IJMMS 28:7 (2001) 403–411
PII. S0161171201006603
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
NEW SERIES INVOLVING THE ZETA FUNCTION
WU YUN-FEI
(Received 29 January 2001)
Abstract. We evaluate sums of certain classes of new series involving the Riemann zeta
function by using the theory of the double gamma function and a property of the gamma
function. Relevant connections with various known results are also pointed out.
2000 Mathematics Subject Classification. 11M99.
1. Introduction. Barnes [1] defined the double gamma function Γ2 = 1/G satisfying
each of the following properties:
(a) G(z + 1) = Γ (z)G(z), for any complex number z;
(b) G(1) = 1;
(c) As n → ∞,
log G(z + n + 2) =
2
n+1+z
n
5
z2
log(2π ) +
+n+
+
+ (n + 1)z log n
2
2
12
2
2
3z
1
1
−
− n(1 + z) − log A +
+O
,
4
12
n
(1.1)
where Γ is the well-known gamma function
Γ −1 (z) = zeγz
∞ 1+
n=1
z −z/n
e
,
n
(1.2)
and A is called Glaisher’s (or Kinkelin’s) constant defined by
log A = lim
n→∞
2
n
1
n2
n
+ +
log n +
,
log 11 · 22 · · · nn −
2
2 12
4
(1.3)
the numerical value of A being 1.282427130 . . . .
From this definition, Barnes deduced
−1
Γ2 (z + 1)
= G(z + 1) + (2π )z/2 e−(1/2)[(1+γ)z
2 +z]
∞ 1+
n=1
z
n
n
e−z+z
2 /2n
,
(1.4)
where γ denotes the Euler-Mascheroni constant given by
1
1
γ = lim 1 + + · · · + − log n 0.577215664 . . . .
n→∞
2
n
(1.5)
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WU YUN-FEI
It is also known that [1]
1
= π 1/2 ,
Γ
2
G
1
= 21/24 · π −1/4 · e1/8 · A−3/2 .
2
(1.6)
The gamma function has the property
m−1
2(m−1)/2 π (m+1)/2 m1/2−ms
1
···Γ s +
Γ (1 − ms) =
.
Γ (s)Γ s +
m
m
sin(mπ s)
(1.7)
The Riemann zeta function ζ(s) is defined by
ζ(s) =
∞
1
,
s
n
n=1
Res(s) > 1.
(1.8)
Indeed it is analytic for all s except for a simple pole at s = 1 with residue 1.
In [2, 3] Choi, Srivastava, and Quine used the theory of the double gamma function
to evaluate some series associated with the zeta function. Now in the present paper,
we use the property of the gamma function and (2.23) and (2.70) in [2] to evaluate new
series
∞
ζ(2k + 1)
α2k ,
(2k
+
1)(2k
+
2)
k=1
∞
ζ(k)
2−k ,
k(k
+
1)(k
+
2)
k=2
∞
ζ(k)
,
k(k
+
1)(k
+ 2)(k + 3)
k=2
∞
(−1)k ζ(k)
2−k ,
k(k
+
1)(k
+
2)
k=2
∞
(−1)k ζ(k)
,
k(k + 1)(k + 2)(k + 3)
k=2
(1.9)
where α can be taken as 1/3, 2/3, 1/4, 3/4, 1/6, 5/6. And relevant connections with
various known results are also pointed out.
2. Series involving the zeta function. Let s = x and making 0 < x < 1 in (1.2), we
have
log Γ (x) + log x + γx =
∞ x
x
− log 1 +
n
n
n=1
=
∞
∞ ∞
∞
(−1)k x k
(−1)k x k 1
=
knk
k
nk
n=1 k=2
n=1
k=2
=
∞
(−1)k x k ζ(k)
.
k
k=2
(2.1)
Replacing x by 1 − x, 1 − 2x, 1 − 3x, x + 1/2, x + 1/3, and x + 2/3, we get
log Γ (1 − x) + log(1 − x) + γ(1 − x) =
∞
(−1)k (1 − x)k ζ(k)
,
k
k=2
0 < x < 1,
(2.2)
log Γ (1 − 2x) + log(1 − 2x) + γ(1 − 2x) =
∞
(−1)k (1 − 2x)k ζ(k)
,
k
k=2
0<x<
1
, (2.3)
2
log Γ (1 − 3x) + log(1 − 3x) + γ(1 − 3x) =
∞
(−1)k (1 − 3x)k ζ(k)
,
k
k=2
0<x<
1
, (2.4)
3
405
NEW SERIES INVOLVING THE ZETA FUNCTION
∞
1
1
(−1)k (x+1/2)k ζ(k)
1
=
,
log Γ x+ +log x+ +γ x+
2
2
2
k
k=2
1
0<x < ,
2
(2.5)
∞
1
1
(−1)k (x+1/3)k ζ(k)
1
=
,
log Γ x+ +log x+ +γ x+
3
3
3
k
k=2
2
0<x < ,
3
(2.6)
∞
2
2
(−1)k (x+2/3)k ζ(k)
2
=
,
log Γ x+ +log x+ +γ x+
3
3
3
k
k=2
1
0<x < .
3
(2.7)
Let s = x, 0 < x < 1 and making m = 1, m = 2, m = 3 in (1.7), we get
Γ (x)Γ (1 − x) =
π
,
sin π x
(2.8)
1
21−2x π 3/2
Γ (x)Γ x +
Γ (1 − 2x) =
,
2
sin(2π x)
1
2
2π 2 31/2−3x
Γ (x)Γ x +
Γ x+
Γ (1 − 3x) =
.
3
3
sin(3π x)
(2.9)
(2.10)
Combining (2.1), (2.2), and (2.8), we obtain
log 2π − log(2 sin π x) + log x(1 − x) + γ
∞
(−1)k x k + (1 − x)k ζ(k)
,
=
k
k=2
0 < x < 1.
(2.11)
Combining (2.1), (2.3), (2.5), and (2.9), we obtain
1
1
3
3
log 2π +
− 2x log 2 − log 2 sin(2π x) + log x x +
(1 − 2x) + γ
2
2
2
2
∞
(−1)k x k + (x + 1/2)k + (1 − 2x)k ζ(k)
1
, 0<x< .
=
k
2
k=2
(2.12)
Combining (2.1), (2.4), (2.6), (2.7), and (2.10), we obtain
1
2
1
−3x log 3−log 2 sin(3π x) +log x x+
x+ (1−3x) +2γ
2
3
3
∞
(−1)k x k + (x + 1/3)k + (x + 2/3)k + (1 − 3x)k ζ(k)
1
, 0<x< .
=
k
3
k=2
2 log 2π +
(2.13)
On the other hand, recalling Euler formula
∞ x2
sin π x
=
1− 2 ,
n
πx
n=1
0 < x < 1,
(2.14)
406
WU YUN-FEI
similarly, we get
log(2 sin π x) − log(2π x) = −
∞
x 2k ζ(2k)
,
k
k=1
∞
(1 − x)2k ζ(2k)
log 2 sin π (1 − x) − log 2π (1 − x) = −
,
k
k=1
(2.15)
(2.16)
∞
(1 − 2x)2k ζ(2k)
log 2 sin π (1 − 2x) − log 2π (1 − 2x) = −
,
k
k=1
(2.17)
∞
(1 − 3x)2k ζ(2k)
log 2 sin π (1 − 3x) − log 2π (1 − 3x) = −
,
k
k=1
(2.18)
∞
1
1
(x + 1/2)2k ζ(2k)
log 2 sin π x +
− log 2π x +
=−
,
2
2
k
k=1
(2.19)
∞
1
1
(x + 1/3)2k ζ(2k)
log 2 sin π x +
− log 2π x +
=−
,
3
3
k
k=1
(2.20)
∞
2
2
(x + 2/3)2k ζ(2k)
log 2 sin π x +
− log 2π x +
=−
.
3
3
k
k=1
(2.21)
Combining (2.15) and (2.16), we obtain
log(2 sin π x) − log 2π −
∞ 2k
x + (1 − x)2k ζ(2k)
1
log x(1 − x) = −
.
2
2k
k=1
(2.22)
Noting that
1
= log 2 sin(2π x) ,
log(2 sin π x) + log 2 sin π x +
2
(2.23)
and by combining (2.15), (2.17), and (2.19), we obtain
3
1
1
(1 − 2x)
log 2 sin(2π x) − log 2π − log x x +
2
2
2
∞ 2k
2k
x + (x + 1/2) + (1 − 2x)2k ζ(2k)
=−
.
2k
k=1
(2.24)
Noting that
2
1
+ log 2 sin π x +
= log 2 sin(3π x) , (2.25)
log(2 sin π x) + log 2 sin π x +
3
3
and by combining (2.15), (2.18), (2.20), and (2.21), we obtain
1
1
2
log 2 sin(3π x) − 2 log 2π − log x x +
x+
(1 − 3x)
2
3
3
∞ 2k
2k
2k
x + (x + 1/3) + (x + 2/3) + (1 − 3x)2k ζ(2k)
.
=−
2k
k=1
(2.26)
407
NEW SERIES INVOLVING THE ZETA FUNCTION
Now combining (2.11) and (2.22), (2.12) and (2.24), (2.13) and (2.26), respectively,
we get
∞ 2k+1
+ (1 − x)2k+1 ζ(2k + 1)
x
1
log x(1 − x) + γ = −
,
2
2k + 1
k=1
0 < x < 1,
1
1
1
3
− 2x log 2 + log x x +
(1 − 2x) + γ
2
2
2
2
2k+1
∞
2k+1
2k+1
x
+ (x + 1/2)
+ (1 − 2x)
ζ(2k + 1)
,
=−
2k
+
1
k=1
1
0<x< ,
2
(2.27)
(2.28)
1
1
1
2
− 3x log 3 + log x x +
x+
(1 − 3x) + 2γ
2
2
3
3
2k+1
∞
x
+ (x + 1/3)2k+1 + (x + 2/3)2k+1 + (1 − 3x)2k+1 ζ(2k + 1)
=−
,
2k + 1
k=1
0<x<
1
.
3
(2.29)
Integrating both sides of (2.27) from x = 0 to x = a, we get
∞ 2k+2
− (1 − a)2k+2 + 1 ζ(2k + 1)
a
1
1
.
a log a − (1 − a) log(1 − a) + a(γ − 1) = −
(2k + 1)(2k + 2)
2
2
k=1
(2.30)
Making a = 1/2, a = 1/3, a = 1/4, and a = 1/6, we have
∞
1
ζ(2k + 1)
= (1 − γ),
(2k
+
1)(2k
+
2)
2
k=1
(2.31)
∞ (2/3)2k+2 − (1/3)2k+2 ζ(2k + 1)
1
1
1
= − log 2 + log 3 + (1 − γ),
(2k
+
1)(2k
+
2)
3
6
6
k=1
(2.32)
∞ (3/4)2k+2 − (1/4)2k+2 ζ(2k + 1)
1
3
1
= log 2 − log 3 + (1 − γ),
(2k
+
1)(2k
+
2)
2
8
4
k=1
(2.33)
∞ (5/6)2k+2 − (1/6)2k+2 ζ(2k + 1)
1
5
1
= log 6 −
log 5 + (1 − γ).
(2k
+
1)(2k
+
2)
3
12
3
k=1
(2.34)
Integrating both sides of (2.28) from x = 0 to x = 1/4, we have
−
3
3
9
log 2 + log 3 − (1 − γ)
16
8
8
∞ (1/4)2k+2 + (3/4)2k+2 − 3(1/2)2k+3 + 1/2 ζ(2k + 1)
.
=−
(2k + 1)(2k + 2)
k=1
(2.35)
Recalling the formula [2, page 113]
∞
ζ(2k + 1)
1
7
2−2k = 1 + log 2 − γ − 6 log A,
(2k
+
1)(2k
+
2)
6
2
k=1
(2.36)
408
WU YUN-FEI
and combining (2.31), (2.35), and (2.36), we get
∞ (1/4)2k+2 + (3/4)2k+2 ζ(2k + 1)
1
3
5
9
= + log 2 − log 3 −
γ − log A.
(2k
+
1)(2k
+
2)
2
8
16
4
k=1
(2.37)
Solving (2.33) and (2.37), we obtain the formulas
∞
ζ(2k + 1)
γ
4−2k = 2 + 4 log 2 − − 18 log A,
(2k
+
1)(2k
+
2)
2
k=1
2k
3
2
γ
ζ(2k + 1)
2 4
= + log 2 − log 3 − − 2 log A.
(2k
+
1)(2k
+
2)
4
3
3
3
2
k=1
∞
(2.38)
Multiplying 1/3 − x on both sides of (2.29), and integrating from x = 0 to x = 1/3,
we have
∞
2 1
ζ(2k + 1)
19
γ
8
− − log 2 +
log 3 + = −
9 9
108
9
9 k=1 (2k + 1)(2k + 2)(2k + 3)
(2.39)
∞ 1 (1/3)2k+2 + (2/3)2k+2 − 1/3 ζ(2k + 1)
.
+
3 k=1
(2k + 1)(2k + 2)
Recalling the formula [2, page 118]
∞
ζ(2k + 1)
3 γ
= − − log A,
(2k
+
1)(2k
+
2)(2k
+
3)
8 6
k=1
(2.40)
and combining (2.31), (2.39), and (2.40), we have
∞ (1/3)2k+2 + (2/3)2k+2 ζ(2k + 1)
1 1
19
5
8
= − log 2 +
log 3 −
γ − log A. (2.41)
(2k
+
1)(2k
+
2)
2
3
36
18
3
k=1
Solving (2.32) and (2.41), we obtain the formulas
∞
γ
3 13
ζ(2k + 1)
3−2k = +
log 3 − − 12 log A,
(2k
+
1)(2k
+
2)
2
8
2
k=1
2k
2
25
γ
3 3
ζ(2k + 1)
log 3 − − 3 log A.
= − log 2 +
(2k
+
1)(2k
+
2)
3
4
4
32
2
k=1
∞
(2.42)
Integrating both sides of (2.29) from x = 0 to x = 1/6, we get
1
1
5
γ
− log 2 +
log 3 +
log 5 +
3
24
12
3
∞
(1/6)2k+2 +(5/6)2k+2 −(1/3)2k+2 −(2/3)2k+2 +(2/3)(1/2)2k+2 +1/3 ζ(2k + 1)
=−
.
(2k + 1)(2k + 2)
k=1
−
(2.43)
NEW SERIES INVOLVING THE ZETA FUNCTION
409
Combining (2.31), (2.36), (2.41), and (2.43), we have
∞ (1/6)2k+2 + (5/6)2k+2 ζ(2k + 1)
(2k + 1)(2k + 2)
k=1
(2.44)
35
5
13
5
1 17
log 2 +
log 3 −
log 5 −
γ − log A.
= +
2 36
72
12
36
3
Solving (2.44) and (2.34), we obtain the formulas
∞
11
γ
5
ζ(2k + 1)
6−2k = 3 + log 2 +
log 3 − − 30 log A,
2
(2k
+
1)(2k
+
2)
2
4
k=1
2k
ζ(2k + 1)
5
59
3
γ 6
3 29
log 2 +
log 3 − log 5 − − log A.
= +
(2k
+
1)(2k
+
2)
6
5
50
100
5
2 5
k=1
∞
(2.45)
Multiplying 1/2 − x on both sides of (2.27), and integrating from x = 0 to x = 1/2,
we have
−
∞
∞ (1/2)2k+2 − 1 ζ(2k + 1)
1
γ
1
ζ(2k + 1)
1
− log 2 + = −
−
.
16 8
8
2 k=1 (2k + 1)(2k + 2) k=1 (2k + 1)(2k + 2)(2k + 3)
(2.46)
Combining (2.31), (2.40), and (2.46), we obtain the formula
∞
γ
ζ(2k + 1)
3 1
2−2k = + log 2 − − 4 log A.
(2k
+
1)(2k
+
2)(2k
+
3)
4
2
6
k=1
(2.47)
Multiplying (1/2−x)2 on both sides of (2.27), and integrating from x = 0 to x = 1/2,
we have
−
∞
∞
γ
1
ζ(2k + 1)
ζ(2k + 1)
1
+
=−
+
18 24
4 k=1 (2k + 1)(2k + 2) k=1 (2k + 1)(2k + 2)(2k + 3)
∞
ζ(2k + 1)
.
−2
(2k
+
1)(2k
+ 2)(2k + 3)(2k + 4)
k=1
(2.48)
Combining (2.31), (2.40), and (2.48), we obtain the formula
∞
11
γ
1
ζ(2k + 1)
=
−
− log A.
(2k
+
1)(2k
+
2)(2k
+
3)(2k
+
4)
72
24
2
k=1
(2.49)
On the other hand, noting that Fourier series
log(2 sin π x) = −
∞
cos(2kπ x)
,
k
k=1
0 < x < 1,
(2.50)
410
WU YUN-FEI
and integrating by parts, we have
1
1
0
∞
cos(2kπ x)
dx
k
0
k=1


∞
sin(2kπ
x)
1 1

(1 − x)2 d
=−
2π 0
k2
k=1
(1 − x)2 log(2 sin π x)dx = −
1
=
2π 2
(1 − x)2
(2.51)


∞
cos(2kπ
x)
 = − 1 ζ(3).
(1−x)d
3
k
2π 2
0
k=1
1
Similarly, we have
1
0
1
log(2 sin π x)dx =
0
1/2 0
1/2
(1 − x) log(2 sin π x)dx =
0
log(2 sin π x)dx = 0,
7
1
− x log(2 sin π x)dx = −
ζ(3).
2
16π 2
(2.52)
(2.53)
Integrating both sides of (2.15) from x = 0 to x = 1, and using (2.52), we get
∞
ζ(2k)
1
= (log 2π − 1).
2k(2k + 1)
2
k=1
(2.54)
Combining (2.31) and (2.54), we obtain the formulas
∞
1
ζ(k)
= (log 2π − γ),
k(k
+
1)
2
k=2
∞
1
γ
(−1)k ζ(k)
= −1 + log 2π + .
k(k
+
1)
2
2
k=2
(2.55)
Integrating both sides of (2.15) from x = 0 to x = 1/2, and using (2.52), we get
∞
1
ζ(2k)
2−2k = (log π − 1).
2k(2k
+
1)
2
k=1
(2.56)
Combining (2.36) and (2.56), we obtain the formulas
∞
7
γ
ζ(k)
1
2−k = log π +
log 2 − − 3 log A,
k(k
+
1)
2
12
4
k=2
∞
7
γ
1
(−1)k ζ(k) −k
2 = −1 + log π −
log 2 + + 3 log A.
k(k
+
1)
2
12
4
k=2
(2.57)
Multiplying 1 − x on both sides of (2.15), and integrating from x = 0 to x = 1, and
using (2.52), we get
∞
3 1
ζ(2k)
= − + log 2π .
2k(2k
+
1)(2k
+
2)
8 4
k=1
(2.58)
NEW SERIES INVOLVING THE ZETA FUNCTION
411
Combining (2.58) and (2.40), we obtain the formulas
∞
1
γ
ζ(k)
= log 2π − − log A,
k(k
+
1)(k
+
2)
4
6
k=2
∞
(2.59)
3 1
γ
(−1)k ζ(k)
= − + log 2π + + log A.
k(k
+
1)(k
+
2)
4
4
6
k=2
Multiplying 1/2 − x on both sides of (2.15), and integrating from x = 0 to x = 1/2,
and using (2.53), we get
∞
7
3 1
ζ(2k)
2−2k = − + log π +
ζ(3).
2
2k(2k
+
1)(2k
+
2)
8
4
8π
k=1
(2.60)
Combining (2.60) and (2.47), we obtain the formulas
∞
ζ(k)
γ
7
1
2−k = log 2π −
+
ζ(3) − 2 log A,
2
k(k
+
1)(k
+
2)
4
12
8π
k=2
∞
(2.61)
π
γ
7
3 1
(−1)k ζ(k)
2−k = − + log +
+
ζ(3) + 2 log A.
2
k(k
+
1)(k
+
2)
4
4
2
12
8π
k=2
Multiplying (1 − x)2 on both sides of (2.15), and integrating from x = 0 to x = 1,
and using (2.51), we get
∞
11
1
1
ζ(2k)
=−
+
log 2π +
ζ(3).
2
2k(2k
+
1)(2k
+
2)(2k
+
3)
72
12
8π
k=1
(2.62)
Combining (2.62) and (2.49), we obtain the formulas
∞
1
γ
1
1
ζ(k)
=
log 2π −
+
ζ(3) − log A,
2
k(k
+
1)(k
+
2)(k
+
3)
12
24
8π
2
k=2
∞
(2.63)
11
1
γ
1
1
(−1)k ζ(k)
=−
+
log 2π +
+
ζ(3) + log A.
2
k(k
+
1)(k
+
2)(k
+
3)
36
12
24
8π
2
k=2
References
[1]
[2]
[3]
E. W. Barnes, The theory of the G-function, Quart. J. Pure Appl. Math. 31 (1899), 264–314.
J. Choi and H. M. Srivastava, Sums associated with the zeta function, J. Math. Anal. Appl.
206 (1997), no. 1, 103–120. MR 97i:11092. Zbl 0869.11067.
J. Choi, H. M. Srivastava, and J. R. Quine, Some series involving the zeta function, Bull.
Austral. Math. Soc. 51 (1995), no. 3, 383–393. MR 96d:11090. Zbl 0830.11030.
Wu Yun-Fei: Department of Mathematics, Ningbo University, Ningbo, Zhejiang
315211, China
E-mail address: yunfeiwu@163.net