arXiv:1606.04850v4 [math.NT] 12 Feb 2017
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND
ζ(2n + 1)
DEREK ORR
Abstract. In this paper, we find rational zeta series with ζ(2n) in terms of
ζ(2k + 1) and β(2k), the Dirichlet beta function. We then develop a certain family
of generalized rational zeta series using the generalized Clausen function and use
those results to discover a second family of generalized rational zeta series. As a
special case of our results from Theorem 3.1, we prove a conjecture given in 2012 by
F.M.S. Lima. Later, we use the same analysis but for the digamma function ψ(x)
and negapolygammas ψ (−m) (x). With these, we extract the same two families of
generalized rational zeta series with ζ(2n+ 1) on the numerator rather than ζ(2n).
Contents
1. Introduction
1.1. Organization of the Paper
2. Rational ζ(2n) series with cot(x)
3. General ζ(2n) series using Clm (x)
4. General ζ(2n) series using cot(x)
5. Rational ζ(2n + 1) series using ψ(x)
6. General ζ(2n + 1) series using ψ (−m) (x)
7. General ζ(2n + 1) series using ψ(x)
References
1
5
5
8
12
16
18
22
26
1. Introduction
In 1734, Leonard Euler proved an amazing result, now known as the celebrated
Euler series:
ζ(2) =
∞
X
1
1 1
1
π2
=
1
+
+
+
+
·
·
·
=
,
2
n
4
9
16
6
n=1
where ζ(s) is the Riemann zeta function, defined as
ζ(s) =
∞
X
1
, ℜ(s) > 1.
s
n
n=1
Key words and phrases. Riemann zeta function, Dirichlet beta function, Clausen integral, negapolygammas, rational zeta series, polygamma function.
2010 Mathematics Subject Classification. Primary 40C10, 11M99. Secondary 41A58.
1
2
DEREK ORR
Later, Euler gave the formula
∞
X
(−1)k+1 B2k (2π)2k
1
=
, k ∈ N0 ,
ζ(2k) =
n2k
2(2k)!
n=1
where Bn are the Bernoulli numbers, defined by
∞
X Bn
z
=
z n , |z| < 2π.
ez − 1 n=0 n!
These Bernoulli numbers also arise in certain power series, namely of the cotangent
function
(1)
cot(x) =
∞
X
(−1)n 22n B2n
n=0
(2n)!
x2n−1 = −2
∞
X
ζ(2n)
n=0
π 2n
x2n−1 , |x| < π.
The odd arguments of ζ(s) are the interesting as they do not have a closed form,
though many mathematicians have studied them in detail. In 1979, Roger Apery
[3] proved that ζ(3) is irrational using the fast converging series
∞
5 X (−1)n−1
.
ζ(3) =
2 n=1 n3 2n
n
It is still unknown whether ζ(5) is irrational however, it is known that at least
one of ζ(5), ζ(7), ζ(9), and ζ(11) is irrational (see [18]). The rational zeta series in
this paper will involve odd arguments of ζ(s) among other things. Another main
function needed is the Clausen function (or Clausen’s integral),
(2)
Cl2 (θ) :=
∞
X
sin(kθ)
k=1
k2
=−
Z
θ
log 2 sin
0
φ 2
dφ.
The Clausen function also has a power series representation which will be used
later in the paper. It is given as
(3)
∞
X
Cl2 (θ)
ζ(2n) θ 2n
= 1 − log |θ| +
, |θ| < 2π.
θ
n(2n + 1) 2π
n=1
There are also higher order Clausen-type function defined as
(4)
Cl2m (θ) :=
∞
X
sin(kθ)
k=1
k 2m
,
Cl2m+1 (θ) :=
∞
X
cos(kθ)
k=1
k 2m+1
.
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND ζ(2n + 1)
3
The Clausen function is widely studied and has many applications in mathematics
and mathematical physics ([5], [6], [9], [12], [14], [15], [17]). We will also discuss the
Dirichlet beta function
∞
X
(−1)n
β(s) =
, ℜ(s) > 0.
(2n + 1)s
n=0
When s = 2,
∞
X
(−1)n
β(2) = G =
(2n + 1)2
n=0
is known as Catalan’s constant. Using this and the Riemann zeta function, we
find
(5)
Cl2m (π) = 0,
Cl2m+1 (π) = −
(4m − 1)ζ(2m + 1)
,
4m
and
(6)
Cl2m (π/2) = β(2m),
Cl2m+1 (π/2) = −
(4m − 1)ζ(2m + 1)
.
24m+1
Also, from (4) we can see
d
Cl2m (θ) = Cl2m−1 (θ),
dθ
(7)
(8)
Z
d
Cl2m+1 (θ) = − Cl2m (θ),
dθ
θ
Cl2m (x) dx = ζ(2m + 1) − Cl2m+1 (θ),
0
Z
θ
Cl2m−1 (x) dx = Cl2m (θ).
0
Using (4) and (7), we find
θ , |θ| < 2π.
Cl1 (θ) = − log 2 sin
2
(9)
Writing others out,
Cl3 (z) = ζ(3) −
Cl4 (z) =
Z
0
z
Cl3 (x) dx = zζ(3) −
Z
0
z
Z
0
Z
z
Cl2 (t) dt,
0
x
Cl2 (t) dt dx = zζ(3) −
Z
0
z
(z − t) Cl2 (t) dt,
4
DEREK ORR
Z
Cl5 (z) = ζ(5) −
z
1
1
Cl4 (x) dx = ζ(5) − z 2 ζ(3) +
2
2
0
Z
z
(z − t)2 Cl2 (t) dt,
0
and by induction, for m ≥ 3,
⌋
⌊ m−1
2
(10) Clm (z) = (−1)
⌊ m−1
⌋
2
X (−1)k z m−2k−1
ζ(2k + 1)
(m − 2k − 1)!
k=1
m−1
(−1)⌊ 2 ⌋
+
(m − 3)!
Z
z
(z − t)m−3 Cl2 (t) dt.
0
In the latter sections of this paper, we will focus on the polygamma function. Its
definition is given by
ψ (n) (z) :=
dn+1
log Γ(z), n ∈ N0 .
dz n+1
This paper will discuss when n = 0. For n = 0, ψ (0) (z) = ψ(z) is called the
digamma
function. It has been shown (see [2]) that there is a closed form of
Z
z
xn ψ(x) dx in terms of sums involving Harmonic numbers, Bernoulli numbers
0
and Bernoulli polynomials. There are also definitions for negative order polygamma
functions, called negapolygammas, given by
ψ (−1) (z) = log Γ(z),
ψ
ψ
(−3)
(z) =
Z
0
z
Z
(−2)
(z) =
x
log Γ(t) dt dx =
0
Z
Z
z
0
z
log Γ(x) dx,
0
Z
z
log Γ(t) dx dt =
t
Z
z
(z − t) log Γ(t) dt,
0
and by induction, for n ≥ 2,
(11)
ψ
(−n)
1
(z) =
(n − 2)!
Z
z
(z − t)n−2 log Γ(t) dt.
0
Note the Taylor series for log Γ(z) is
(12)
log Γ(z) = − log z − γz +
∞
X
(−1)k ζ(k)
k=2
k
z k , |z| < 1,
where γ is the Euler-Mascheroni constant. The last function to introduce is the
Hurwitz zeta function, defined by
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND ζ(2n + 1)
ζ(s, a) =
∞
X
k=0
5
1
, a ∈ R\ − N, ℜ(s) > 1.
(k + a)s
The negapolygammas are related to the derivative of the Hurwitz zeta function
with respect to the first variable (see [2]).
R πz
1.1. Organization of the Paper. We begin by investigating 0 xp cot(x) dx for
|z| < 1, which is studied in [15]. We then compute this same integral using the
power series for the cotangent and obtain a rational zeta series representation with
ζ(2n) on the numerator for z = 1/2 and z = 1/4. Afterwords, we work on a
generalized rational zeta series with ζ(2n) on the numerator and an arbitrary number
of monomials on the denominator using the generalized Clausen function Clm (z).
As a special case, we immediately prove a conjecture given in 2012 by F.M.S. Lima.
After doing this, we go back to the cotangent function and discover a separate class of
generalized ζ(2n) series. Lastly, we perform the same analysis but for the digamma
function ψ(x) instead of cot(x) and instead of Clm (x), we use the negapolygammas
ψ (−m) (x). Using the ζ(2n) sums from earlier, we extract the same rational zeta
series representations with ζ(2n + 1) on the numerator for z = 1/2 and z = 1/4.
Acknowledgements. I would like to thank Cezar Lupu, Tom Hales, and George
Sparling for their interest in my paper as well as their advice which led to some
improvements of the paper.
2. Rational ζ(2n) series with cot(x)
Theorem 2.1. For p ∈ N and |z| < 1,
(13)
Z πz
0
p
k+3
p
X
p!(−1)⌊ 2 ⌋
p!(−1) 2
x cot(x) dx = (πz)
Clk+1 (2πz)+δ⌊ p2 ⌋, p2
ζ(p+1),
k
(p
−
k)!(2πz)
2p
k=0
p
p
where δj,k is the Kronecker delta function.
Proof. Let f (z) be the left hand side of the equation and let g(z) be the right
hand side. Note that f ′ (z) = π p+1 z p cot(πz). Using (7) and (9),
k+3
p
1 X p!(−1)⌊ 2 ⌋ (p − k)(2π)p−k z p−k−1
(πz)p 2π cos(πz)
g (z) = p
Clk+1 (2πz) +
2 k=0
(p − k)!
2 sin(πz)
′
k+3
p
o
1 X p!(−1)⌊ 2 ⌋ (2πz)p−k n
(−1)k+1 2π Clk (2πz)
+ p
2 k=1
(p − k)!
6
DEREK ORR
=π
p+1 p
z cot(πz) + (πz)
p
p
X
k=1
p! Clk (2πz)
⌋
⌊ k+3
⌋
k
⌊ k+2
2
2
+
(−1)
(−1)
.
(−1)
(p − k)!z k+1 (2π)k−1
So indeed, g ′ (z) = π p+1z p cot(πz) = f ′ (z). Clearly, f (0) = 0. For g(z), note that
all terms in the sum are zero except when k = p. So we have
p
p+3
1
p!(−1) 2
g(0) = p p!(−1)⌊ 2 ⌋ Clp+1 (0) + δ⌊ p2 ⌋, p2
ζ(p + 1).
2
2p
From (4), Clp+1 (0) = δ⌊ p2 ⌋, p2 ζ(p + 1). So we see g(0) = 0. Since f (0) = g(0) and
f ′ (z) = g ′(z), f (z) = g(z). Using (5), (6), and (9), setting z = 1/2 and z = 1/4, we find
(14)
Z π/2
p
p
x cot(x) dx =
0
π p 2
log 2+
⌊2⌋
X
p!(−1)k (4k − 1)
k=1
p
p!(−1) 2 ζ(p + 1)
ζ(2k+1) +δ⌊ p2 ⌋, p2
,
(p − 2k)!(2π)2k
2p
and
Z
(15)
π/4
0
⌊ p2 ⌋
π p X
p!(−1)k (4k − 1)
1
log 2 +
ζ(2k + 1)
xp cot(x) dx =
2 4
(p − 2k)!(2π)2k
k=1
⌊ p+1
⌋
2
−
X
k=1
p!(−4)k β(2k)
(p + 1 − 2k)!π 2k−1
p
+δ
⌊ p2 ⌋, p2
p!(−1) 2
ζ(p + 1).
2p
We can also integrate xp cot(x) using (1) and Fubini’s theorem. Doing so, we
obtain
Z
πz
p
x cot(x) dx = −2
0
Z
0
∞
πz X
n=0
∞
X
ζ(2n)z 2n
ζ(2n) 2n−1+p
p
x
dx
=
−2(πz)
,
π 2n
2n + p
n=0
and using (13),
(16)
−2
p
k+3
p
X
p!(−1)⌊ 2 ⌋
p!(−1) 2
=
Clk+1(2πz) + δ⌊ p2 ⌋, p2
ζ(p + 1).
k
p
2n + p
(p
−
k)!(2πz)
(2πz)
k=0
∞
X
ζ(2n)z 2n
n=0
So for z = 1/2 and z = 1/4, we have
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND ζ(2n + 1)
7
(17)
−2
∞
X
n=0
p
⌊2⌋
p
X
ζ(2n)
p!(−1)k (4k − 1)ζ(2k + 1)
p!(−1) 2 ζ(p + 1)
= log 2 +
+ δ⌊ p2 ⌋, p2
,
2k
p
(2n + p)4n
(p
−
2k)!(2π)
π
k=1
and
(18)
−2
∞
X
n=0
⌊p⌋
2
1
1X
p!(−1)k (4k − 1)ζ(2k + 1)
ζ(2n)
=
log
2
+
(2n + p)16n
2
2 k=1
(p − 2k)!(2π)2k
⌋
⌊ p+1
2
p
π X p!(−4)k β(2k)
p!(−1) 2 2p ζ(p + 1)
−
+ δ⌊ p2 ⌋, p2
.
2
(p + 1 − 2k)!π 2k
πp
k=1
Plugging in p = 1 into both of the equations yields two nice series representations
for log 2,
(19)
−2
∞
X
n=0
ζ(2n)
= log 2,
(2n + 1)4n
and
∞
X
4G
ζ(2n)
−
−4
= log 2.
n
π
(2n
+
1)16
n=0
(20)
Other series from (17) and (18) are
∞
X
n=0
∞
X
n=0
∞
X
n=0
∞
X
n=0
7ζ(3)
ζ(2n)
=
−
log
2
+
,
(n + 1)4n
2π 2
1
9ζ(3)
ζ(2n)
=
−
log
2
+
,
(2n + 3)4n
2
4π 2
9ζ(3) 93ζ(5)
ζ(2n)
= − log 2 +
−
,
n
(n + 2)4
π2
2π 4
1
35ζ(3) 4G
ζ(2n)
= − log 2 +
−
,
n
(n + 1)16
2
4π 2
π
and
∞
X
n=0
ζ(2n)
1
9ζ(3) 3G 24β(4)
= − log 2 +
−
+
.
n
(2n + 3)16
4
8π 2
π
π3
8
DEREK ORR
3. General ζ(2n) series using Clm (x)
Theorem 3.1. For p ∈ N0 , m ∈ N and |z| < 1,
Z
(21)
2πz
m+p+1
X (2πz)p+m+1−k p!(−1)⌊ m2 ⌋ (−1)⌊ k2 ⌋
x Clm (x) dx = −
Clk (2πz)
(p + m + 1 − k)!
p
0
k=m+1
m
+ δ⌊ p+m ⌋, p+m (−1)⌊ 2 ⌋ p!(−1)
2
p+m
2
2
ζ(p + m + 1).
Proof. Similar to the proof of (13), we will call the left hand side f (z) and the
right hand side g(z). Note f (0) = g(0) and f ′ (z) = (2π)p+1z p Clm (2πz). Using (7),
m+p
X p!(2π)p+m+1−k z p+m−k (−1)⌊ m2 ⌋+⌊ k2 ⌋
g (z) = −
Clk (2πz)
(p + m − k)!
k=m+1
′
m+p+1
o
X p!(2πz)p+m+1−k (−1)⌊ m2 ⌋+⌊ k2 ⌋ n
(−1)k 2π Clk−1 (2πz)
−
(p + m + 1 − k)!
k=m+1
m+p
= −(2π)
X p!(2πz)p+m−k (−1)⌊ m2 ⌋
z (−1)
∗
(−1)
Clm (2πz) − 2π
(p + m − k)!
k=m+1
o
n
k
k+1
Clk (2πz) (−1)⌊ 2 ⌋ (−1)k+1 + (−1)⌊ 2 ⌋ = (2π)p+1z p Clm (2πz).
p+1 p
⌊m
⌋+⌊ m+1
⌋
2
2
m+1
So indeed f ′ (z) = g ′(z) and f (0) = g(0), which implies f (z) = g(z). We can see if p = 0, we recover (8). Now letting z = 1/2 and z = 1/4, we find
Z
(22)
⌋
⌊ p+m
2
π
xp Clm (x) dx = (−1)
⌊m
⌋
2
X
p!π p+m
0
k=⌊ m+1
⌋
2
(−1)k (4k − 1)ζ(2k + 1)
(p + m − 2k)!(2π)2k
m
+ δ⌊ p+m ⌋, p+m (−1)⌊ 2 ⌋ p!(−1)
2
p+m
2
2
ζ(p + m + 1),
and
(23)
Z
π/2
xp Clm (x) dx = (−1)
⌋
⌊m
2
0
π
−
2
⌊ p+m+1
⌋
2
X
⌋
k=⌊ m+2
2
(−1)k 4k β(2k)
(p + m + 1 − 2k)!π 2k
π p+m 1
p!
2
2
⌋
⌊ p+m
2
X
k=⌊ m+1
⌋
2
m
(−1)k (4k − 1)ζ(2k + 1)
(p + m − 2k)!(2π)2k
+ δ⌊ p+m ⌋, p+m (−1)⌊ 2 ⌋ p!(−1)
2
2
p+m
2
ζ(p + m + 1).
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND ζ(2n + 1)
9
Now, we will integrate the left hand side of (21) using (3) and (10), giving us
Z
2πz
p
x Clm (x) dx =
0
0
⌋
⌊ m−1
2
=
Z
x
k=1
m−1
(−1)⌊ 2 ⌋+k xm−2k−1
ζ(2k + 1)
(m − 2k − 1)!
m−1 Z
(−1)⌊ 2 ⌋ x
m−3
+
(x − t)
Cl2 (t) dt dx
(m − 3)! 0
m−1
m−1
k=1
⌊ m−1
⌋
2
X
k=1
⌋
⌊ m−1
2
=
p
m−1
⌋
⌊X
2
(−1)⌊ 2 ⌋
(−1)⌊ 2 ⌋+k (2πz)m+p−2k
ζ(2k+1)+
(m − 1 − 2k)!(m + p − 2k)
(m − 3)!
X
=
2πz
X
k=1
Z
2πz
0
Z
x
xp (x−t)m−3 Cl2 (t) dt dx
0
m−1
m−1
(−1)⌊ 2 ⌋
(−1)⌊ 2 ⌋+k (2πz)m+p−2k
ζ(2k + 1) +
∗
(m − 1 − 2k)!(m + p − 2k)
(m − 3)!
Z 2πz Z x
∞
X
ζ(2n)t2n+1
p
m−3
dt dx
t − t log t +
x (x − t)
n(2n + 1)(2π)2n
0
0
n=1
m−1
m−1
(−1)⌊ 2 ⌋+k (2πz)m+p−2k
(−1)⌊ 2 ⌋
ζ(2k+1)+
(m − 1 − 2k)!(m + p − 2k)
(m − 3)!
2πz
xm−1 (Hm−1 − log x)
(m − 1)(m − 2)
0
∞
X
2ζ(2n)Γ(m − 2)xm+2n−1
dx,
+
2n(2n + 1)(2n + 2) . . . (2n + m − 1)(2π)2n
n=1
Z
p
x
where Hk is the k-th harmonic number and H0 := 0. Integrating again and
simplifying, we arrive at
(24)
Z
0
⌋
⌊ m−1
2
+
X
k=1
2πz
(2πz)p+m (−1)⌊
x Clm (x) dx =
(m − 1)!
p
m−1
⌋
2
Hm−1 − log(2πz)
1
+
(p + m)
(p + m)2
∞
X
(m − 1)!ζ(2n)z 2n
(m − 1)!(−1)k ζ(2k + 1)
.
+
(m − 1 − 2k)!(p + m − 2k)(2πz)2k n=1 n(2n + 1) . . . (2n + m − 1)(2n + p + m)
Using (21), we can rearrange this and find
10
DEREK ORR
(25)
∞
X
n=1
m+p
log(2πz) − Hm−1
ζ(2n)z 2n
=
n(2n + 1) . . . (2n + m − 1)(2n + m + p)
(m − 1)!(p + m)
p+m
⌋
X (−1)m p!(−1)⌊ k+1
2
(−1)m+1 p!(−1) 2
p+m
p+m
+
Clk+1(2πz) + δ⌊
ζ(p + m + 1)
⌋, 2
2
(p + m − k)!(2πz)k
(2πz)p+m
k=m
m−1
⌊ 2 ⌋
X
1
(−1)k ζ(2k + 1)
−
−
.
(m − 1)!(p + m)2
(m − 1 − 2k)!(m + p − 2k)(2πz)2k
k=1
Note when p = 0, we have the very special formula
(26)
∞
X
n=1
(−1)m (−1)⌊
ζ(2n)z 2n
=
n(2n + 1) . . . (2n + m)
(2πz)m
m+1
⌋
2
Clm+1 (2πz)
m−1
⌊ 2 ⌋
3m
X (−1)k ζ(2k + 1)
log(2πz) − Hm
(−1) 2
ζ(m
+
1)
−
+
.
− δ⌊ m2 ⌋, m2
m
2k
(2πz)
(m − 2k)!(2πz)
m!
k=1
Letting z = 1/2 and z = 1/4, we see that
(27)
∞
X
n=1
ζ(2n)
(−1)
= δ⌊ m2 ⌋, m2
n
n(2n + 1) . . . (2n + m)4
3m+2
2
(2m+1 − 1)ζ(m + 1)
(2π)m
⌋
⌊ m−1
2
−
X (−1)k ζ(2k + 1) log π − Hm
+
,
(m − 2k)!π 2k
m!
k=1
and
(28)
∞
X
n=1
ζ(2n)
(−1)
= δ⌊ m+1 ⌋, m+1
n
2
2
n(2n + 1) . . . (2n + m)16
1
(−1)
− δ⌊ m2 ⌋, m2
2
3m
2
3m+1
2
2m β(m + 1)
πm
m−1
⌊ 2 ⌋
(−4)k ζ(2k + 1) log(π/2) − Hm
(22m+1 + 2m − 1)ζ(m + 1) X
−
+
,
2k
(2π)m
(m
−
2k)!π
m!
k=1
both of which were conjectured in 2012 and only for m odd (see [13]). For general
p, we have
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND ζ(2n + 1)
(29)
∞
X
n=1
11
log π − Hm−1
ζ(2n)
=
n
n(2n + 1) . . . (2n + m − 1)(2n + m + p)4
(m − 1)!(p + m)
p+m
(−1) 2
m+1
+ (−1)
p! δ⌊ p+m ⌋, p+m
ζ(p + m + 1) +
2
2
π p+m
⌊ m−1
⌋
2
−
X
k=1
⌊ p+m
⌋
2
X
⌋
k=⌊ m+1
2
(−1)k (4k − 1)ζ(2k + 1)
(p + m − 2k)!(2π)2k
1
(−1)k ζ(2k + 1)
−
,
(m − 1 − 2k)!(p + m − 2k)π 2k (m − 1)!(p + m)2
and
(30)
∞
X
n=1
⌋
⌊ p+m+1
2
X
k=⌊ m+2
⌋
2
⌋
⌊ p+m
2
X
k=⌊ m+1
⌋
2
ζ(2n)
log(π/2) − Hm−1 π
=
+ ∗
n
n(2n + 1) . . . (2n + m − 1)(2n + m + p)16
(m − 1)!(p + m)
2
(−1)m+1 p!(−1)
p!(−1)m+k 4k β(2k)
p+m p+m
+δ
(p + m + 1 − 2k)!π 2k ⌊ 2 ⌋, 2
π p+m
p+m
2
2p+m
ζ(p+m+1)−
p!(−1)m
∗
2
⌋
⌊ m−1
2
(−1)k 4k ζ(2k + 1)
(−1) (4 − 1)ζ(2k + 1) X
1
−
−
.
2k
2k
(p + m − 2k)!(2π)
(m − 1 − 2k)!(p + m − 2k)π
(m − 1)!(p + m)2
k
k
k=1
Remark. When m = 1 and p = 0, one has
∞
X
n=1
ζ(2n)
= log π − 1,
n(2n + 1)4n
and
∞
X
n=1
π 2G
ζ(2n)
,
=
−
1
+
log
n(2n + 1)16n
π
2
the first of which is a famous series (see [16]) and both are immediate from (3)
with θ = π and θ = π/2, respectively. Below we compute other sums for certain m
and p.
∞
X
n=1
∞
X
n=1
ζ(2n)
3ζ(3) 1
1
=
−
+
log
π
−
n(2n + 3)4n
2π 2
3
9
ζ(2n)
7ζ(3)
3
=
− log π −
n
2
n(2n + 1)(n + 1)4
2π
2
12
DEREK ORR
∞
X
n=1
∞
X
n=1
∞
X
n=1
2ζ(3) 1
11
ζ(2n)
=
+ log π −
n
2
n(2n + 1)(n + 1)(2n + 3)4
π
3
18
ζ(2n)
2ζ(3) 31ζ(5) 1
7
=
+
+ log π −
n
2
4
n(2n + 1)(n + 1)(n + 2)4
π
4π
2
8
ζ(3) ζ(5)
1
137
ζ(2n)
=
−
+
log
π
−
n(2n + 1) . . . (2n + 5)4n
6π 2
π4
120
7200
∞
X
n=1
∞
X
n=1
∞
X
n=1
∞
X
n=1
π 1
35ζ(3) 4G
ζ(2n)
−
=
−
+
+
log
n(n + 1)16n
4π 2
π
2
2
π 3
ζ(2n)
35ζ(3)
−
=
+ log
n(2n + 1)(n + 1)16n
4π 2
2
2
π 4
3ζ(3) 8β(4) 1
ζ(2n)
−
=
+
+
log
n(2n + 1)(2n + 3)16n
8π 2
π3
3
2
9
π 25
2ζ(3) 527ζ(5)
1
ζ(2n)
−
=
−
+
log
n
2
4
n(2n + 1) . . . (2n + 4)16
π
32π
24
2
288
Now, we will revisit the cotangent function and investigate a general zeta series
using its power series.
4. General ζ(2n) series using cot(x)
Similar to above, we will investigate the double integral
Z πz Z x
xp (x − t)m t cot(t) dt dx.
0
0
Using the binomial theorem, (13), and a change of variables among other things,
Z
πz
0
=
Z
0
x
m
X
Z πz
Z x
m
p+m−j
x
(−1)
tj+1 cot(t) dt dx
x (x − t) t cot(t) dt dx =
j
0
0
j=0
p
m
j
k+3 Z
j+1 m X
X
m (−1)j (j + 1)!(−1)⌊ 2 ⌋
j=0 k=0
(j + 1 − k)!2k
j
m
X
πz
xp+m+1−k Clk+1 (2x) dx
0
(−1)j (j + 1)!(−1)
δ⌊ j+1 ⌋, j+1
+
2
2
2j+1
j=0
j+1
2
Z πz
m
ζ(j + 2)
xp+m−j dx
j
0
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND ζ(2n + 1)
=
Z
m X
m (−1)j+1
2m+p+2
j
j=0
2πz
u
p+m+1
up+m−k Clk+2 (u) du −
0
k
m X
m X
m (−1)j (j + 1)!(−1)⌊ 2 ⌋
k=0 j=k
⌋
⌊ m+1
2
2πz
Z
Cl1 (u) du +
0
X
k=1
13
(j − k)!2p+m+2
j
∗
(2k)(−1)k m!(πz)p+m+2−2k
ζ(2k + 1)
22k (m + 1 − 2k)!(p + m + 2 − 2k)
Z
m
(−1)⌊ 2 ⌋ m! 2πz p+1
u Cl1 (u) du − (1 − δm,0 ) p+m+2
u Clm+1 (u) du
2
0
0
⌊ m+1
⌋
m+1 Z
2
X
(m + 1)!(−1)⌊ 2 ⌋ 2πz p
(2k)(−1)k+1 m!(πz)p+m+2−2k ζ(2k + 1)
.
u
Cl
(u)
du+
+
m+2
2p+m+2
22k (m + 1 − 2k)!(p + m + 2 − 2k)
0
k=1
δm,0
= − p+2
2
Z
2πz
p+1
where we’ve evaluated the sum on j and k. The terms with δm,0 cancel each other.
We can use (21) to evaluate the other integrals. Rearranging and simplifying, we
will arrive at
(31)
πz
Z
0
p+m+3
x
Z
x (x − t) t cot(t) dt dx = (p + m + 2) 2(−1)m (πz)m+p+3 p!m!∗
p
0
m
k
(−1)⌊ 2 ⌋ Clk (2πz)
p!(−1)
+ δ⌊ p+m ⌋, p+m
k
2
2
(p + m + 3 − k)!(2πz)
k=m+3
X
−
(−1)⌊
m+1
⌋
2
p+m
2
(−1)m m!ζ(p + m + 3)
2m+p+2
m+1
⌊ 2 ⌋
X (2k)(−1)k+1m!(πz)p+m+2−2k ζ(2k + 1)
m!(πz)p+1 Clm+2 (2πz)
+
.
2k (m + 1 − 2k)!(p + m + 2 − 2k)
2m+1
2
k=1
Another way to evaluate this double integral is to use (1) and Fubini’s theorem,
similar to the first section. Doing so, we see
Z
πz
0
Z
x
p
m
x (x − t) t cot(t) dt dx = −2
0
= −2
∞
X
n=0
Z
πz
p
x
0
ζ(2n)m!Γ(2n + 1)
π 2n Γ(m + 2n + 2)
= −2
∞
X
n=0
Z
∞
X
ζ(2n)
n=0
Z
π 2n
x
(x − t)m t2n dt dx
0
πz
x2n+m+p+1 dx
0
ζ(2n)m!(πz)m+p+2 z 2n
.
(2n + 1) . . . (2n + m + 1)(2n + m + p + 2)
Putting this and (31) together, we find
14
DEREK ORR
∞
X
ζ(2n)z 2n
(32)
= (p + m + 2) (−1)m+1 p!∗
(2n + 1) . . . (2n + m + 1)(2n + m + p + 2)
n=0
p+m
k
p+m+3
X
πz(−1)⌊ 2 ⌋ Clk (2πz)
p!(−1) 2 (−1)m ζ(p + m + 3)
− δ⌊ p+m ⌋, p+m
2
2
(p + m + 3 − k)!(2πz)k
2(2πz)m+p+2
k=m+3
+
(−1)⌊
m+1
⌊ 2 ⌋
X
Clm+2 (2πz)
k(−1)k ζ(2k + 1)
+
.
2k (m + 1 − 2k)!(m + p + 2 − 2k)
2(2πz)m+1
(2πz)
k=1
m+1
⌋
2
When p = 0, this formula simplifies to
∞
X
m+1
(−1)⌊ 2 ⌋ Clm+2 (2πz)
ζ(2n)z 2n
(33)
=
(2n + 1) . . . (2n + m + 2)
2(2πz)m+1
n=0
m
⌊ m+1
⌋
2
X
(−1)⌊ 2 ⌋ Clm+3 (2πz) − δ⌊ m2 ⌋, m2 ζ(m + 3) (m + 2)
k(−1)k ζ(2k + 1)
+
.
+
2k (m + 2 − 2k)!
2(2πz)m+2
(2πz)
k=1
Setting z = 1/2 and z = 1/4, we obtain
(34)
∞
X
n=0
(−1)
ζ(2n)
m+1 m+1
=
−δ
⌋,
⌊
2
2
(2n + 1) . . . (2n + m + 2)4n
m+1
2
(2m+1 − 1)ζ(m + 2)
2(2π)m+1
m+1
⌊ 2 ⌋
m
X
(−1) 2 (2m+3 − 1)(m + 2)ζ(m + 3)
k(−1)k ζ(2k + 1)
+
,
− δ⌊ m2 ⌋, m2
2(2π)m+2
(2πz)2k (m + 2 − 2k)!
k=1
and
(35)
∞
X
n=0
m
ζ(2n)
= (−1)⌊ 2 ⌋ (m + 2)∗
n
(2n + 1) . . . (2n + m + 2)16
2m+1 β(m + 2)
(22m+5 + 2m+2 − 1)ζ(m + 3)
⌊ m+1
⌋
2
m m
−
δ
+
(−1)
∗
δ⌊ m+1 ⌋, m+1
⌊
⌋,
m+2
m+2
2
2
2
2
π
4(2π)
m+1
⌋
⌊X
2
k(−4)k ζ(2k + 1)
2m β(m + 2)
(2m+1 − 1)ζ(m + 2)
m
m
m+1 m+1
−
δ
+
.
δ⌊ 2 ⌋, 2
⌊ 2 ⌋, 2
π m+1
4(2π)m+1
π 2k (m + 2 − 2k)!
k=1
For general p,
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND ζ(2n + 1)
15
∞
X
(−1)m p!(m + p + 2)
ζ(2n)
=
∗
(2n + 1) . . . (2n + m + 1)(2n + m + p + 2)4n
2
n=0
!
⌊ p+m+2
⌋
p+m
2
X
(−1)k (4k − 1)ζ(2k + 1)
(−1) 2 ζ(p + m + 3)
− δ⌊ p+m ⌋, p+m
2
2
(p + m + 2 − 2k)!(2π)2k
π m+p+2
m+3
(36)
k=⌊
−δ⌊ m+1 ⌋, m+1
2
2
⌋
(−1)
m+1
2
2
m+1
⌊ 2 ⌋
(2m+1 − 1)ζ(m + 2) X
k(−1)k ζ(2k + 1)
+
,
2(2π)m+1
π 2k (m + 1 − 2k)!(m + p + 2 − 2k)
k=1
and
∞
X
(37)
n=0
(−1)m p!(m + p + 2)
ζ(2n)
=
∗
(2n + 1) . . . (2n + m + 1)(2n + m + p + 2)16n
4
⌊ p+m+2
⌋
2
X
⌋
k=⌊ m+3
2
− δ⌊ p+m ⌋, p+m
2
−δ⌊ m+1 ⌋, m+1
2
2
(−1)k (4k − 1)ζ(2k + 1)
−
(p + m + 2 − 2k)!(2π)2k
(−1)
p+m
2
2
(−1)
m+1
2
⌊ p+m+3
⌋
2
X
⌋
k=⌊ m+4
2
ζ(p + m + 3)2p+m+3
π m+p+2
!
(−1)k β(2k)4k
(p + m + 3 − 2k)!π 2k−1
+ δ⌊ m2 ⌋, m2
(−1)⌊
m+1
⌋
2
2m β(m + 2)
π m+1
m+1
⌊ 2 ⌋
(2m+1 − 1)ζ(m + 2) X
k(−1)k ζ(2k + 1)4k
+
.
4(2π)m+1
π 2k (m + 1 − 2k)!(m + p + 2 − 2k)
k=1
Remark. For m = 0 and p = 0, we have
∞
X
n=0
ζ(2n)
7ζ(3)
=−
,
n
(2n + 1)(n + 1)4
2π 2
and
∞
X
n=0
2G 35ζ(3)
ζ(2n)
=
−
,
n
(2n + 1)(n + 1)16
π
4π 2
the first of which was rediscovered by Ewell (see [11]). Below we compute other
sums for certain m and p.
∞
X
n=0
∞
X
n=0
9ζ(3)
ζ(2n)
=
−
(2n + 1)(2n + 3)4n
8π 2
3ζ(3) 31ζ(5)
ζ(2n)
=− 2 +
n
(2n + 1)(n + 2)4
π
2π 4
16
DEREK ORR
∞
X
n=0
∞
X
ζ(2n)
ζ(3) 31ζ(5)
=− 2 −
n
(2n + 1)(n + 1)(n + 2)4
2π
2π 4
n=0
∞
X
n=0
∞
X
n=0
∞
X
n=0
ζ(3) 49ζ(5)
ζ(2n)
=− 2 +
n
(2n + 1) . . . (2n + 5)4
6π
32π 4
ζ(2n)
9ζ(3) G 12β(4)
=−
+ −
n
(2n + 1)(2n + 3)16
16π 2
π
π3
61ζ(3) 12β(4)
ζ(2n)
=
−
+
(2n + 1)(2n + 2)(2n + 3)16n
16π 2
π3
n=0
∞
X
5ζ(3)
ζ(2n)
=−
n
(2n + 1)(2n + 2)(2n + 3)4
8π 2
ζ(2n)
2ζ(3) 527ζ(5) 4β(4)
=− 2 +
−
n
(2n + 1) . . . (2n + 4)16
π
16π 4
π3
Now, we will use these results to establish the same families of general rational
zeta series but for ζ(2n + 1).
5. Rational ζ(2n + 1) series using ψ(x)
Theorem 5.1. For p ∈ N and |z| < 1,
Z
(38)
z
p
x ψ(x) dx =
0
p
X
p!(−1)k ψ (−k−1) (z)
(p − k)!
k=0
z p−k .
Proof. Let f (z) be the left hand side and g(z) be the right hand side. It is clear
that f (0) = g(0) = 0 and f ′ (z) = z p ψ(z). Taking the derivative of g(z), we find
′
g (z) =
p
X
p!(−1)k ψ (−k) (z)
k=0
(p − k)!
z
p−k
+
p−1
X
p!(−1)k ψ (−k−1) (z)
(p − k − 1)!
k=0
z p−k−1
p−1
X p!(−1) ψ
(z) p−k X p!(−1)k ψ (−k−1) (z) p−k−1
z
+
z
.
= z ψ(z) +
(p − k)!
(p − k − 1)!
k=0
k=1
p
p
k
(−k)
Reindexing the first sum, one can see the two sums cancel each other out. So,
g (z) = f ′ (z) for all z. Since they are equal at z = 0, then f (z) = g(z). ′
We can also compute this integral using (12). So we will have
Z
0
z
p
x ψ(x) dx =
Z
0
z
p
x
∞
1 X
(−1)k ζ(k)xk−1
−γ− +
x
k=2
dx
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND ζ(2n + 1)
17
∞
γ p+1 z p X (−1)k ζ(k) k+p
z
−
+
z .
=−
p+1
p
k
+
p
k=2
Setting these two results equal to each other, one has
∞
X
(−1)k ζ(k)
(39)
k=2
k+p
p
zk =
X p!(−1)k ψ (−k−1) (z)
1
γz
+
+
.
p p + 1 k=0
(p − k)!z k
Now we can split this up as follows:
∞
X
(−1)k ζ(k)z k
k=2
=
k+p
∞
X
ζ(2n)z 2n
n=1
2n + p
−
∞
X
ζ(2n + 1)z 2n+1
n=1
2n + p + 1
.
Using (13) and rearranging, we see
∞
X
ζ(2n + 1)z 2n
(40)
n=1
p
k+1
X p!(−1)⌊ 2 ⌋ π
1
γ
=−
−
+
Clk+1 (2πz)
2n + p + 1
2zp p + 1 k=0 (p − k)!(2πz)k+1
p
p
X
p!(−1)k ψ (−k−1) (z)
p!(−1) 2 π
− δ⌊ p2 ⌋, p2
ζ(p
+
1)
−
.
(2πz)p+1
(p − k)!z k+1
k=0
For z = 1/2 and z = 1/4, we find
∞
X
(41)
n=1
⌊p⌋
2
X
p!(−1)k (4k − 1)ζ(2k + 1)
ζ(2n + 1)
1
γ
=
−
−
log
2
−
−
(2n + p + 1)4n
p
p + 1 k=1
(p − 2k)!(2π)2k
−δ
⌊ p2 ⌋, p2
p
p
X
p!(−2)k ψ (−k−1) (1/2)
p!(−1) 2 ζ(p + 1)
−
2
,
πp
(p − k)!
k=0
and
(42)
∞
X
n=1
⌊ p+1
⌋
2
+π
X
k=1
⌊p⌋
2
X
ζ(2n + 1)
2
γ
p!(−1)k (4k − 1)ζ(2k + 1)
= − − log 2 −
−
(2n + p + 1)16n
p
p + 1 k=1
(p − 2k)!(2π)2k
p
p
X
p!(−4)k ψ (−k−1) (1/4)
p!(−1) 2 2p+1 ζ(p + 1)
p!(−4)k β(2k)
p
p
−δ
−4
.
⌊
⌋,
2 2
(p + 1 − 2k)!π 2k
πp
(p − k)!
k=0
Below we compute a few examples for specific p. Note that A is the Glaisher1
Kinkelin constant, defined by log A = 12
− ζ ′ (−1).
18
DEREK ORR
∞
X
ζ(2n + 1)
n=1
(n + 1)4n
= −2 − γ + 12 log A −
1
log 2
3
∞
X
ζ(2n + 1)
1 γ
1
= − − + 4 log A − log 2
n
(2n + 3)4
2 3
3
n=1
∞
X
199 γ
3
ζ(2n + 1)
=−
− + 8 log A − 56ζ ′(−3) − log 2
n
(2n + 5)4
180 5
5
n=1
∞
1
X
ζ(2n + 1)
γ
2
=
−2
−
+
18
log
A
+
log
2π
−
4
log
Γ
(2n + 2)16n
2
4
n=1
∞
1
X
ζ(2n + 1)
1 1
γ 3ζ(3)
′
2
+
−
2,
=
−64ζ
+
4
log
A
+
log
2π
−
+
−
4
log
Γ
(2n + 3)16n
4
2
3
2π 2
4
n=1
6. General ζ(2n + 1) series using ψ (−m) (x)
Theorem 6.1. For p ∈ N0 , m ∈ N and |z| < 1,
Z
(43)
z
p
xψ
(−m)
(x) dx =
0
p
X
p!(−1)k ψ (−k−m−1) (z)
(p − k)!
k=0
z p−k .
Proof. The proof is exactly the same as the proof of (38) as there was no dependence on ψ(x) = ψ (0) (x). Now, let us compute this integral using (11) and (12). Doing so, we see
Z
z
p
xψ
0
=
(−m)
(x) dx =
Z
0
Z
z
0
z
xp
(m − 2)!
Z
0
x
(x−t)
m−2
∞
X
(−1)k ζ(k) k
−γt−log t+
t dt dx
k
k=2
∞
γxm
xm−1 (log x − Hm−1 ) X (−1)k ζ(k)Γ(m − 1)xm+k−1
−xp
dx
+
−
(m − 2)! m(m − 1)
m−1
k(k + 1) . . . (k + m − 1)
k=2
=−
γz m+p+1
z m+p (Hm−1 − log z)
z m+p
+
+
m!(m + p + 1)
(m − 1)!(m + p)
(m − 1)!(m + p)2
∞
X
(−1)k ζ(k)z m+p+k
.
+
k(k
+
1)
.
.
.
(k
+
m
−
1)(k
+
m
+
p)
k=2
Setting this result equal to (43) and simplifying, one has the nice result
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND ζ(2n + 1)
(44)
∞
X
k=2
19
(−1)k ζ(k)z k
γz
=
k(k + 1) . . . (k + m − 1)(k + m + p)
m!(m + p + 1)
p
X p!(−1)k ψ (−k−m−1) (z)
log z − Hm−1
1
+
−
+
.
(m − 1)!(m + p) (m − 1)!(m + p)2 k=0
(p − k)!z m+k
In the special case of p = 0, we have
(45)
∞
X
k=2
(−1)k ζ(k)z k
γz
log z − Hm ψ (−m−1) (z)
=
+
+
.
k(k + 1) . . . (k + m)
(m + 1)!
m!
zm
Again, we can split (44) into
∞
X
k=2
∞
X
ζ(2n)z 2n
(−1)k ζ(k)z k
=
k(k + 1) . . . (k + m − 1)(k + m + p) n=1 2n(2n + 1) . . . (2n + m − 1)(2n + m + p)
−
∞
X
n=1
ζ(2n + 1)z 2n+1
,
(2n + 1) . . . (2n + m)(2n + 1 + m + p)
and the first sum has been computed earlier. Using (25), we find
p
X
ζ(2n + 1)z 2n
p!(−1)k ψ (−k−m−1) (z)
=−
(2n + 1) . . . (2n + m)(2n + m + 1 + p)
(p − k)!z m+k+1
n=1
k=0
m+p
p+m
k+1
(−1)⌊ 2 ⌋
(−1)m p! X
(−1) 2 ζ(p + m + 1)
Clk+1 (2πz)−δ⌊ p+m ⌋, p+m
+
2
2
2z
(p + m − k)!(2πz)k
(2πz)p+m
k=m
(46)
∞
X
m−1
⌊ 2 ⌋
log(2π/z) + Hm−1
(−1)k ζ(2k + 1)
1 X
+
−
2k
2z k=1 (m − 1 − 2k)!(m + p − 2k)(2πz)
2z(m − 1)!(p + m)
+
1
γ
−
.
2
2z(m − 1)!(p + m)
m!(p + m + 1)
If p = 0, we have the nice representation
(47)
∞
X
n=1
m−1
⌊ 2 ⌋
ζ(2n + 1)z 2n
ψ (−m−1) (z)
1 X
(−1)k ζ(2k + 1)
=−
−
(2n + 1) . . . (2n + m + 1)
z m+1
2z k=1 (m − 2k)!(2πz)2k
m
(−1)⌊ 2 ⌋ (Clm+1 (2πz) − δ⌊ m2 ⌋, m2 ζ(m + 1)) log(2π/z) + Hm
γ
+
+
−
.
m
2z(2πz)
2zm!
(m + 1)!
20
DEREK ORR
and for z = 1/2 and z = 1/4, we have
(48)
∞
X
log 4π + Hm
γ
ζ(2n + 1)
=
−
n
(2n + 1) . . . (2n + m + 1)4
m!
(m + 1)!
∞
X
ζ(2n + 1)
2 log 8π + 2Hm
γ
=
−
n
(2n + 1) . . . (2n + m + 1)16
m!
(m + 1)!
n=1
m−1
⌊ 2 ⌋
m
X (−1)k ζ(2k + 1)
(−1) 2 (2m+1 − 1)ζ(m + 1)
m
m
−
− 2m+1 ψ (−m−1) (1/2),
− δ⌊ 2 ⌋, 2
2k
(2π)m
(m
−
2k)!π
k=1
and
(49)
n=1
− δ⌊ m+1 ⌋, m+1
2
(−1)
2
m+1
2
m
(−1) 2 (22m+1 + 2m − 1)ζ(m + 1)
2m+1 β(m + 1)
m m
−
δ
⌊
⌋,
2
2
πm
(2π)m
⌊ m−1
⌋
2
−
X 2(−4)k ζ(2k + 1)
− 4m+1 ψ (−m−1) (1/4).
2k
(m
−
2k)!π
k=1
For general p,
∞
X
(50)
n=1
+
ζ(2n + 1)
log 4π + Hm−1
=
(2n + 1) . . . (2n + m)(2n + 1 + m + p)4n
(m − 1)!(m + p)
⌊ p+m
⌋
2
X (−1)k (4k − 1)ζ(2k + 1)
γ
1
m
−
−
(−1)
p!
(m − 1)!(m + p)2 m!(m + p + 1)
(p + m − 2k)!(2π)2k
m+1
k=⌊
+ δ⌊ p+m ⌋, p+m
2
2
(−1)m+1 p!(−1)
π p+m
p+m
2
p
ζ(p + m + 1) − 2m+1
X p!(−2)k ψ (−k−m−1) (1/2)
k=0
⌋
⌊ m−1
2
−
X
k=1
and
⌋
2
(p − k)!
(−1)k ζ(2k + 1)
,
(m − 1 − 2k)!(m + p − 2k)π 2k
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND ζ(2n + 1)
∞
X
(51)
n=1
+
log 8π + Hm−1
2
ζ(2n + 1)
=
(2n + 1) . . . (2n + m)(2n + 1 + m + p)16n
(m − 1)!
(m + p)
1
+
(m + p)2
⌊ p+m+1
⌋
2
X
⌋
k=⌊ m+2
2
− δ⌊ p+m ⌋, p+m
2
⌋
⌊ m−1
2
−
21
X
k=1
2
m+k k
⌊ p+m
⌋
2
X p!(−1)k+m (4k − 1)ζ(2k + 1)
p!(−1)
4 β(2k)
−
(p + m + 1 − 2k)!π 2k−1
(p + m − 2k)!(2π)2k
m+1
p!(−1)m (−1)
k=⌊
2
⌋
p+m
2
2p+m+1 ζ(p + m + 1)
γ
−
p+m
π
m!(m + p + 1)
p
X
2(−4)k ζ(2k + 1)
p!(−4)k ψ (−k−m−1) (1/4)
m+1
−
4
.
(m − 1 − 2k)!(m + p − 2k)π 2k
(p
−
k)!
k=0
Below we compute some sums for certain m and p.
∞
X
n=1
∞
X
n=1
∞
X
n=1
n=1
∞
X
n=1
∞
X
ζ(2n + 1)
3 γ
= −8 log A + − + log 2
n
(2n + 1)(n + 1)(2n + 3)4
2 3
53 γ 121
ζ(2n + 1)
= −8 log A − 20ζ ′(−3) +
− +
log 2
n
(2n + 1)(n + 1)(n + 2)4
36 2
90
2
8 ′
551
γ
1
ζ(2n + 1)
=
−
log
A
+
ζ
(−3)
+
−
+
log 2
(2n + 1) . . . (2n + 5)4n
3
3
4320 120 24
∞
X
n=1
∞
X
n=1
1 γ 2
ζ(2n + 1)
= −2 log A + − + log 2
n
(2n + 1)(2n + 3)4
4 3 3
19 γ 89
ζ(2n + 1)
′
=
−4
log
A
+
20ζ
(−3)
+
− −
log 2
(2n + 1)(n + 2)4n
36 2 90
n=1
∞
X
7
ζ(2n + 1)
= −12 log A + 2 − γ + log 2
n
(2n + 1)(n + 1)4
3
ζ(2n + 1)
= −36 log A + 4 − γ + 8 log 2
(2n + 1)(n + 1)16n
3
1
1 γ
ζ(2n + 1)
′
− 2 log A − 2 ζ(3) − − + 2 log 2
−
2,
=
32ζ
n
(2n + 1)(2n + 3)16
4
4π
4 3
∞
X
n=1
ζ(2n + 1)
187
γ
41
′
=
−8
log
A
+
45ζ
(−3)
+
−
+
log 2
(2n + 1) . . . (2n + 4)16n
144 24 60
To conclude, we will revisit the digamma function and find another general ζ(2n+
1) series.
22
DEREK ORR
7. General ζ(2n + 1) series using ψ(x)
Similar to section 4, we will investigate the double integral
Z
0
z
Z
x
xp (x − t)m tψ(t) dt dx.
0
Using the binomial theorem and (38) among other things, we find
Z
z
0
Z
x
0
m
X
Z z
Z x
m
p+m−j
x
tj+1 ψ(t) dt dx
(−1)
x (x − t) tψ(t) dt dx =
j
0
0
j=0
p
=
m
j
Z
j+1 m X
X
m (−1)j (j + 1)!(−1)k
j=0 k=0
j
(j + 1 − k)!
z
xp+m+1−k ψ (−k−1) (x) dx
0
Z z
m X
m
j
(−1)
xp+m+1 ψ (−1) (x) dx
=
j
0
j=0
Z
m X
m X
m (−1)j (j + 1)!(−1)k+1 z p+m−k (−k−2)
+
x
ψ
(x) dx
j
(j
−
k)!
0
k=0 j=k
= δm,0
Z
z
p+1
x
ψ
(−1)
(x) dx+(1−δm,0 )m!
0
Z
z
p+1
x
ψ
(−m−1)
(x) dx−(m+1)!
0
Z
z
xp ψ (−m−2) (x) dx
0
Simplifying and using (43), we find
(52)
Z zZ
0
x
p
m
x (x−t) tψ(t) dt dx = m!z
p
0
zψ
(−m−2)
p
X
p!(−1)k ψ (−k−m−3) (z)
.
(z)−(m+p+2)
k
(p
−
k)!z
k=0
Now we will evaluate the same integral using the power series for ψ(x) and Fubini’s
theorem once again. Doing so, we see
Z
0
z
Z
x
p
m
x (x − t) tψ(t) dt dx = −
0
=−
Z
0
Z
p
x
Z
x
∞
X
k
k
(−1) ζ(k)t dt dx
(x − t) tγ + 1 −
m
0
k=2
∞
z
xp
0
z
X (−1)k ζ(k)m!xk+m+1
xm+1
xm+2 γ
+
−
(m + 1)(m + 2) m + 1
(k + 1) . . . (k + m + 1)
k=2
!
dx
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND ζ(2n + 1)
=
23
−z p+m+3 γ
−z p+m+2
+
(m + 1)(m + 2)(m + p + 3) (m + 1)(m + p + 2)
∞
X
(−1)k ζ(k)m!z p+m+2+k
+
(k + 1) . . . (k + m + 1)(k + m + p + 2)
k=2
Using this result and (52), we arrive at
∞
X
(53)
k=2
(−1)k ζ(k)z k
γz
=
(k + 1) . . . (k + m + 1)(k + m + p + 2)
(m + 2)!(p + m + 3)
p
X
p!(−1)k ψ (−k−m−3) (z)
ψ (−m−2) (z)
1
+
−
(m
+
p
+
2)
.
+
(m + 1)!(p + m + 2)
z m+1
(p − k)!z k+m+2
k=0
Note when p = 0,
∞
X
(54)
k=2
(−1)k ζ(k)z k
γz + m + 3 ψ (−m−2) (z) (m + 2)ψ (−m−3) (z)
=
+
−
.
(k + 1) . . . (k + m + 2)
(m + 3)!
z m+1
z m+2
As before, we will split the sum in (53) as
∞
X
k=2
∞
X
ζ(2n)z 2n
(−1)k ζ(k)z k
=
(k + 1) . . . (k + m + 1)(k + m + p + 2) n=1 (2n + 1) . . . (2n + m + 1)(2n + m + p + 2)
−
∞
X
n=1
ζ(2n + 1)z 2n+1
(2n + 2) . . . (2n + m + 2)(2n + m + p + 3)
and using (32), we have
(55)
∞
X
n=1
ζ(2n + 1)z 2n
= (p + m + 2) (−1)m+1 p!∗
(2n + 2) . . . (2n + m + 2)(2n + m + p + 3)
p+m+3
p+m
k
p!(−1) 2 (−1)m ζ(p + m + 3)
π(−1)⌊ 2 ⌋ Clk (2πz)
p+m p+m
−
δ
⌊ 2 ⌋, 2
(p + m + 3 − k)!(2πz)k
2z(2πz)m+p+2
k=m+3
p
X
p!(−1)k ψ (−k−m−3) (z)
γ
1
+
−
−
k+m+3
(p − k)!z
(m + 2)!(p + m + 3) 2z(m + 1)!(p + m + 2)
k=0
X
m+1
⌊ 2 ⌋
m+1
k(−1)k ζ(2k + 1)
(−1)⌊ 2 ⌋ Clm+2 (2πz) ψ (−m−2) (z) X
−
+
.
+
2k (m + 1 − 2k)!(m + p + 2 − 2k)
2z(2πz)m+1
z m+2
z(2πz)
k=1
If p = 0, this simplifies to
24
DEREK ORR
∞
X
m+1
(−1)⌊ 2 ⌋ Clm+2 (2πz)
ζ(2n + 1)z 2n
=
(56)
(2n + 2) . . . (2n + m + 3)
2z(2πz)m+1
n=1
m
⌋
⌊ m+1
2
X
(−1)⌊ 2 ⌋ Clm+3 (2πz) − δ⌊ m2 ⌋, m2 ζ(m + 3) (m + 2)
k(−1)k ζ(2k + 1)
+
+
2z(2πz)m+2
z(2πz)2k (m + 2 − 2k)!
k=1
−
(2γz + m + 3) zψ (−m−2) (z) − (m + 2)ψ (−m−3) (z)
−
.
2z(m + 3)!
z m+3
For z = 1/2 and z = 1/4, we see
(57)
∞
X
n=1
ζ(2n + 1)
(−1)
= −δ⌊ m+1 ⌋, m+1
n
2
2
(2n + 2) . . . (2n + m + 3)4
m+1
2
(2m+1 − 1)ζ(m + 2)
(2π)m+1
m+1
⌊ 2 ⌋
m
X 2k(−1)k ζ(2k + 1)
(−1) 2 (2m+3 − 1)(m + 2)ζ(m + 3)
+
− δ⌊ m2 ⌋, m2
(2π)m+2
π 2k (m + 2 − 2k)!
k=1
− 2m+2 ψ (−m−2) (1/2) + (m + 2)2m+3 ψ (−m−3) (1/2) −
1
γ
−
,
(m + 3)! (m + 2)!
and
∞
X
ζ(2n + 1)
m
(58)
= (m + 2) 4m+3 ψ (−m−3) (1/4) + (−1)⌊ 2 ⌋ ∗
n
(2n + 2) . . . (2n + m + 3)16
n=1
m+1
2m+3 β(m + 3)
(22m+5 + 2m+2 − 1)ζ(m + 3)
δ⌊ m+1 ⌋, m+1
− δ⌊ m2 ⌋, m2
+ (−1)⌊ 2 ⌋ ∗
m+2
m+2
2
2
π
(2π)
m+2
m+1
2
β(m + 2)
(2
− 1)ζ(m + 2)
δ⌊ m2 ⌋, m2
− δ⌊ m+1 ⌋, m+1
− 4m+2 ψ (−m−2) (1/4)
m+1
m+1
2
2
π
(2π)
⌋
⌊ m+1
2
+4
For general p,
X k(−4)k ζ(2k + 1)
γ
2
−
−
.
2k
π (m + 2 − 2k)! (m + 3)! (m + 2)!
k=1
GENERALIZED RATIONAL ZETA SERIES FOR ζ(2n) AND ζ(2n + 1)
∞
X
(59)
n=1
25
ζ(2n + 1)
= (−1)m p!(m + p + 2)∗
(2n + 2) . . . (2n + m + 2)(2n + m + p + 3)4n
⌋
⌊ p+m+2
2
p+m
(−1) 2 ζ(p + m + 3)
(−1)k (4k − 1)ζ(2k + 1)
p+m p+m
−
δ
+ 8(−2)m ∗
⌊
⌋,
2k
m+p+2
2
2
(p
+
m
+
2
−
2k)!(2π)
π
k=⌊ m+3
⌋
2
!
m+1
p
X
(−2)k ψ (−k−m−3) (1/2)
(−1) 2 (2m+1 − 1)ζ(m + 2) m+2 (−m−2)
−δ⌊ m+1 ⌋, m+1
−2
ψ
(1/2)
2
2
(p − k)!
(2π)m+1
X
k=0
⌋
⌊ m+1
2
+
X
k=1
γ
1
2k(−1)k ζ(2k + 1)
−
−
,
2k
π (m + 1 − 2k)!(m + p + 2 − 2k) (m + 2)!(p + m + 3) (m + 1)!(p + m + 2)
and
∞
X
(60)
n=1
ζ(2n + 1)
= (−1)m p!(m + p + 2)∗
(2n + 2) . . . (2n + m + 2)(2n + m + p + 3)16n
⌊ p+m+2
⌋
2
+δ
⌊m
⌋, m
2
2
k
(−1) (4 − 1)ζ(2k + 1)
−
(p + m + 2 − 2k)!(2π)2k
(−1)⌊
−δ⌊ m+1 ⌋, m+1
2
k
⌊ p+m+3
⌋
2
(−1)k β(2k)4k
+ 64(−4)m ∗
2k−1
(p + m + 3 − 2k)!π
⌋
⌋
k=⌊ m+3
k=⌊ m+4
2
2
!
p+m
p
X
(−4)k ψ (−k−m−3) (1/4)
(−1) 2 ζ(p + m + 3)2p+m+3
−δ⌊ p+m ⌋, p+m
−4m+2 ψ (−m−2) (1/4)
m+p+2
2
2
(p
−
k)!
π
k=0
X
X
m+1
⌋
2
2m+2 β(m + 2)
γ
2
−
−
m+1
π
(m + 2)!(p + m + 3) (m + 1)!(p + m + 2)
(−1)
m+1
2
2
m+1
⌊ 2 ⌋
k(−1)k ζ(2k + 1)4k+1
(2m+1 − 1)ζ(m + 2) X
+
.
2k (m + 1 − 2k)!(m + p + 2 − 2k)
(2π)m+1
π
k=1
Below we compute some sums for certain m and p.
∞
X
n=1
∞
X
n=1
∞
X
n=1
ζ(2n + 1)
γ 1
= 4 log A − 1 − + log 2
n
(n + 1)(2n + 3)4
3 3
5
γ 19
ζ(2n + 1)
= 60ζ ′(−3) −
− +
log 2
n
(n + 1)(n + 2)4
12 2 30
4
112ζ ′(−3)
19
γ 13
ζ(2n + 1)
=
−
log
A
+
+
− +
log 2
n
(n + 1)(2n + 5)4
3
3
270 5 45
26
DEREK ORR
∞
X
n=1
∞
X
n=1
19 γ
1
ζ(2n + 1)
= 8 log A − 60ζ ′(−3) −
− +
log 2
n
(n + 1)(2n + 3)(n + 2)4
12 6 30
4
28ζ ′(−3)
289
γ
1
ζ(2n + 1)
=
log
A
−
−
−
+
log 2
n
(2n + 2)(2n + 3)(2n + 5)4
3
3
1080 30 90
∞
X
n=1
ζ(2n + 1)
2
17ζ ′(−3)
277
γ
1
=
log
A
−
−
−
−
log 2
n
(2n + 2) . . . (2n + 5)4
3
3
2160 120 360
∞
X
n=1
3ζ(3)
ζ(2n + 1)
1
γ
′
+ 28 log A −
=
128ζ
−
2,
−5−
n
2
(n + 1)(2n + 3)16
4
π
3
∞
X
9ζ(3) 41 γ 1
1
ζ(2n + 1)
′
+24 log A+540ζ ′(−3)− 2 − − + log 2
=
384ζ
−2,
n
(n + 1)(n + 2)16
4
π
12 2 5
∞
X
ζ(2n + 1)
3ζ(3) 79 γ 1
1
′
+8 log A−135ζ ′(−3)+
−2,
=
−32ζ
− − − log 2
n
(2n + 2) . . . (2n + 4)16
4
4π 2 48 24 20
n=1
n=1
References
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University of Pittsburgh, Department of Mathematics, 301 Thackeray Hall,
Pittsburgh, PA 15260, USA
E–mail address: [email protected]