1
The evaluation of character Euler double sums
J.M. Borwein*, I.J. Zucker† and J. Boersma‡
*Faculty of Computing Science, Dalhousie University, Halifax, NS, B3H 3J5 Canada.
Research supported by NSERC and by the Canada Research Chairs programme.
† Wheatstone Physics Laboratory, King’s College, Strand, London WC2R 2LS, UK.
‡Dept of Mathematics and Computing Science, Eindhoven University of Technology.
5600 MB Eindhoven, The Netherlands.
Abstract
Euler considered sums of the form
∞
m−1
∑
1 ∑ 1
.
s
t
m
n
m=1
n=1
Here natural generalisations of these sums namely
∞
∑ χd (n)
∑
χd1 (m) m−1
2
[d1 , d2 ] := [d1 , d2 ](s, t) =
,
s
m
nt
n=1
m=1
have been investigated, where χd1 and χd2 are characters, and s and t are positive
integers. The cases when d1 and d2 were either 1, 2a, 2b or −4 have been examined
in detail, and closed form expressions have been found for t = 1 and general s in
terms of the Riemann zeta function and the Catalan zeta function—the Dirichlet series
L−4 (s) = 1−s − 3−s + 5−s − 7−s . . .. Some results for arbitrary d1 and d2 have also been
obtained.
2
1. Introduction
In 1742 Goldbach wrote to Euler posing the problem of finding a closed form for the
double sum
∞
m−1
∑
1 ∑ 1
ms n=1 nt
m=1
(1.1)
in which the first sum is unrestricted, whilst the second is bounded. We refer to [1a]
for a description of this exchange. We shall refer to the first sum as the U-sum and the
second as the B-sum. Any double sum so constructed will be referred to as an Euler or
E-sum.
The motivation for this investigation was an attempt to generalise the E-sums such that
the U- and B-sums were general Dirichlet series, i.e. to consider E-sums of the form
∞
∑
∑ χd (n)
χd1 (m) m−1
2
[d1 , d2 ] := [d1 , d2 ](s, t) =
,
s
m
nt
m=1
n=1
(1.2)
where χd1 and χd2 are characters. This proved too ambitious a project, and here we
have concentrated on E-sums involving the primitive characters χ1 and χ−4 , and two
imprimitive forms of χ1 , namely χ2a (counting odd terms) and χ2b (counting all terms
alternatingly). In Appendix 1 some properties of characters and Dirichlet series relevant
to this communication are summarised. In section 6 some properties of such general
E-sums will be considered. The four series considered here in detail are
∞
∞
∑
χ1 (n) ∑
1
ζ(s) =
=
= L1 (s),
s
n
ns
n=1
n=1
(1.3)
∞
∞
∑
χ2a (n) ∑
1
λ(s) =
=
= L2a (s) = (1 − 2−s )L1 (s),
s
s
n
(2n
−
1)
n=1
n=1
(1.4)
∞
∞
∑
χ2b (n) ∑
(−1)n−1
η(s) =
=
= L2b (s) = (1 − 21−s )L1 (s),
s
s
n
n
n=1
n=1
(1.5)
β(s) =
∞
∑
χ−4 (n)
n=1
ns
∞
∑
(−1)n−1
=
= L−4 (s).
(2n − 1)s
n=1
(1.6)
Thus, we will consider double series with any one of the four above as the U-sum or
B-sum yielding the following sixteen E-sums to be considered. These are displayed in
Table 1.
3
Table 1
[1, 1](s, t) =
∞
m−1
∑
1 ∑ 1
.
ms n=1 nt
m=1
[1, 2a]*(s, t) =
∞
m
∑
1 ∑
1
.
ms n=1 (2n − 1)t
m=1
[1, 2b](s, t) =
∞
m−1
∑
1 ∑ (−1)n−1
.
ms n=1
nt
m=1
[1, −4]*(s, t) =
∞
m
∑
1 ∑ (−1)n−1
.
ms n=1 (2n − 1)t
m=1
[2a, 1](s, t) =
[2a, 2a](s, t) =
[2a, 2b](s, t) =
[2a, −4](s, t) =
[2b, 1](s, t) =
[2b, 2a]*(s, t) =
[2b, 2b](s, t) =
[2b, −4]*(s, t) =
[−4, 1](s, t) =
∞
∑
m−1
∑ 1
1
.
s
t
(2m
−
1)
n
m=1
n=1
∞
∑
m−1
∑
1
1
.
s
(2m − 1) n=1 (2n − 1)t
m=1
∞
∑
m−1
∑ (−1)n−1
1
.
(2m − 1)s n=1
nt
m=1
∞
∑
m−1
∑ (−1)n−1
1
.
s
t
(2m
−
1)
(2n
−
1)
n=1
m=1
m−1
∞
∑
(−1)m ∑ 1
.
ms n=1 nt
m=1
m
∞
∑
1
(−1)m−1 ∑
.
s
m
(2n − 1)t
n=1
m=1
m−1
∞
∑
(−1)m ∑ (−1)n−1
.
ms n=1
nt
m=1
m
∞
∑
(−1)m−1 ∑ (−1)n−1
.
ms
(2n − 1)t
n=1
m=1
m−1
∞
∑
(−1)m−1 ∑ 1
.
s
t
(2m
−
1)
n
n=1
m=1
∞
∑
[−4, 2a](s, t) =
m−1
1
(−1)m ∑
.
s
(2m − 1) n=1 (2n − 1)t
m=1
[−4, 2b](s, t) =
∞
m−1
∑
(−1)m−1 ∑ (−1)n−1
.
(2m − 1)s n=1
nt
m=1
[−4, −4](s, t) =
∞
m−1
∑
(−1)m−1 ∑ (−1)n−1
.
s
t
(2m
−
1)
(2n
−
1)
m=1
n=1
4
The sum [1,1] is the original E-sum considered by Euler. Euler[2] responded to Goldbach’s query (somewhat tardily) and wrote a memoir on the topic in 1775. He showed
that [1,1] could be evaluated in terms of the Riemann ζ-function for all s + t ≤ 6, and
s + t ≤ 13 when s + t is odd. He then extrapolated his results to obtain a general
formula for all s + t odd. He also proved a formula [1, 1](s, 1) for general s. Some 130
years later Nielson[5] extended Euler’s investigation to consider [1,2b], [2b,1] and [2b,2b]
and derived a formula for [1, 2b](s, 1). More recently Rao[6] found general formulae for
[2b, 1](2s, 1) and [2b, 2b](2s, 1). Borwein, Borwein and Girgensohn (see [1b]) also investigated [1,1], [1,2b], [2b,1] and [2b,2b]. They proved Euler’s general result for [1,1]
with s + t odd and also Nielson’s and Rao’s results. Jordan[3] in a paper unconnected
with E-sums per se gave formulae for [1, 2a](2s, 1) and [2a, 2a](2s, 1) but they were only
correct for s = 1. Otherwise, we have found very little on general double E-sums in the
literature.
Throughout we proceed formally—noting that the sums in question are all uniformly
convergent for s > 1 and do not explicitly mention the routine justification of rearrangements.
It will be noted that some of the E-sums in Table 1 have B-sums in which the upper
limit is n = m not n = m − 1. This is because one may transform any E-sum into a
completely unrestricted double sum—referred to as a UD-sum, and in these cases an
upper limit of n = m yields a simpler form. Such E-sums will be denoted [d1 , d2 ]*.
As an example of how to turn an E-sum into a UD-sum consider [1, 1](s, t)] and expand
[1, 1](s, t) =
∞
m−1
∑
1 ∑ 1
. = 2−s (1−t ) + 3−s (1−t + 2−t ) + 4−s (1−t + 2−t + 3−t ) . . .
s
t
m n=1 n
m=1
(1.7)
We rearrange as
1−t (ζ(s) − 1−s ) + 2−t (ζ(s) − 1−s − 2−s ) + 3−t (ζ(s) − 1−s − 2−s − 3−s ) . . . ,
(1.8)
and this may be written as the UD-sum
∞ ∑
∞
∑
m=1 n=1
1
.
+ n)s
mt (m
(1.9)
5
Alternatively we may change variables directly in the sum.
From this it is possible to derive an integral representation. Thus,
∞ ∞ ∫
1
(−1)s−1 ∑ ∑ ∞ us−1 e−(m+n)u
=
du
mt (m + n)s
Γ(s) m=1 n=1 0
mt
m=1 n=1
∫
∞
(−1)s−1 ∞ s−1 e−u ∑ e−mu
=
u
du.
Γ(s)
1 − e−u n=1 mt
0
∞ ∑
∞
∑
(1.10)
Substituting x = e−u and recalling that
∞
∑
xn
= Lit (x),
t
n
n=1
(1.11)
is the polylogarithm of Lewin[4], one has
∞ ∑
∞
∑
1
(−1)s−1
[1, 1](s, t) =
=
mt (m + n)s
Γ(s)
m=1 n=1
∫
0
1
(log x)s−1 Lit (x)
dx.
(1 − x)
(1.12)
In this way all sixteen E-sums of Table 1 can be converted into UD-sums with integral
representations and these are all given in Table 2.
6
Table 2a
Euler Sum
[1, 1](s, t)
[1, 2a]*(s, t)
[1, 2b](s, t)
[1, −4]*(s, t)
[2a, 1](s, t)
[2a, 2a](s, t)
[2a, 2b](s, t)
[2a, −4](s, t)
[2b, 1](s, t)
[2b, 2a]*(s, t)
[2b, 2b](s, t)
[2b, −4]*(s, t)
Transformed Sum
Integral Representation
∞ ∑
∞
∑
∞ ∑
∞
∑
1
t
(2m − 1) (m + n − 1)s
m=1 n=1
∞ ∑
∞
∑
(−1)m−1
mt (m + n)s
m=1 n=1
∞ ∑
∞
∑
(−1)m−1
(2m − 1)t (m + n − 1)s
m=1 n=1
s (−1)
2
1
t
(2m − 1) (2m + 2n − 1)s
m=1 n=1
∞ ∑
∞
∑
(−1)m−1
mt (2m + 2n − 1)s
m=1 n=1
2
∞ ∑
∞
∑
(−1)(m+n−1)
(2m − 1)t (m + n − 1)s
m=1 n=1
s (−1)
(−1)n
mt (m + n)s
m=1 n=1
∞ ∑
∞
∑
(−1)n
(2m − 1)t (m + n − 1)s
m=1 n=1
∫
∫
1
x(log x)s−1 T hit (x)
dx
(1 − x2 )
0
−(log x)s−1 Lit (−x2 )
dx
1 − x2
1
0
∫
1
−(log x)s−1 Lit (−x)
dx
(1 + x)
1
0
∫
s−1
s−1
Γ(s)
x(log x)s−1 T it (x)
dx
(1 − x2 )
0
∫
(log x)s−1 T it (x)
dx
1 − x2
(log x)s−1 Lit (x2 )
dx
(1 − x2 )
0
(−1)s−1
Γ(s)
2
1
1
Γ(s)
s (−1)
(log x)s−1 T hit (x)
dx
1 − x2
0
∫
(−1)s−1
Γ(s)
∞ ∑
∞
∑
∫
Γ(s)
(−1)s−1
Γ(s)
2
1
−(log x)s−1 Lit (−x)
dx
(1 − x)
1
0
(−1)s−1
Γ(s)
(−1)m−1
(2m − 1)t (2m + 2n − 1)s
m=1 n=1
∞ ∑
∞
∑
(−1)(m+n−1)
mt (m + n)s
m=1 n=1
∫
s−1
(−1)s−1
Γ(s)
(log x)s−1 Lit (x)
dx
(1 − x)
0
(−1)s−1
Γ(s)
∞ ∑
∞
∑
∫
Γ(s)
s (−1)
1
t
m (2m + 2n − 1)s
m=1 n=1
1
0
s−1
(−1)s−1
Γ(s)
∞ ∑
∞
∑
∞ ∑
∞
∑
∫
(−1)s−1
Γ(s)
1
t
m (m + n)s
m=1 n=1
1
0
∫
1
0
∫
0
1
(log x)s−1 T it (x)
dx
1 + x2
(log x)s−1 Lit (x)
dx
(1 + x)
(log x)s−1 T hit (x)
dx
1 + x2
7
Table 2b
Euler Sum
Transformed Sum
∞ ∑
∞
∑
(−1)m+n
mt (2m + 2n − 1)s
m=1 n=1
[−4, 1](s, t)
[−4, 2a](s, t)
(−1)s−1
Γ(s)
∞ ∑
∞
∑
(−1)m+n
(2m − 1)t (2m + 2n − 1)s
m=1 n=1
∫
(−1)s−1
Γ(s)
∞ ∑
∞
∑
(−1)n−1
mt (2m + 2n − 1)s
m=1 n=1
[−4, 2b](s, t)
[−4, −4](s, t)
Integral Representation
(−1)s−1
Γ(s)
∞ ∑
∞
∑
(−1)n−1
(2m − 1)t (2m + 2n − 1)s
m=1 n=1
(−1)s−1
Γ(s)
−(log x)s−1 Lit (−x2 )
dx
(1 + x2 )
1
0
∫
1
x(log x)s−1 T it (x)
dx
(1 + x2 )
1
(log x)s−1 Lit (x2 )
dx
1 + x2
1
x(log x)s−1 T hit (x)
dx
(1 + x2 )
0
∫
0
∫
0
The related functions T it (x) and T hit (x) are defined [4] as
∞
∑
(−1)n−1 x2n−1
T it (x) =
,
(2n − 1)t
n=1
T hit (x) =
∞
∑
x2n−1
.
(2n − 1)t
n=1
(1.13)
The following properties of these functions are noted.
Li1 (x) = − log(1 − x),
T i1 (x) = arctan(x),
T hi1 = arctanh(x).
(1.14)
All these functions obey the general rule
∫
fn (x) =
0
x
fn−1 (t)
dt,
t
(1.15)
and
Lin (1) = ζ(n),
Lin (−1) = −η(n),
T in (1) = β(n),
and T hin (1) = λ(n).
(1.16)
2. Some general results
Euler found a reflection formula for [1,1], namely
[1, 1](s, t) + [1, 1](t, s) = ζ(s)ζ(t) − ζ(s + t).
(2.1)
8
We illustrate how this may be proved.
∞
∞
∞
∞
m−1
∞
∞
∞
∑
∑
∑
1 ∑ 1
1 ∑ 1 ∑ 1 ∑ 1
1 ∑ 1
=
−
= ζ(s)ζ(t)−
. (2.2)
ms n=1 nt
ms n=1 nt m=1 ms n=m nt
ms n=m nt
m=1
m=1
m=1
Now
∞
∞
∞
∞
∑
∑
∑
1 ∑ 1
1
1
=
+
ms n=m nt
ms+t m=1 ms
m=1
1
∞
∑
∞
∑
1
1
=
ζ(s
+
t)
+
nt
ms
n=m+1
m=1
∞
∑
1
. (2.3)
t
n
n=m+1
Expanding the second term of (2.3)
∞
∑
1
ms
m=1
∞
∑
1
= 1−s (ζ(t)−1−t )+2−s (ζ(t)−1−t −2−t )+3−s (ζ(t)−1−t −2−t −3−t ) . . . ,
t
n
n=m+1
(2.4)
and comparing with (1.6), it is seen that is just [1, 1](t, s), and hence (2.1). Similarly,
Nielson[5] found reflection formulae involving [1,2b], [2b,1] and [2b,2b], and we have
computed the corresponding results involving all the E-sums considered here. These
are given below. The general case is described in section six.
[1, 2b](s, t) − [2b, 1](t, s) =ζ(s)η(t) − η(s + t).
(2.5)
[2b, 2b](s, t) + [2b, 2b](t, s) = − η(s)η(t) + ζ(s + t).
(2.6)
[2a, 2a](s, t) + [2a, 2a](t, s) =λ(s)λ(t) − λ(s + t).
(2.7)
[2a, −4](s, t) − [−4, 2a](t, s) =λ(s)β(t) − β(s + t).
(2.8)
[−4, −4](s, t) + [−4, −4](t, s) =β(s)β(t) − λ(s + t).
(2.9)
[1, 2a]*(s, t) + [2a, 1](t, s) = ζ(s)λ(t),
(2.10)
[1, −4]*(s, t) − [−4, 1](t, s) = ζ(s)β(t),
(2.11)
[2b, 2a]*(s, t) + [2a, 2b](t, s) = η(s)λ(t)
(2.12)
[2b, −4]*(s, t) − [−4, 2b](t, s) = η(s)β(t)
(2.13)
Now, in the UD form it is simple to split sums over all integers into sums over all even
integers plus sums over all odd integers. Thus, for example, [1, 1](s, t) yields
∞ ∑
∞
∑
m=1 n=1
1
+ n)s
mt (m
9
=
∞ ∑
∞
∑
∞ ∑
∞
∑
1
1
+
(2m)t (2m + n)s m=1 n=1 (2m − 1)t (2m − 1 + n)s
m=1 n=1
=
∞ ∑
∞
∑
1
1
+
(2m)t (2m + 2n)s m=1 n=1 (2m − 1)t (2m − 1 + 2n)s
m=1 n=1
∞ ∑
∞
∑
∞ ∑
∞
∑
∞ ∑
∞
∑
1
1
+
t
s
t
(2m) (2m + 2n − 1)
(2m − 1) (2m − 1 + 2n − 1)s
m=1 n=1
m=1 n=1
(2.14)
This gives
[1, 1] = 2−s−t [1, 1] + 2−t [2a, 1] + 2−s [1, 2a]* + [2a, 2a].
(2.15)
Similarly we find
[1, 2b] = − 2−s−t [1, 1] − 2−t [2a, 1] + 2−s [1, 2a]* + [2a, 2a].
(2.16)
[2b, 1] =2−s−t [1, 1] − 2−t [2a, 1] + 2−s [1, 2a]* − [2a, 2a].
(2.17)
[2b, 2b] = − 2−s−t [1, 1] + 2−t [2a, 1] + 2−s [1, 2a]* − [2a, 2a].
(2.18)
3. Exact expressions for E-sums
We first must make explicit what is meant here by ’exact’. For the E-sums considered
previously it clearly meant expressing the result in terms of the constants of classical
analysis ζ(n), λ(n) and η(n), i.e., essentially the zeta function at integer arguments.
However, since we have now added β(n) as an extra factor in the E-sums considered
here, it would be expected that β(n) should appear as an element in any solution. Hence
if we are able to express a general E-sum in terms of the four functions given by (1.3)(1.6)—or by like quantities—it will be termed exact, and said to be given in terms of
well-known (WK) constants of analysis. Later it will seen that the adjunction of less
familiar constants of analysis will allow other hitherto unknown E-sums to be expressed
’exactly’.
Our initial attempts at finding E-sums not previously evaluated were concentrated on
evaluating their integral representations as given in Table 2. This was haphazard at best,
and generally only for very small values of s and t could the evaluations be accomplished.
With the aid of considerable experimental mathematical insight, however, a method was
subsequently developed for finding the integrals for general s and t = 1.
10
A generating function G[d1 , d2 ](w) was established for each integral representation. It
was found that the generating functions were connected two-by-two in the following
pairs.
([1, 1] ↔ [2b, 1]), ([1, 2a]* ↔ [2b, 2a]*), ([1, 2b] ↔ [2b, 2b]), ([1, −4]* ↔ [2b, −4]*),
([2a, 1] ↔ [−4, 1]), ([2a, 2a] ↔ [−4, 2a]), ([2a, 2b] ↔ [−4, 2b]), ( [2a, −4] ↔ [−4, −4]),
In general our generating functions are defined by
G[d1 , d2 ](w) =
∞
∑
[d1 , d2 ](s, 1) ws−1 .
(3.1)
s=1 or 2
Here, the lower limit in the summation is s = 1 or s = 2 dependent on whether
[d1 , d2 ](1, 1) is convergent or divergent. An integral representation for G readily follows
from the integral representation for [d1 , d2 ](s, 1) in Table 2. The even and odd parts of
G(w) (even or odd in w) are denoted by
∞
∑
1
GE[d1 , d2 ](w) = [G[d1 , d2 ](w) + G[d1 , d2 ](−w)] =
[d1, d2](2s + 1, 1) w2s ,(3.2)
2
s=0 or 1
∞
∑
1
GO[d1 , d2 ](w) = [G[d1 , d2 ](w) − G[d1 , d2 ](−w)] =
[d1 , d2 ](2s, 1) w2s−1 .(3.3)
2
s=1
A typical approach to the problem of determining [d1 , d2 ](s, 1) will be illustrated by the
specific example of the pair ([2a, 1] ↔ [−4, 1]).
Starting from their integral expressions, given in Table 2 above, and using (3.1) we have
G[2a, 1](w) =
∞
∑
∫
G[−4, 1](w) =
∫
1
x−w
log(1 + x2 )dx.
1 + x2
(3.5)
[2a, 1](s, 1)w =
s=2
and
(3.4)
0
1 − x−w
log(1 − x2 )dx.
1 − x2
1
s
∞
∑
s
[−4, 1](s, 1)w =
0
s=1
To evaluate (3.4) we first evaluate G[1,1] which is given by the expression
G[1, 1](w) =
∞
∑
s=2
∫
1
s
[1, 1](s, 1)w =
0
1 − x−w
log(1 − x)dx.
1−x
(3.6)
11
We note it can be expressed as the derivative of an Euler beta function thus
[∫ 1
]
[
]
d
d 1 Γ(1 − w)Γ(q)
−w
q−1
G[1, 1](w) =
(1 − x )(1 − x) dx
=
−
. (3.7)
dq 0
dq q
Γ(1 − w + q) q↓0
q↓0
This is expanded in powers of q yielding
1 Γ(1 − w)Γ(q)
1 1
q2
−
= − {1 + q [ψ(1) − ψ(1 − w)] + [ψ 2 (1) + ψ ′ (1)
q
Γ(1 − w + q)
q
q
2
2
′
− 2ψ(1)ψ(1 − w) + ψ (1 − w) − ψ (1 − w)] + O(q 3 )},
(3.8)
where ψ(z) = Γ′ (z)/Γ(z) is the psi function. Hence G[1, 1](w) is found to be
1
1
2
G[1, 1](w) = − [ψ(1 − w) − ψ(1)] + [ψ ′ (1 − w) − ψ ′ (1)] .
2
2
(3.9)
Finding Taylor expansions of ψ(1 − w) and ψ ′ (1 − w) about the point w = 1, and using
the well-known result
ψ (k) = (−1)k+1 k!ζ(k + 1);
we have
ψ(1 − w) =
∞
∑
(−1)k
k=0
ψ ′ (1 − w) =
∞
∑
k=0
k!
ψ
(k)
(k = 1, 2, 3, . . .),
(1)w = ψ(1) −
k
∞
∑
ζ(k + 1)wk ,
(3.10)
(3.11)
k=1
(−1)k (k+1)
ψ
(1)wk = ψ ′ (1) +
k!
∞
∑
(k + 1)ζ(k + 2)wk .
(3.12)
k=1
Substituting (3.11) and (3.12) into (3.9), [1, 1](s, 1) is found as the coefficient of ws−1
in the expansion of G[1, 1](w). We thus obtain Euler’s original result
1
1∑
[1, 1](s, 1) = sζ(s + 1) −
ζ(k + 1)ζ(s − k).
2
2
s−2
(3.13)
k=1
Now substitute u = x2 in (3.4) and we have
∫
1 1 u−1/2 − u−w/2−1/2
G[2a, 1](w) =
log(1 − u)du
2 0
1−u
]
1[
G[1, 1]( 12 + w/2) − G[1, 1]( 12 ) .
=
2
(3.14)
Thus from (3.9) we have
]2 [
]
1[ 1
ψ( 2 − w/2) − ψ( 12 ) + ψ( 21 − w/2) − ψ( 12 ) log 2
4
]
1[ ′ 1
+
ψ ( 2 − w/2) − ψ ′ ( 12 ) .
4
G[2a, 1](w) = −
(3.15)
12
Then similar to (3.11) and (3.12)
ψ( 12
− w/2) =
∞
∑
(−1)k
k!
k=0
ψ ′ ( 12
∞
∑
(−1)k
− w/2) =
k!
k=0
ψ (k) ( 12 )(w/2)k
ψ (k+1) ( 12 )(w/2)k
=
ψ( 12 )
−2
∞
∑
λ(k + 1)wk ,
(3.16)
k=1
=
ψ ′ ( 12 )
+4
∞
∑
(k + 1)λ(k + 1)wk , (3.17)
k=1
Putting (3.16) and (3.17) into (3.15) and expanding in powers of w, G[2a, 1](s, 1) is
found as the coefficient of ws−1 .
[2a, 1](s, 1) = sλ(s + 1) − 2λ(s) log 2 −
s−2
∑
λ(k + 1)λ(s − k).
(3.18)
k=1
This is a new result to add to those of Euler[2] and Nielson[5], that might have been
obtained from Euler and Nielsons’ results by subtracting (2.16) from (2.15) giving
21−t [2a, 1](s, t) = (1 − 21−s−t )[1, 1](s, t) − [1, 2b](s, t).
(3.19)
4. The evaluation of G[−4, 1](w)
Let C be the quarter circle contour made up of three parts: the line segment z = x, 0 ≤
x ≤ 1; z = eiθ , 0 ≤ θ ≤ π/2; and the line segment z = iy, 1 ≥ y ≥ 0. We start from the
contour integral
∫
C
(z e−πi/2 )−w − 1
log(1 + z 2 )dz = 0.
1 + z2
(4.1)
Evaluating this integral along the three parts of C yields
πiw/2
e
−
1
2
∫
i
G[−4, 1](w) − G[−4, 1](1, 1) +
2
π/2
0
−1
log(2 sin θ)dθ
sin θ
π/2 iwθ
e
−1
(π/2 − θ)dθ + iG[2a, 1](w) = 0.
sin θ
iwθ
e
0
∫
(4.2)
Thus—having previously found G[2a, 1](w)—if the integrals in (4.2) can be evaluated
then we shall have G[−4, 1](w). Taking the integrals in (4.2) and expanding in powers
of w, we have
∫
π/2
0
∞
∑
eiwθ − 1
(iw)n
log(2 sin θ)dθ =
K[−4, 1](n),
sin θ
n!
n=1
(4.3)
13
and
∫
π/2
0
∞
∑
eiwθ − 1
(iw)n
(π/2 − θ)dθ =
[(π/2)Jn − Jn+1 ] ,
sin θ
n!
n=1
where
∫
π/2
K[−4, 1](n) =
0
and
∫
θn
log(2 sin θ)dθ,
sin θ
π/2
Jn =
0
θn
dθ.
sin θ
(4.4)
(4.5)
(4.6)
Expanding the left-hand side of (4.2) in powers of w and denoting the coefficient of the
real part ws−1 by As , for s = 2, 3, 4 . . ., the following recurrence relations may be found
1
[−4, 1](2, 1) = K[−4, 1](1),
2
π2
[−4, 1](3, 1) = [−4, 1](1, 1) −
8
π2
[−4, 1](4, 1) = [−4, 1](2, 1) −
8
π2
[−4, 1](5, 1) = [−4, 1](3, 1) −
8
(4.7)
π
1
J2 + J3 ,
8
4
1
K[−4, 1](3),
12
π4
π
1
[−4, 1](1, 1) + J4 − J5 .
384
96
48
(4.8)
(4.9)
(4.10)
Now [−4, 1](1, 1) may be evaluated directly from its integral representation thus
∫
[−4, 1](1, 1) ==
0
1
log(1 + x2 )
π
dx = log 2 − β(2),
2
1+x
2
(4.11)
and it is straightforward to evaluate Jn in terms of WK. Thus it may be seen from the
pattern exhibited by (4.7)–(4.10) that [−4, 1](2s + 1) may be found in WK. However,
the integrals K[−4, 1](n) seemingly cannot so be found.
These examples illustrate what has been obtained for the 16 E-sums in Table 1, namely
that a few can be found for all s with t = 1 but most can only be found in terms of WK
either for even s, or odd s as above. In the cases where only even or odd s E-sums can
be found the analysis yields integrals which can apparently only be evaluated in WK
for either even or odd s.
In Table 3 all the general results found have been displayed. The evaluation of the
general result for [−4, 1](2s + 1, 1) is given in detail in Appendix 2.
14
Table 3
1
1∑
[1, 1](s, 1] = sζ(s + 1) −
ζ(k)ζ(s − k + 1).
2
2
s−1
k=2
[
[1, 2a]*(2s, 1] =22s
]
s−1
∑
1
2−2k ζ(2k)λ(2s − 2k + 1) .
λ(2s + 1) −
2
k=1
1
1∑
[1, 2b][s, 1] = − sζ(s + 1) + η(1)ζ(s) +
η(k)η(s − k + 1).
2
2
s
k=1
[1, −4]*(2s + 1, 1] =2 β(1)λ(2s + 1) −
2s
s
∑
22k+1 β(2k + 2)η(2s − 2k).
k=0
[2a, 1][s, 1] =sλ(s + 1) − 2η(1)λ(s) −
s−1
∑
λ(k)λ(s − k + 1).
k=2
∑
1
[2a, 2a](2s, 1) = − λ(2s + 1) + η(1)λ(2s) −
22k−2s−1 λ(2k)ζ(2s − 2k + 1).
2
s−1
k=1
[2a, 2b][s, 1] = − sλ(s + 1) + η(1)λ(s) −
s
∑
β(k)β(s − k + 1).
k=1
[
]
s
∑
1
−2s
β(1)λ(2s + 1) +
β(2k + 1)η(2s − 2k + 1) .
[2a, −4](2s + 1, 1] = β(2s + 2) + 2
2
k=1
∑
1
1
[2b, 1](2s, 1] = ζ(2s + 1) − (2s − 1)η(2s + 1) +
η(2k)ζ(2s − 2k + 1).
2
2
s−1
k=1
[
[2b, 2a]*(2s, 1] =22s
]
s−1
∑
1
2−2k η(2k)λ(2s − 2k + 1) .
β(1)β(2s) − λ(2s + 1) −
2
k=1
15
[2b, 2b](2s, 1) =λ(2s + 1) + sη(2s + 1) − η(1)η(2s) −
s
∑
ζ(2k)η(2s − 2k + 1).
k=1
[2b, −4]*(2s + 1, 1] = −
s
∑
22k+1 β(2k + 2)ζ(2s − 2k).
k=0
[−4, 1](2s + 1, 1] = − (2s + 1)β(2s + 2) + 2η(1)β(2s + 1) + 2
s
∑
β(2k − 1)λ(2s − 2k + 3).
k=1
1
[−4, 2a](2s + 1, 1] = β(2s + 2) − η(1)β(2s + 1)
2
[
+2
−2s−1
β(1)η(2s + 1) +
s
∑
]
2k−2
2
β(2k − 1)ζ(2s − 2k + 3) .
k=1
[−4, 2b](2s + 1, 1] =(2s + 1)β(2s + 2) − η(1)β(2s + 1) − 2
s
∑
β(2k)λ(2s − 2k + 2).
k=1
∑
1
[−4, −4](2s, 1) = λ(2s + 1) −
2−2k+1 η(2k − 1)λ(2s − 2k + 2).
2
s
k=1
5. E-sums likely not to be expressible in WK
The six sums [1, −4]*, [2a, −4], [2b, −4]*, [−4, 1], [−4, 2a], [−4, 2b] —we refer to this set
as A(s, 1)—seem only to be expressible in WK for odd values of s when t = 1. This was
borne out by our experiments with Integer relation methods (see [1a,1b]). By attempting
to solve the integrals involved for the particular case A(2, 1) it was found that a solution
could be found provided one extra constant of analysis was added to the set of WK.
This extra constant is
∫
(
π/2
ℓ2 = −
log
2
0
θ
2 sin
2
With this extra constant it can be shown that
[1, −4]*(2, 1) = −2ℓ2 −
π3
12
)
dθ.
(5.1)
16
ℓ2
7π 3
β(2) log 2
−
+
2
192
2
3
7π
[2b, −4]*(2, 1) = 4ℓ2 +
24
5π 3
[−4, 1](2, 1) = −ℓ2 −
+ 2β(2) log 2
48
17π 3
[−4, 2a](2, 1) = ℓ2 +
− β(2) log 2
192
π3
[−4, 2b](2, 1) = −ℓ2 −
− β(2) log 2
24
[2a, −4](2, 1) = −
(5.2)
However, so far all attempts at finding any like result for A(4, 1) have failed. It seems
unlikely that just one single extra constant, ℓ2 , will suffice to accommodate all the even
s in A, and perhaps higher order examples of ℓn will appear as s increases.
The sums [1, 2a]*, [2a, 2a], [2b, 1], [2b, 2a]*, [2b, 2b], [−4, −4] —call them B(s, 1)—can only
be expressed in WK for even s when t = 1 with the exception of the case when s = 1.
In this instance the first two members of B are divergent, but the others represented by
the integrals given in Table 2 are particularly simple to evaluate and we have
log2 2
1
2
,
[2b, 1](1, 1) = η(1) =
2
2
π2
[2b, 2a]*(1, 1) = β 2 (1) =
,
16
1
π2
log2 2
[2b, 2b](1, 1) = η(2) − η 2 (1) =
−
,
2
12
2
π2
β 2 (1)
[−4, −4](1, 1) =
=
.
2
32
(5.3)
In determining B(3, 1) a similar situation to that of finding A(2, 1) was encountered,
namely the addition of one extra constant to WK allowed all B(3, 1) to be denoted.
This time the extra constant is c(4) where c(n) is defined by
c(n) =
n−2
∑
k=0
( 1 ) (log 2)n
(n − 2)!
k
(log 2) Lin−k 2 +
.
k!
n
Thus
53 4
π ,
720
1
7
53 4
[2a, 2a](3, 1) = − c(4) + ζ(3) log 2 +
π ,
2
8
5760
[1, 2a](3, 1) = 4c(4) −
(5.4)
17
[2b, 1](3, 1) = c(4) −
1 4
π ,
48
[2b, 2a]*(3, 1) = −2c(4) +
51 4
π ,
2880
7
1 4
[2b, 2b](3, 1) = c(4) − ζ(3) log 2 −
π ,
4
180
1
7
31 4
[−4, −4](3, 1) = c(4) − ζ(3) log 2 −
π .
4
16
23040
(5.5)
It should be noted that for small values of n, c(n) is expressible in WK, thus
c(1) = log 2,
c(2) =
1 2
π ,
12
c(3) =
7
ζ(3),
8
(5.6)
so all the results in (5.5) might be viewed as combinations of c(1) to c(4), e.g.
4
[2b, 2b](3, 1) = c(4) − 2c(1)c(3) − c2 (2).
5
(5.7)
suggesting that B(5, 1) might be various combinations of c(1)–c(6). However, as with
A(4, 1), no single value of B(5, 1) has been evaluated in any form, despite much relation
hunting. It remains an open question as to whether A(2s, 1) or B(2s+1, 1) is expressible
in any reasonable type of closed form.
6. General character E-sums
A reflection formula is accessible for general double E-sums. Write
[d1 , d2 ](s, t) =
∞
∑ χd (n)
∑
χd1 (m) m−1
2
m=1
as
[d1 , d2 ](s, t) =
Then
[d2 , d1 ](t, s) =
ms
nt
n=1
,
(6.1)
∑ χd (m) χd (n)
1
2
.
s
t
m
n
m>n
∑ χd (n) χd (m)
2
1
m<n
nt
ms
(6.2)
.
(6.3)
Adding (6.2) to (6.3) we get
∑ χd (n) χd (m)
2
1
m̸=n
nt
ms
=
∑
m≥1,n≥1
χd1 (m) χd2 (n)
ms
nt
−
∑ χd (m)χd (m)
1
2
m=n
ms+t
,
(6.4)
and since χd1 (m)χd2 (m) = χd1 d2 (m) we have the reflection formula
[d1 , d2 ](s, t) + [d2 , d1 ](t, s) = Ld1 (s)Ld2 (t) − Ld1 d2 (s + t).
(6.5)
18
When d1 = d2 = d it is simple to show that
Ld2 (s) = (1 − d−s )ζ(s).
(6.6)
It is also straightforward to establish [d1 , d2 ](s, t) as a UD-sum thus
[d1 , d2 ](s, t) =
∞ ∑
∞
∑
χd2 (m) χd1 (m + n)
mt
m=1 n=1
(m + n)s
,
(6.7)
Sadly, however, we are as of yet only able to put this into an integral form when
χd1 (m + n) can be split multiplicatively as f (m)g(n). This can be done when d1 = 1
when χ1 (m + n) = 1 for all m, n, and for χ2b (m + n) = (−1)m+n .
So for these cases we have
(−1)s−1
[1, d](s, t) =
Γ(s)
(−1)s−1
[2b, d](s, t) =
Γ(s)
∫
∫
1
0
1
0
(log x)s−1 Ld (t : x)
dx,
1−x
(6.8)
(log x)s−1 Ld (t : −x)
dx,
1+x
(6.9)
where Ld (t : −x) is defined in Appendix 1. It is from the integrals that we might hope
to find some analytic results. The difficulty of doing this even in the simplest of cases
was illustrated when finding [2b, −3](1, 1) and [2b, 5](1, 1). We have
L−3 (s) =
∞
∑
χ−3 (n)
n=1
L5 (s) =
∞
∑
χ5 (n)
n=1
ns
=
∞ [
∑
n=1
ns
∞ [
∑
=
n=1
]
1
1
−
,
(3n − 2)s
(3n − 1)s
]
1
1
1
1
−
−
+
,
(5n − 4)s
(5n − 3)s
(5n − 2)s
(5n − 1)s
Naturally the set of WK has to be enlarged to include L−3 (s)
Using (6.9) we have
∫
1
[2b, −3](1, 1) =
0
It may be shown that
(6.10)
(6.11)
and L5 (s).
L−3 (1 : −x)
dx.
1+x
√
2
3x
L−3 (1 : −x) = √ arctan
.
x−2
3
(6.12)
(6.13)
Thus, integrating (6.12) by parts gives
2π log 2
+
[2b, −3](1, 1) = − √
3 3
∫
1
0
log(1 + x)
dx.
1 − x + x2
(6.14)
19
The evaluation of the integral in (6.14) may be accomplished using a result given by
Lewin[4], namely equation (8.18) in his book. The value of the integral is found to be
∫
1
0
π log 3 1
log(1 + x)
dx = √ − L−3 (2).
2
1−x+x
4
3 3
√
Now from A(1.11) it is found that L−3 (1) = π/(3 3), so finally we have
L−3 (2)
[2b, −3](1, 1) = −
− log
4
( )
4
L−3 (1).
3
(6.15)
Similarly we find
√
5L5 (1 : −x) = log(x2 − ωx + 1) − log(x2 + x/ω + 1)
(6.16),
√
where ω = ( 5 + 1)/2. Evaluating the integral obtained via (6.15) is by no means
trivial. The result we eventually obtain is the striking
[2b, 5](1, 1) =
1
√
2 5
{
(
Li2
}
(
√ )
√ )
5+ 5
5− 5
6π 2
− Li2
−
,
8
8
25
(6.17)
and it suggests general results will be very difficult to obtain. We note that L−10b (2) =
√
−6π 2 /(25 5).
Of course, using the reflection formula we can find [−3, 2b](1, 1) and [5, 2b](1, 1). Also
from the reflection formula (6.5), and using (6.6) we have for d1 = d2 = d and s = t
2[d, d](s, s) = L2d (s) − (1 − d−2s )ζ(2s).
(6.18)
So for example for d = 5 and s = 1
√
1 2
1
2
5
2
2 1+
−2
[5, 5](1, 1) = L5 (1) − (1 − 5 )ζ(2) = log
− π2 .
2
2
5
2
25
(6.19)
These are the only results we have so far obtained for more general character Euler
double sums.
Acknowledgements. The encouragement and support of Geoff Joyce and Richard
Delves at Kings College, London, is much appreciated.
20
Appendix 1
Elementary Dirichlet series are given by
∞
∑
χd (n)xn
Ld (s : x) =
ns
n=1
A(1.1)
where χd is multiplicative and is called a character. Usually we are only concerned with
x = 1, and we write Ld (s : 1) := Ld (s). Real primitive characters—meaning that all
others may be reduced to linear combinations of scalings thereof—are defined as follows.
If p is a prime > 2 then
χp (n) = χp = (n | p)
A(1.2)
where (n|p) is the Legendre-Jacobi-Kronecker symbol. The suffix p will be signed according to whether χp (p − 1) = ±1. Thus, for example
χ−3 (n) = +1
n ≡ 1(mod3)
χ5 (n) = +1
n ≡ 1, 4(mod5)
= −1
n ≡ 2(mod3)
= −1
n ≡ 2, 3(mod5)
= 0
n ≡ 0(mod3)
= 0
n ≡ 0mod5)
A(1.3)
In addition to these, there are three other primitive characters. These are
χ−4 = +1
n ≡ 1(mod4)
= −1
n ≡ 3(mod4)
= 0
(n, 4) ̸= 1
A(1.4)
and
χ−8 (n) = +1
n ≡ 1, 3(mod8)
χ8 (n) = +1
n ≡ 1, 7(mod8)
= −1
n ≡ 5, 7(mod8)
= −1
n ≡ 3, 5(mod8)
= 0
(n, 8) ̸= 1
= 0
(n, 8) ̸= 1
A(1.5)
Further primitive characters may be formed from square-free products of prime characters, and those multiplied by χ−4 , χ−8 and χ8 . Thus χP = χp1 χp2 χp3 .... and χ−4 χP ,
χ−8 χP and χ8 χP are all primitive characters. The character series
LP = LP (s) =
∞
∑
χP (n)
n=1
ns
A(1.6)
21
each yield an independent Dirichlet L-function. For example
L1 (s) =
∞
∑
χ1 (n)
n=1
∞
∑
L−3 (s) =
ns
= 1−s + 2−s + 3−s ... = ζ(s) the Riemann zeta functionA(1.7)
χ−3 (n)
n=1
ns
= 1−s − 2−s + 4−s − 5−s ...
A(1.8)
Two other types of character are encountered in our work. These are as follows
χka (n) = +1
= 0
if
χkb (n) = +1
(k, n) = 1
(k, n) ̸= 1
if
if
(k, n) = 1
= −(k − 1) n ≡ 0(modk) A(1.9)
These latter non-primitive characters are essentially modifications of the principal character, χ1 (n)=+1 for all n. When these non-primitive characters multiply primitive
characters, the latter are modified in such a way that the character series produced are
L-series of the primitive character multiplied by a factor. Thus it may be shown
∞
∑
χP χka
n=1
∞
∑
n=1
ns
χP χkb
ns
[
]
= 1 − (k | P )k −s LP
A(1.10)
[
]
= 1 − (k | P )k 1−s LP
A(1.11)
Of these non-primitive characters two occur often in this communication namely χ2a and
χ2b and when they multiply primitive characters the combination may be represented
by χ2P a and χ2P b .
It may also be shown[7] that
Ld1 (2s) =
d1
s−1 2s−1 2s ∑
(−1)
2
√
d1
π
(
χd1 (n)
(−1)
d2
s−1 2s−2 2s−1 ∑
2
√
π
d2
χ−d2 (n)
n=1
,
(2s)!
n=1
and
L−d2 (2s − 1) =
B2s 1 −
)
n
d1
(
B2s−1 1 −
n
d2
(2s − 1)!
A(1.12)
)
,
A(1.13)
where Bs (x) are the Bernoulli polynomials. As both n and d are positive integers,
Bs ((1 − n/d) are rational numbers, hence for s a positive integer
√
d1 π 2s ,
√
L−d2 (2s − 1) =R(d2 ) d2 π 2s−1 ,
Ld1 (2s) =R(d1 )
A(1.14)
A(1.15)
22
where R(d1 ) and R(d2 ) are rational numbers. It is also known that
2h(d1 )
Ld1 (1) = √
log ϵ0 ,
d1
A(1.16)
where h(d1 ) is the class number of the binary quadratic form of discriminant d1 and ϵ0
√
is the fundamental unit of the number field Q( d1 ).
Appendix 2
Evaluation of [−4, 1](2s + 1, 1) A(2.16).
Let w be real, then taking real parts in (4.2) yields
cos
+
( πw )
1
2
∫
2
π/2
0
1
G[−4, 1](w) − G[−4, 1](1, 1) −
2
∫
π/2
0
sin(wθ)
log(2 sin θ)dθ
sin θ
1 − cos(wθ)
(π/2 − θ)dθ = 0.
sin θ
A(2.1)
Next, by taking the part even in w in A(2.1) we find
cos
( πw )
2
1
GE[−4, 1](w) − G[−4, 1](1, 1) +
2
∫
0
π/2
1 − cos(wθ)
(π/2 − θ)dθ = 0. A(2.2)
sin θ
where
∞
∑
1
GE[−4, 1](w) = [G[−4, 1](w) + G[−4, 1](−w)] =
[−4, 1](2s + 1, 1)w2s .
2
s=0
A(2.3)
We thus require the value of the integral
∫
1 − cos(wθ)
(π/2 − θ)dθ =
sin θ
0
∫
∫ π/2
∫ π/2
π π/2 1 − cos(wθ)
d
sin(wθ)
θ
dθ +
dθ −
dθ
2 0
sin θ
dw 0
sin θ
sin θ
0
π/2
A(2.4)
The last integral in A(2.4) is 2β(2). The first two integrals are found from
∫
C
e−πiw/2 z w + eπiw/2 z −w − 2
dz = 0,
1 + z2
A(2.5)
23
∫
C
e−πiw/2 z w − eπiw/2 z −w
dz = 0,
1 + z2
A(2.6)
where C is the quarter circle contour given in section four. The integrals in A(2.5) and
A(2.6) vanish as the integrands are analytic in C; in particular, they have a removable
singularity at z = i. Evaluation of A(2.5) and A(2.6) yield
( πw ) ∫ 1 xw + x−w
( πw ) ∫ 1 x−w − xw
π
cos
dx + i sin
dx −
2
2
2
1+x
2
1+x
2
0
0
∫ π/2
∫ 1 w
1 − cos(wθ)
y + y −w − 2
−i
dθ − i
dy = 0,
sin θ
1 − y2
0
0
A(2.7)
and
− cos
∫
( πw ) ∫
π/2
+
0
( πw ) ∫ 1 xw + x−w
x−w − x−w
π
dx − i sin
dx −
2
2
2
1+x
2
1+x
2
0
0
∫ 1 w
sin(wθ)
y − y −w − 2
dθ − i
dy = 0.
sin θ
1 − y2
0
1
A(2.8)
In both A(2.7) and A(2.8), in the third integral the original integration variable θ was
replaced by π/2 − θ. From the imaginary part of A(2.7) we obtain
∫ π/2
∫ 1 w
( πw ) ∫ 1 x−w − xw
1 − cos(wθ)
y + y −w − 2
dθ = −
dy + sin
dx,
sin θ
1 − y2
2
1 + x2
0
0
0
whereas the real part of A(2.8) gives
∫ π/2
( πw ) ∫ 1 x−w − x−w
sin(wθ)
dθ = cos
dx.
sin θ
2
1 + x2
0
0
A(2.9)
A(2.10)
Now after expanding in powers of w we have
∫ 1 −w
∫ 1
∞
∞
∑
∑
x − x−w
w2n+1
log2n+1 (x)
dx
=
−2
dx
=
2
β(2n + 1)w2n+1 ,
2
2
1
+
x
(2n
+
1)!
1
+
x
0
0
n=0
n=0
A(2.11)
and
∫
1
0
∫ 1
∞
∞
∑
∑
y w + y −w − 2
w2n
log2n (y)
dy = 2
dy = 2
λ(2n + 1)w2n .
2
1 − y2
(2n)!
1
−
y
0
n=1
n=1
A(2.12)
Using the results A(2.9)–A(2.12) in A(2.4) gives us
∫ π/2
1 − cos(wθ)
(π/2 − θ)dθ =
sin θ
0
∞
∞
( πw ) ∑
∑
2 cos
(2n + 1)β(2n + 2)w2n − 2β(2) − π
λ(2n + 1)w2n . A(2.13)
2 n=0
n=1
24
Putting A(2.13) into A(2.2) and using [−4, 1](1, 1) = −β(2)+ π2 log 2 (section 4) we have
[
]
∞
∞
( πw )
∑
∑
π
GE[−4, 1](w) +
(2n + 1)β(2n + 2)w2n = sec
log 2 +
λ(2n + 1)w2n .
2
2
n=0
n=1
A(2.14)
Finally using the expansion
∞
( πw )
∑
π
sec
=2
β(2n + 1)w2n ,
2
2
n=0
A(2.15)
[−4, 1](2s + 1, 1) is found as the coefficient of w2s in the expansion of A(2.14) namely
[−4, 1](2s + 1, 1] = − (2s + 1)β2s + 2 + 2 log 2 β2s + 1
+2
s−1
∑
β(2k + 1)λ(2s − 2k + 1).
A(2.16)
k=0
References
1a. J. BORWEIN and D. BAILEY, Mathematics by Experiment: Plausible Reasoning
in the 21st Century. (AK Peters, Natick Mass, 2003).
1b. J. BORWEIN, D. BAILEY and R. GIRGENSOHN Experimentation in Mathematics: Computational Paths to Discovery. (AK Peters, Natick Mass, 2004).
2. L. EULER, Meditationes circa singulare serierum genus, Novi commentarii acad.
scientiarum Petripolitanae 20 (1775), 140–186.
3. P.F. JORDAN, Infinite sums of psi functions. Bull. Amer. Math.Soc. 79 (1973),
681–683.
4. L. LEWIN, Dilogarithms and associated functions. (MacDonald, London, 1958).
5. N. NIELSON, Die Gammafunction. (Chelsea, New York, 1965).
6. R. SITRAMACHANDRA RAO, A formula of S.Ramanujan, J. Number Theory 25
(1987), 1–19
7. I.J. ZUCKER and M.M. ROBERTSON, Some properties of Dirichlet L-series, J.
Phys. A: Math. Gen. 9 (1976), 1207–1225.
January 23, 2004