Clj+1 (π)
Γ(n
+
1
−
j)
j=0
(111)
Provided a = b we readily determine that
1 sin(a − b)x sin(a + b)x
−
+c
sin ax sin bx dx =
2
a−b
a+b
and hence we have
0
t
1 sin(p − 2n − 1)t sin(p + 2n + 1)t
sin px sin(2n + 1)x dx =
−
2
p − 2n − 1
p + 2n + 1
and more specifically
0
π
2
(−1)n+1 cos(pπ/2)
1
1
sin px sin(2n + 1)x dx =
+
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
π
2
0
57
∞ π
2
sin px
sin px sin(2n + 1)x dx
dx = 2
sin x
0
n=0
so that
π
2
0
∞
sin px
(−1)n
dx = 2p cos(pπ/2)
sin x
(2n + 1)2 − p2
n=0
(112)
and, using L’Höpital’s rule, we see that in the limit as p → 1
π
2
dx = lim
p→1
0
π
−pπ sin(pπ/2) + 2 cos(pπ/2)
=
−2p
2
With p = 2 we obtain
∞
n=0
1
(−1)n+1
=
2
2
(2n + 1) − 4
2
(113)
We may write (112) as
∞
n=0
(−1)n
1
=
(2n + 1)2 − p2
2p cos(pπ/2)
π
2
0
sin px
dx
sin x
and, using L’Höital’s rule, we obtain as p → 0
π
∞
2 x cos px
1
(−1)n
= lim
dx
2
p→0 −pπ sin(pπ/2) + 2 cos(pπ/2) 0
(2n
+
1)
sin
x
n=0
Hence we obtain (99) again
π
∞
2
(−1)n
x
2
=
dx
2
(2n + 1)
0 sin x
n=0
We note from Fettis [56] that
Donal F. Connon
58
0
π
2
pπ 3 + p sin px
π 1
1+p
ψ
dx = − cos
−ψ
sin x
2 2
2
4
4
(114)
and
pπ 3 + p 1+p
1
1
1+p
ψ
−ψ
− sin
−ψ
2
2
2
2
4
4
0
(115)
Comparing (112) with (114) gives us
π
2
1 − cos px
dx = ψ
sin x
pπ 3 + p π 1
1+p
(−1)n
ψ
2p cos(pπ/2)
= − cos
−ψ
(2n + 1)2 − p2
2 2
2
4
4
n=0
(116)
Differentiating (114) results in
∞
π
2
x cos px
dx
sin x
0
pπ 3 + p pπ 3 + p 1+p
π
1+p
1
ψ
−ψ
+ sin
ψ
−ψ
= − cos
8
2
4
4
4
2
4
4
and with p = 0 we obtain
π
2
0
1
1 3
x
dx = − ψ
−ψ
sin x
8
4
4
Using (99) we see that
1
1 3
2G = − ψ
−ψ
8
4
4
which concurs with Kölbig [58].
With p = 1 we have the well known integral
(117)
Some infinite series involving the Riemann zeta function
59
π
1
π
x cot x dx =
ψ (1) − ψ
= log 2
4
2
2
0
With p = 2 we obtain
π
2
1
3
x
1 − 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
pπ 3 + p 1+p
1
− sin
ψ
−ψ
8
2
4
4
0
pπ 3 + p 1+p
π
ψ
−ψ
− cos
4
2
4
4
and with p = 0 we obtain
π
3
1
1 1
ψ
=
ψ
−ψ
2
2
4
4
4
so that
1
π2
ψ
=
(118)
2
2
π
2
1
x sin px
dx = ψ sin x
2
1+p
2
in accordance with Kölbig’s paper [58].
With p = 1 we obtain
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 = 2n + 1
0
t
1 cos(p − 2n − 1)t cos(p + 2n + 1)t
2n + 1
cos px sin(2n+1)x dx =
−
− 2
2
p − 2n − 1
p + 2n + 1
p − (2n + 1)2
Donal F. Connon
60
and thus
0
π
2
1
1
2n + 1
(−1)n+1
+
− 2
cos px sin(2n+1)x dx =
2
p − 2n − 1 p + 2n + 1
p − (2n + 1)2
=
(−1)n+1 p − (2n + 1)
p2 − (2n + 1)2
Therefore letting p(x) = 1 − cos px in (96) gives us
0
π
2
∞ 1
(−1)n p + (2n + 1)
1 − cos px
dx = 2
+
sin x
2n
+
1
p2 − (2n + 1)2
n=0
(119)
and comparing this with (115) results in
∞ 1
(−1)n p + (2n + 1)
2
+
2n + 1
p2 − (2n + 1)2
n=0
pπ 3 + p 1+p
1
1
1+p
=ψ
−ψ
− sin
ψ
−ψ
2
2
2
2
4
4
pπ Multiplying across by cos 2 gives us
2 cos
∞ pπ 2
n=0
(−1)n p + (2n + 1)
1
+
2n + 1
p2 − (2n + 1)2
pπ 1 + p pπ pπ 3 + p 1
1
1+p
= cos
ψ
−ψ
− cos
ψ
−ψ
sin
2
2
2
2
2
2
4
4
We have
2 cos
∞ pπ 2
n=0
1
(−1)n p + (2n + 1)
+
2n + 1
p2 − (2n + 1)2
Some infinite series involving the Riemann zeta function
61
∞ pπ 3 + p pπ 1
π 1
1+p
2n + 1
= − cos
ψ
−ψ
+2 cos
+
2 2
2
4
4
2 n=0 2n + 1 p2 − (2n + 1)2
∞ pπ pπ 3 + p π 1
1
1+p
1
2
ψ
= − cos
−ψ
+2p cos
2 2
2
4
4
2 n=0 2n + 1 p2 − (2n + 1)2
and therefore we see that
2p2
∞
n=0
=ψ
1
1
2
2n + 1 p − (2n + 1)2
(120)
pπ 3 + p pπ 1
1
1+p
π
1+p
−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
= ψ(1 + p) + ψ(1 − p) + 2γ
2p
2
2
n
p
−
n
n=1
2
and separating the even and odd terms gives us
∞
∞
∞
1
1
1
1
1
1
=
+
2
2
2
2
2
np −n
2n p − 4n
2n + 1 p − (2n + 1)2
n=1
n=1
n=0
Letting p → p/2 in (121) we see that
∞
p
p
1
1
2
ψ 1+
+ψ 1−
+ 2γ = 2p
2
2
2
n p − 4n2
n=1
and hence we obtain
p
p
1 ψ 1+
+ψ 1−
+ 2γ
ψ(1 + p) + ψ(1 − p) + 2γ −
2
2
2
(121)
Donal F. Connon
62
=ψ
pπ 3 + p pπ 1+p
1
1
1+p
π
−ψ
+ 1 − sin
ψ
−ψ
− / cos
2
2
2
2
4
4
2
2
which may be written as
p
p 1 ψ 1+
+ψ 1−
ψ(1 + p) + ψ(1 − p) −
2
2
2
=ψ
(122)
pπ 3 + p pπ 1+p
1
1+p
π
+2 log 2+ 1 − sin
ψ
−ψ
− / cos
2
2
2
4
4
2
2
For example, with p = 1/2 we obtain
1
1
1
1
3
ψ 1+
+ψ
−
ψ 1+
+ψ
2
2
2
4
4
√
3
7
1
1√
3
π 2
=ψ
2 ψ
+ 2 log 2 +
1−
−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])
(m)
where Hn
sin
−1
y
4
3 Hn−1
=
2 n=1 n2
∞
(2)
2n
n
−1
(2y)2n , −1 ≤ y ≤ 1
is the generalized harmonic number
Hn(m)
n
1
=
km
k=1
Therefore upon letting x = sin−1 y we get
(123)
Some infinite series involving the Riemann zeta function
3 Hn−1
x =
2 n=1 n2
∞
4
(2)
2n
n
−1
63
22n sin2n x, −π/2 ≤ x ≤ π/2
(124)
Integration results in
π5
3 Hn−1
=
160
2 n=1 n2
∞
(2)
2n
n
−1
2n
π
2
2
sin2n x dx
0
and hence we obtain the Euler sum
∞
(2)
π5
3π Hn−1
=
160
4 n=1 n2
or equivalently
Hn−1 Hn
π4
=
− ζ(4)
=
120 n=1 n2
n2
n=1
∞
Since ζ(4) =
example [19])
π4
90
∞
(2)
(2)
this may be written in its more familiar form as (see for
Hn
7
ζ(4) =
4
n2
n=1
∞
(2)
Alternatively, we now multiply equation (124) by cot x and integrate to obtain
0
t
3 Hn−1
x cot x dx =
2 n=1 n2
∞
4
(2)
2n
n
−1
2n
2
t
sin2n−1 x cos x dx
0
which results in
0
t
3 22n Hn−1 sin2n t
x cot x dx =
4 n=1
n3
4
∞
(2)
2n
n
−1
(125)
Donal F. Connon
64
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
3 22n Hn−1
4t(t − 6) cos t + (t − 12t + 24) sin t =
2 n=1 (2n + 1)n2
2
4
∞
2
(2)
2n
n
−1
sin2n+1 t
and with t = π/2 we obtain
π4
3 22n Hn−1
2
− 3π + 24 =
16
2 n=1 (2n + 1)n2
∞
(2)
2n
n
−1
(126)
We showed in [19] that for 0 ≤ t < 1
0
t
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
∞
ζ(2n)x2n , (|x| < 1)
(128)
n=0
(where ζ(0) = −1/2). This gives us
∞
ζ(2n) 2n+1
t
= ζ (−1, t) − ζ (−1, 1 − t) + t log(2 sin πt)
−2
2n + 1
n=0
(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
0
t
[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)] +
0
t
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 πt
this becomes
= t log sin πt −
t
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
2
0
Let us now differentiate (129) to obtain
−2
∞
n=0
ζ(2n)t2n =
d [ζ (−1, t) − ζ (−1, 1 − t)] + log(2 sin πt) + π cot πt
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
Donal F. Connon
66
∂ ∂
∂ ∂
∂
ζ(s, t) =
ζ(s, t) = − [sζ(s + 1, t)]
∂t ∂s
∂s ∂t
∂s
and we then see that
∂ ∂
∂
ζ(s, t) = −ζ(s + 1, t) − s ζ(s + 1, t)
∂t ∂s
∂s
Hence with s = −1 we see that
(131)
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
and we then obtain
−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
log(2π)
2
and the well-known result [7, p. 264] involving the Bernoulli polynomials
Bn (t)
ζ(1 − n, t) = − Bnn(t) for n ≥ 1
this becomes
ζ (0, t) = log Γ(t) −
−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
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
∞
ζ(2n) t2n = πt cot πt
n=0
Dividing this by t, dealing separately with the n = 0 term, and integrating
results in
∞
ζ(2n)
n=1
t2n
= log t − log sin πt + c
n
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
= log(πt) − log sin πt
n
We now multiply this by t and integrate to obtain
∞
n=0
ζ(2n)
u2n+2 = u log(πu) − u −
n(2n + 2)
u
t log sin πt dt
0
and, in the case u = 1/2, we obtain using (31)
∞
2π 2
1 ζ(2n)
ζ(3) =
log π − −
7
2 n=1 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 form
of Euler’s infinite product identity for the sine function
x2
log sin x = log x +
log 1 − 2 2
nπ
n=1
∞
Upon integrating (131) with respect to t we see that
v
v
v
∂ ∂
ζ (s + 1, t)dt =
ζ(s + 1, t) dt
ζ(s, t)dt +
−s
0
0 ∂t ∂s
0
Donal F. Connon
68
We therefore get
v
ζ (s + 1, t)dt = ζ (s, v) − ζ (s, 0) +
−s
0
v
ζ(s + 1, t) dt
0
and with s = −n we have
n
0
v
ζ (1 − n, t)dt = ζ (−n, v) − ζ (−n, 0) +
0
v
ζ(1 − n, t) dt
Then using the well-known formula
ζ(1 − n, v) = − Bnn(v) for n ≥ 1
we obtain
n
0
v
ζ (1 − n, t) dt =
Bn+1 − Bn+1 (v)
+ ζ (−n, v) − ζ (−n, 0)
n(n + 1)
(132)
This identity was originally derived by Adamchik [1] in a different manner
in 1998. With n = 2 we obtain
v
1
1
ζ(3)
ζ (−1, t) dt = − B3 (v) + ζ (−2, v) +
(133)
12
2
8π 2
0
since ζ (−n, 0) = ζ (−n) and ζ (−2) = − ζ(3)
.
4π 2
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
Sr (x) = exp
= exp
xr−1
r−1
xr−1
r−1
∞
Pr
x nr−1
n=−∞,n=0
∞ Pr
69
n
(135)
x
x nr−1
Pr −
n
n
n=1
where
Pr (u) = (1 − u) exp
u u2
ur
+
+ ··· +
1
2
r
For example, the triple sine function is defined by
n2 ∞
x2
x2
2
1− 2
ex
S3 (x) = e 2
n
n=1
(136)
and we then have
∞ 1 2 2
x2
2
log S3 (x) = x +
n log 1 − 2 + x
2
n
n=1
where upon differentiation results in
∞ S3 (x)
2n2 x
=x+
+ 2x
S3 (x)
x 2 − n2
n=1
= x + x2
= x2
∞
n=1
2x
x 2 − n2
1 2x
+
x n=1 x2 − n2
∞
We have the well-known decomposition formula [12, p.131]
1 2x
π cot πx = +
x n=1 x2 − n2
∞
(137)
Donal F. Connon
70
and, since S3 (0) = 1, we then see that
x
πt2 cot πt dt
log S3 (x) =
(138)
0
The double sine function is defined by
S2 (x) = e
n
∞ 1 − x/n
x
1 + x/n
n=1
2x
e
(139)
It is easily seen that
log S2 (x) = x +
∞ x
x n log 1 −
− log 1 −
+ 2x
n
n
n=1
and, in the same manner as before, we easily find that
x
log S2 (x) =
πt cot πt dt
(140)
(141)
0
The function S1 (x) is defined by
∞ x2
S1 (x) = 2πx
1− 2
= 2 sin πx
n
n=1
(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]
∞
1
x
x2
2
2
G(1 + x) = (2π) exp[− (γx + x + x)] [(1 + )k exp( − x)]
2
k
2k
k=1
x
2
(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
x
1
1 2 1 2
log G(1+x) = x log(2π)− x(1+x)− γx =
x − x + n log 1 +
2
2
2
2n
n
n=1
(144)
and letting x → −x gives us
∞ x
1
1 2 1 2
1
x + x + n log 1 −
log G(1−x) = − x log(2π)+ x(1−x)− γx =
2
2
2
2n
n
n=1
Subtraction results in
log G(1 − x) − log G(1 + x)
= x − x log(2π) +
∞ n=1
x
x 2x + n log 1 −
− n log 1 +
n
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 by
Kinkelin in 1860 [43, p. 30]
x
G(1 + x)
πt cot πt dt
(146)
log
= x log(2π) −
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
Γn (u)[Γn (−u)]
= exp[−π
xn−1 cot πxdx]
0
for n ≥ 2.
(147)
Donal F. Connon
72
where we have followed the Vignéras 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
x
log Sn (x) =
πtn−1 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
∞ 1
2n + 1
− 1 = (1 − log 2)
n log
2n − 1
2
n=1
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
x log 1 +
log Γ(x + 1) =
− log 1 +
n
n
n=1
and this may be written as
log Γ(x + 1) =
∞
1
x log 1 +
n
+
∞
(−1)k xk
knk
∞
∞
1
1
(−1)k xk
x log 1 +
=
−
+
n
n
knk
n=1
k=2
n=1
k=1
∞
∞ ∞ 1
1
(−1)k xk
log 1 +
−
x+
=
n
n
knk
n=1
n=1 k=2
∞ ∞
∞
1
1
(−1)k xk 1
log 1 +
=
−
x+
n
n
k
nk
n=1
n=1
k=2
Hence we have the well-known Maclaurin series [43, p. 160] for log Γ(x + 1)
Some infinite series involving the Riemann zeta function
log Γ(x + 1) = −γx +
∞
(−1)k ζ(k)
k=2
k
xk
73
(149)
Now referring back to (144)
∞ 1
x
1
1 2 1 2
log G(1+x) = x log(2π)− x(1+x)− γx +
x − x + n log 1 +
2
2
2
2n
n
n=1
and applying the same procedure as with log Γ(x + 1) above this becomes
∞
∞
1
1 2 1 2
(−1)k xk
1
x −x−
= x log(2π) − x(1 + x) − γx +
2
2
2
2n
knk−1
n=1
k=1
∞ ∞
1
1
1 2 (−1)k xk
= x log(2π) − x(1 + x) − γx −
2
2
2
knk−1
n=1 k=3
∞ ∞
1
1
1 2 (−1)m+1 xm+1
= x log(2π) − x(1 + x) − γx −
2
2
2
(m + 1)nm
n=1 m=2
∞
∞
1
1
1 2 (−1)m+1 xm+1 1
= x log(2π) − x(1 + x) − γx −
2
2
2
m+1
nm
m=2
n=1
∞
1
1 2 (−1)m+1 ζ(m)xm+1
1
= x log(2π) − x(1 + x) − γx −
2
2
2
m+1
m=2
Hence, as reported in [43, p. 211], we obtain
∞
1
1 2 (−1)k+1 ζ(k)xk+1
1
(150)
log G(1 + x) = x log(2π) − x(1 + x) − γx −
2
2
2
k+1
k=2
Donal F. Connon
74
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)k+1 ζ(k) k+1
1
1 2 (−1)k+1 ζ(k) k+1
1
2
x log(2π)− x(1+x)− γx −
x +γx +
x
2
2
2
k+1
k
k=2
k=2
= ζ (−1) − ζ (−1, 1 + x)
This may be written as
∞
1
1
1
(−1)k+1 ζ(k) k+1
x
ζ (−1)−ζ (−1, 1+x) = x log(2π)− x(1+x)+ γx2 +
2
2
2
k(k
+
1)
k=2
(152)
which is contained in a slightly disguised form in [43, p. 222].
Differentiation results in
−
∞
1
d 1
(−1)k+1 ζ(k) k
ζ (−1, 1 + x) = log(2π) − − x + γx +
x
dx
2
2
k
k=2
and as shown previously we have
d ζ (−1, 1 + x) = ζ (0, 1 + x) − ζ(0, 1 + x)
dx
= log Γ(1 + x) −
1
1
log(2π) + + x
2
2
Therefore we simply recover (149)
log Γ(x + 1) = −γx +
∞
(−1)k ζ(k)
k=2
k
xk
Some infinite series involving the Riemann zeta function
75
We have
∞ 1 2 2
x2
2
n log 1 − 2 + x
log S3 (x) = x +
2
n
n=1
∞
∞
1 2 2 x2k
= x +
x −
2
kn2k−2
n=1
k=1
∞ ∞
∞
∞
1
1 2 x2k 1
x2k
= x2 −
=
−
x
2
kn2k−2
2
k n=1 n2k−2
n=1 k=2
k=2
∞
1 2 ζ(2k − 2) 2k
= x −
x
2
k
k=2
This gives us
∞
∞
1 2 ζ(2k) 2k+2
ζ(2k) 2k+2
log S3 (x) = x −
=−
x
x
2
k+1
k+1
k=1
k=0
(153)
The following identity for |x|<1 appears in the book by Srivastava and Choi
[43, p. 216]
∞
ζ(2k)
k=1
k+1
x
2k+2
−x
x
1
G(1 + x)
2
= [1−log(2π)]x +x log
log G(1+t) dt−
log G(1+t) dt
−
2
G(1 − x) 0
0
We note that
0
−x
log G(1 + t) dt = −
x
0
log G(1 − t) dt
and integration by parts shows that
−x
x
x
d
G(1 + t)
G(1 + x)
log G(1+t)dt−
log G(1+t)dt =
t log
−
dt
x log
G(1 − x) 0
dt
G(1 − t)
0
0
Donal F. Connon
76
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
x
d
G(1 + t)
1
2
t log
πt2 cot πt dt
dt = log(2π)x −
dt
G(1 − t)
2
0
0
This results in
∞
ζ(2k)
k=1
k+1
x
2k+2
1
= x2 −
2
x
πt2 cot πt dt
(154)
0
We have the following well-known identity
t cot t =
∞
(−1)n
n=0
22n B2n 2n
t , (|t| < π)
(2n)!
(155)
Combining this with Euler’s formula
2n−1 2n
π B2n
n+1 2
ζ(2n) = (−1)
, (n ≥ 1)
(156)
ζ(2n) t2n , (|t| < 1)
(157)
(2n)!
and, letting t → πt, we obtain
πt cot πt = −2
∞
n=0
Since the first term of the series (155) is equal to B0 = 1, to be consistent
with (157), we define ζ(0) = − 12 (which in fact also coincides with the value
determined by the analytic continuation of ζ(s) at s = 0.) We may now
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
x
log G(1+t)dt =
0
77
1
1
1
− 2 log A x+ log(2π)x2 − x3 −(1−x) log G(1+x)
4
4
6
+ 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 and
triple gamma functions. This type of representation naturally arises from
the combination of the facts that
x
πtn−1 cot πt dt
log Sn (x) =
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]
Γ3 (1 + x) = exp(c1 x + c2 x2 + c3 x3 )F (x)
where
F (x) =
∞ k=1
=
x − 12 k(k+1)
1+
exp
k
∞ k=1
x − 12 k(k+1)
1+
exp
k
1
1
(k + 1)x −
2
4
1
1+
k
1
1+
k
1 x3 1 x3
x +
+
6 k
6 k2
2
1 x3
1
1
kx − x2 +
2
4
6 k
and
1
3 1
1
γ + log(2π) +
c1 = − log(2π) − log A, c2 =
8 4
4
2
1
3
c3 = − γ + ζ(2) +
6
2
78
Donal F. Connon
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