Some evaluation of parametric Euler sums

0.18 MB·

by  Ce Xu

Some evaluation of parametric Euler sums
Ce Xu∗
School of Mathematical Sciences, Xiamen University
7
Xiamen 361005,P.R. China
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0
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n Abstract In this paper, by using the method of Contour Integral Representations and the
a Theorem of Residues and integral representations of series, we discuss the analytic representa-
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tions of parametric Euler sums that involve harmonic numbers through zeta values and rational
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function series, either linearly or nonlinearly. Furthermore, we give explicit formulae for several

] parametric quadraticandcubicsumsin termsof zeta values andrational series. Moreover, some
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interesting new consequences and illustrative examples are considered.
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. Keywords Harmonic number; Euler sum; Riemann zeta function; Hurwitz zeta function.
h
t
a AMS Subject Classifications (2010): 11M06; 11M32.
m
[
1 Introduction

1
v
Throughoutthisarticle wewillusethefollowing definitionsandnotations. LetN := {1,2,3,...}
1
2 bethesetofnaturalnumbers,andN := N {0},N\{1} := {2,3,4,···}. Inthispaper,harmonic
0
7 numbers, alternating harmonic numbers and their generalizations are classically defined by
3 S
0 n 1 n (−1)j−1
1. ζn(k) := jk, Ln(k) := jk ,k ∈ N, (1.1)
0 j=1 j=1
X X
7
1 n
1
: where H := ζ (1) = is classical harmonic number and the empty sum ζ (m) is conven-
v n n j 0
i j=1
X X
tionally understood to be zero. The subject of this paper is Euler sums, which are the infinite
r
sums whose general term is a product of harmonic numbers and alternating harmonic numbers
a
of index n and a power of n−1 or ((−1)n−1n−1). Hence, more generally we can define the Euler
sums by the series ([18,19])
m1 ζqi (k ) m2 Llj (h ) m1 ζqi (k ) m2 Llj (h )
∞ n i n j ∞ n i n j
i=1 j=1 , i=1 j=1 (−1)n−1, (1.2)
Q npQ Q npQ
n=1 n=1
X X
m1 m2
wherem ,m ,q ,k ,h ,l ,p(p ≥ 2)arepositiveintegers. Thequantityw := (k q )+ (h l )+
1 2 i i j j i i j j
i=1 j=1
X X
m1 m2
p is called the weight, the quantity k := q + l is called the degree.
i j
i=1 j=1
X X
∗Corresponding author. Email: [email protected]
1
Apartfromtheactualevaluation oftheseries, oneofthemainquestionsthatonesetsoutto
solve is whether or not a given series can be expressed in terms of a linear rational combination
of known constants. When this is the case, we say that the series is reducible to these values.
It has been discovered in the course of the years that many Euler sums admit expressions in-
volving finitely the zeta values, that is to say value of the Riemann zeta function,
∞
1
ζ(s):= ,ℜ(s)> 1
ns
n=1
X
with positive integer arguments. Note that the alternating Riemann zeta function is defined
respectively by
∞ (−1)n−1
ζ¯(s) = = (1−21−s)ζ(s), ℜ(s)≥ 1.
ns
n=1
X
For a pair (p,q) of positive integers with q ≥ 2, the classical linear Euler sum is defined by
∞ n
1 1
S := . (1.3)
p,q nq kp
n=1 k=1
X X
The study of these Euler sums was started by Euler. The earliest results on Euler sums are due
toEulerwhoelaborated amethodtoreducelinearsumsofsmallweighttocertain rationallinear
combinations of products of zeta values. In 1742, Goldbach proposed to Euler the problem of
expressing the S in terms of values at positive integers of the Riemann zeta function ζ(s).
p,q
Euler showed this problem in the case p = 1 and gave a general formula for odd weight p+q in
1775. Moreover, he conjectured that the double linear sums would be reducible to zeta values
whenp+q isodd,andevengavewhathehopedtoobtainthegeneralformula. In[4],D.Borwein,
J.M. Borwein and R. Girgensohn proved conjecture and formula, and in [1], D.H. Bailey, J.M.
Borwein and R. Girgensohn conjectured that the double linear sums S when p+q > 7,p+q
p,q
is even, are not reducible.
Letπ = (π ,...,π )beapartitionofintegerpandp = π +···+π withπ ≤ π ≤ ··· ≤ π .
1 k 1 k 1 2 k
The classical nonlinear Euler sum of index π,q is defined as follows (see [9])
∞
ζ (π )ζ (π )···ζ (π )
n 1 n 2 n k
S := ,
π,q nq
n=1
X
where the quantity π +···+π +q is called the weight, the quantity k is called the degree.
1 k
The relationship between the values of the Riemann zeta function and nonlinear Euler sums
has been studied by many authors, for example see [1-11,13-19]. Euler sums may be studied
through a profusion of methods: combinatorial, analytic and algebraic. Philippe Flajolet and
Bruno Salvy informed us about some ongoing work of theirs ([9]) to evaluate Euler sums in an
entirely different way, namely using contour integration and the residue theorem. In this way
they manage to prove, for example, that the cubic sums
∞ H3
S13,q = nqn (q = 2,3,4,6)
n=1
X
can beevaluated interms ofRiemann zeta values. Furthermore, theyproved thequadraticsums
∞
ζ (p )ζ (p )
n 1 n 2
S =
p1p2,q nq
n=1
X
2
can be expresses as a rational linear combination of products of linear sums and zeta values
whenever p +p +q is even, and p > 1,p > 1,q > 1. In [19], we showed that all quadratic
1 2 1 2
Euler sums of the form
∞
H ζ (m)
n n
S = (3≤ m+p ≤ 8)
1m,p np
n=1
X
can be reduced to polynomials in zeta values and linear sums.
So far, surprisingly little work has been done on parametric Euler sums. Similarly to the
definition of (1.3), the parametric linear Euler sum is defined by the series
∞
ζ (p)
n
S (a ,··· ,a ) := , (1.4)
p,q 1 p
R (a ,a ,...,a )
n 1 2 q
n=1
X
R (a ,a ,...,a ) := nq +a nq−1+···+a ,
n 1 2 q 1 q
where R (a ,a ,...,a )6= 0 is a rational function and deg(R (a ,a ,...,a ) = q ≥ 2 .
n 1 2 q n 1 2 q
Similarly to the definition of (1.2), the generalized parametric Euler sums is defined as
m1 ζqi (k ) m2 Llj (h ) m1 ζqi (k ) m2 Llj (h )
∞ n i n j ∞ n i n j
i=1 j=1 , i=1 j=1 (−1)n−1. (1.5)
QR (a ,aQ,...,a ) QR (a ,aQ,...,a )
n 1 2 q n 1 2 q
n=1 n=1
X X
In the paper, we will consider the following type of parametric linear sums involving harmonic
numbers
∞ ∞ ∞
H ζ (2m−1) ζ (2m)
n n n
, , . (1.6)
(n+a)s+1 n2s(n2−a2) n(n2−a2)
n=1 n=1 n=1
X X X
and parametric quadratic, cubic Euler sums of the form
∞ H2 ∞ H3
n , n . (1.7)
(n+b)s+1 (n+b)s+1
n=1 n=1
X X
where m,s are positive integers and a 6= ±1,±2,··· , b 6= −1,−2,··· .
We prove that the sums of (1.6) can be expressed as a rational linear combination of several
givenrationalseriesandtheparametricquadraticEulersumsof(1.7)arereducibletoparametric
linear sums. For example, we prove the identity
∞ s−2
H s 1
n
= ζ(s+1,a+1)− ζ(s−j,a+1)ζ(j +1,a+1)
(n+a)s 2 2
n=1 j=1
X X
∞ ∞
1 1
+aζ(s,a+1) + , (1.8)
n(n+a) n(n+a)s
n=1 n=1
X X
wheres ∈ N\{1}. ζ(s,a+1)andζ¯(s,a+1)standfortheHurwitzzetafunctionandalternating
Hurwitz zeta function defined by
∞ 1 ∞ (−1)n−1
ζ(s,a+1) = ,ζ¯(s,a+1) = , ℜ(s) > 1, a 6= −1,−2,··· . (1.9)
(n+a)s (n+a)s
n=1 n=1
X X
Similarly, the parametric harmonic number (also called partial sums of Hurwitz zeta function)
ζ (p,a) for p ≥ 1 is defined as
n
n
1
ζ (p,a+1)= ,a 6= −1,−2,....
n (k+a)p
k=1
X
3
2 Parametric linear Euler sums
Inthissection weconsiderthefollowingtypeofparametriclinearEulersumsinvolving harmonic
numbers by the method of contour integration
∞ ∞
ζ (2m−1) ζ (2m)
n , n ,s ∈ N .
n2s(n2−a2) n(n2−a2) 0
n=1 n=1
X X
Contour integration is a classical technique for evaluating infinite sums by reducing them to a
finite number of residue computations. This summation mechanism is formalized by a lemma
that goes back to Cauchy and is nicely developed throughout [9]. Next, we give two lemmas.
The following lemma will be useful in the development of the main theorems.
Lemma 2.1 ([9]) Let ξ(s) be a kernel function and let r(s) be a rational function which is
O(s−2) at infinity. Then
Res(r(s)ξ(s)) + Res(r(s)ξ(s)) = 0. (2.1)
s=α s=β
α∈O β∈S
X X
where S is the set of poles of r(s) and O is the set of poles of ξ(s) that are not poles r(s) . Here
Res(r(s)) denotes the residue of r(s) at s = α. The kernel function ξ(s) is meromorphic in
s=α
the whole complex plane and satisfies ξ(s) = o(s) over an infinite collection of circles |z| = ρ
k
with ρ → ∞.
k
Lemma 2.2 For integer n ∈ N, then the following relations holds
n
s(j,p) s(n+1,p+1)
= , p ∈ N, (2.2)
j! n!
j=1
X
n−1
H s(n−j,p−1) s(n+1,p+1)
j = p , p ∈ N\{1}. (2.3)
(n−j)! n!
j=1
X
where s(n,k) is called (unsigned) Stirling number of the first kind ([12]) defined by
n
x x
n!(1+x) 1+ ··· 1+ = s(n+1,k+1)xk,
2 n
(cid:16) (cid:17) (cid:16) (cid:17) Xk=0
when n < k, s(n,k)= 0; when n,k > 1, s(n,0) = s(0,k) = 0; when n= k = 0, s(0,0) = 1.
Proof. From [12], we have the generating function of (unsigned) Stirling number of the first
kind:
∞ xn
lnp+1(1−x) = (−1)p+1(p+1)! s(n,p+1) , p ∈N ,−1≤ x < 1. (2.4)
0
n!
n=p+1
X
Differentiating this equality, we obtain
lnp(1−x) ∞ xn
= (−1)pp! s(n+1,p+1) ,p ∈ N. (2.5)
1−x n!
n=p
X
4
On the other hand, using the cauchy product of power series, we have
lnp(1−x) ∞ s(n,p) ∞ ∞ n s(j,p)
= (−1)pp! xn xn−1 = (−1)pp! xn. (2.6)
1−x n!  j! 
n=p n=1 n=1 j=1
X X X X
 
Thus, comparing the coefficients of xn in (2.5) and (2.6), we can deduce (2.2). Similarly, using
the cauchy product of power series again, we get
lnp(1−x) ln(1−x)
= lnp−1(1−x)
1−x 1−x
∞ ∞
s(n,p−1)
=− H xn(−1)p−1(p−1)! xn
n
n!
n=1 n=p−1
X X
∞ n
H s(n−j+1,p−1)
=(−1)p(p−1)! j xn+1. (2.7)
 (n−j +1)! 
n=1 j=1
X X
 
Combining (2.5) and (2.7), we can obtain (2.3). The proof of Lemma 2.2 is thus completed. (cid:3)
Moreover, by the definition s(n,k), we can rewrite it as
n n
x
s(n+1,k+1)xk = n!exp ln 1+
 j 
Xk=0 Xj=1 (cid:18) (cid:19)
 n ∞ xk
= n!exp (−1)k−1
 kjk
Xj=1Xk=1 
∞ ζ (k)xk
= n!exp (−1)k−1 n  .
k
( )
k=1
X
Therefore, we know that s(n,k) is a rational linear combination of products of harmonic num-
bers. The following identities is easily derived
(n−1)!
s(n,1) = (n−1)!,S(n,2) = (n−1)!H ,S(n,3) = H2 −ζ (2) ,
n−1 2 n−1 n−1
(n−1)! (cid:2) (cid:3)
s(n,4) = H3 −3H ζ (2)+2ζ (3) ,
6 n−1 n−1 n−1 n−1
(n−1)!(cid:2) (cid:3)
s(n,5) = H4 −6ζ (4)−6H2 ζ (2)+3ζ2 (2)+8H ζ (3) .
24 n−1 n−1 n−1 n−1 n−1 n−1 n−1
(cid:2) (cid:3)
Therefore, putting p = 2 in (2.2) and (2.3), we can give the identities
n−1 H n H H2+ζ (2)
j = H2−ζ (2), k = n n . (2.8)
n−j n n k 2
j=1 k=1
X X
We make here an essential use of kernels involving the ψ function. The ψ function is the
logarithmic derivative of the Gamma function,
∞
d 1 1 1
ψ(s) = lnΓ(s)= −γ− + − (2.9)
ds s n n+s
n=1(cid:18) (cid:19)
X
5
and it satisfies the complement formula
1
ψ(s)−ψ(−s) = − −πcot(πs), (2.10)
s
as well as an expansion at s = 0 that involves the zeta values:
∞
1
ψ(−s)+γ = − ζ(k)sk−1. (2.11)
s
k=2
X
Using differentiation n times, we obtain
ψ(n)(−s) 1 ∞ n+k−1
= −(−1)n ζ(k+n)sk−1,s → 0. (2.12)
n! sn+1 k−1
k=1(cid:18) (cid:19)
X
In their paper, Euler Sums and Contour Integral Representations, Philippe Flajolet and Bruno
Salvy gave the following formulae
∞
πcot(πs)s→=n 1 −2 ζ(2k)(s−n)2k−1,
s−n
k=1
X
∞
ψ(−s)+γs→=n 1 +H + (−1)kζ (k+1)−ζ(k+1) (s−n)k, n ≥ 0
n n
s−n
Xk=1(cid:16) (cid:17)
∞
ψ(−s)+γs→=−nH + (ζ (k+1)−ζ(k+1))(s+n)k, n > 0
n−1 n−1
k=1
X
ψ(p−1)(−s)s→=n 1 1+(−1)p i−1 ζ(i)+(−1)iζ (i) (s−n)i , n≥ 0, p > 1
(p−1)! (s−n)p  p−1 n 
Xi≥p(cid:18) (cid:19)(cid:16) (cid:17)
ψ(p−1)(−s)s→=−n(−1)p  p−1+i (ζ(p+i)−ζ (p+i))(s+n)i, n > 0, p > 1.
(p−1)! p−1 n−1
i≥0(cid:18) (cid:19)
X
Table 1. Local expansions of basic kernels
Nielsen [14], elaborating on Euler’s work, proved by a method based on partial fraction
expansions that every linear sum S whose weight p+q is odd is expressible as a polynomial
p,q
in zeta values. We give explicit formula for several classes of parametric Euler sums in terms of
Riemann zeta values and rational function series. Next we evaluate the sums in (1.6).
Theorem 2.3 Let m be positive integers with a is a real and a 6= 0,−1,−2,···. Then the
following parametric linear sums are reducible to zeta values and rational function series,
∞ ∞ ∞
ζ (2m) 1 1 πcot(πa) 1 1
n
= − −
n(n2−a2) 2 n2m+1(n2−a2) 4a2 (n−a)2m (n+a)2m
n=1 n=1 n=1(cid:26) (cid:27)
X X X
m ∞
1 1 1 2
+ ζ(2k) + −
2a2 ((n+a)2m−2k+1 (n−a)2m−2k+1 n2m−2k+1)
k=1 n=1
X X
∞
m 1 1 1 2
+ ζ(2m+1)− + − . (2.13)
a2 4a2 (n+a)2m+1 (n−a)2m+1 n2m+1
n=1(cid:26) (cid:27)
X
6
Proof. The theorem results from applying the kernel function
πcot(πz)ψ(2m−1)(−z)
ξ(z)=
(2m−1)!
to the base function r(z) = z−1 z2−a2 −1. The only singularities are poles at the integers and
±a. At a negative integer −n the pole is simple and the residue is
(cid:0) (cid:1)
ζ (2m)−ζ(2m) 1
n
− .
n(n2−a2) n2m+1(n2−a2)
At a positive integer n, the pole has order 2m+1 and the residue is
ζ (2m)+ζ(2m) 1 1 1 2
n
+ + −
n(n2−a2) 2a2 (n+a)2m+1 (n−a)2m+1 n2m+1
(cid:26) (cid:27)
m
1 1 1 2
− ζ(2k) + − .
a2 ((n+a)2m−2k+1 (n−a)2m−2k+1 n2m−2k+1)
k=1
X
The residue of the pole at ±a is
∞
πcot(πa) 1 1
− .
2a2 (n−a)2m (n+a)2m
n=1(cid:26) (cid:27)
X
Finally the residue of the pole of order 2m+2 at 0 is found to be
2m
− ζ(2m+1).
a2
Summing these four contributions yields the statement of the theorem. (cid:3)
Theorem 2.4 Let m,s > 0 be integers with a is a real and a 6= 0,−1,−2,···. The following
parametric linear sums are reducible to zeta values and rational function series,
∞ ∞
ζ (2m+1) 1 1
n
=
n2s(n2−a2) 2 n2s+2m+1(n2−a2)
n=1 n=1
X X
s n
2m+2k−1 ζ(2m+2k−1)ζ(2n−2k+2)
+
2k−1 a2s−2n+2
n=1k=1(cid:18) (cid:19)
XX
∞
1 1 1
+ζ(2m+1) −
n2−a2 n2s a2s
!
n=1 (cid:18) (cid:19)
X
s+1
1 2m+2k−2 ζ(2m+2k−1)
−
2 2k−2 a2s−2k+4
k=2(cid:18) (cid:19)
X
∞
πcot(πa) 1 1 2
+ + −
4a2s+1 (n−a)2m+1 (n+a)2m+1 n2m+1
n=1(cid:26) (cid:27)
X
s
1 ζ(2m+2j +1) 2m+2j
−
2 a2s+2−2j 2j −1
j=1 (cid:18) (cid:19)
X
7
m s
ζ(2k) 2m−2k+2j
+ ζ(2m+2j −2k+1)
a2s+2−2j 2j −1
k=1j=1 (cid:18) (cid:19)
XX
m ∞
1 1 1
− ζ(2k) − , (2.14)
2a2s+1 ((n−a)2m−2k+2 (n+a)2m−2k+2)
k=0 n=1
X X
1
where the value ζ(0) = − should be used and ζ(1) should be replaced by 0 whenever it occurs.
2
Proof. Similarly to the proof of Theorem 2.3. The theorem results from applying the kernel
function
πcot(πz)ψ(2m)(−z)
ξ(z) =
(2m)!
to the base function r(z) = z−2s z2−a2 −1. Note that r(z) can be rewritten as
(cid:0) (cid:1) s
1 1 1 1 1 1
r(z) = = − − · . (2.15)
z2s(z2−a2) 2a2s+1 z−a z+a a2s+2−2j z2j
(cid:18) (cid:19) j=1
X
From (2.12), (2.15) and Table 1, we can find that, at a positive integer n, the pole has order
2m+2 and the residue is
m
ζ (2m+1)−ζ(2m+1) 1 1 1
n
+ ζ(2k) −
n2s(n2−a2) a2s+1 ((n−a)2m−2k+2 (n+a)2m−2k+2)
k=1
X
m s
ζ(2k) 2m−2k+2j 1 1 1 1
−2 − −
a2s+2−2j 2j −1 n2m+2j−2k+1 2a2s+1 (n−a)2m+2 (n+a)2m+2
k=1j=1 (cid:18) (cid:19) (cid:26) (cid:27)
XX
s
1 2m+2j 1
+ .
a2s+2−2j 2j −1 n2m+2j+1
j=1 (cid:18) (cid:19)
X
At a negative integer −n the pole is simple and residue is
ζ (2m+1)−ζ(2m+1) 1
n
− .
n2s(n2−a2) n2s+2m+1(n2−a2)
The residue of the pole at ±a is
∞
πcot(πa) 1 1
− + .
2a2s+1 (n−a)2m+1 (n+a)2m+1
n=1(cid:26) (cid:27)
X
Finally the residue of the pole of order 2m+2s+2 at 0 is found to be
s+1 s n
2m+2k−2 ζ(2m+2k−1) 2m+2k−1 ζ(2m+2k−1)ζ(2n−2k+2)
−2 .
2k−2 a2s−2k+4 2k−1 a2s−2n+2
k=1(cid:18) (cid:19) n=1k=1(cid:18) (cid:19)
X XX
Summing these four contributions yields the statement of the theorem. (cid:3)
Taking s = 0 in Theorem 2.4, we have the following Corollary.
8
Corollary 2.5 ([3]) For integers m ∈ N with a is a real and a 6= 0,−1,−2,···, we have
0
∞ ∞
ζ (2m+1) 1 1
n
=
n2−a2 2 n2m+1(n2−a2)
n=1 n=1
X X
m ∞
1 1 1
− ζ(2k) −
2a ((n−a)2m−2k+2 (n+a)2m−2k+2)
k=0 n=1
X X
∞
πcot(πa) 1 1 2
+ + − . (2.16)
4a (n−a)2m+1 (n+a)2m+1 n2m+1
n=1(cid:26) (cid:27)
X
Note that formula above was also proved in [3] by another method.
NextweconsiderthefollowingtypeofparametricEulerSumsbythemethodofconstructing
function T (x,y)
s,t
∞ ∞
H L (1)
n n
, , a 6= −1,−2,··· .
(n+a)s (n+a)s
n=1 n=1
X X
Theorem 2.6 If a is a real number, s is a positive integer and a 6= −1,−2,···, then
n−1xn−j n−1yn−j
yn +xn
∞ j j s
j=1 j=1
X X =sLi (a,xy)− Li (a,x)Li (a,y)
(n+a)s s+1 j s+1−j
n=1 j=1
X X
+Li (a,xy)(Li (x)+Li (y)). (2.17)
s 1 1
where x,y ∈ [−1,1) and the parametric polylogarithm function Li (a,x) is defined by
s
∞ xn
Li (a,x) = , ℜ(s)≥ 1,−1 ≤ x < 1.
s (n+a)s
n=1
X
If a = 0, then the function Li (a,x) reduces to the classical polylogarithm function Li (x) which
s s
is defined by
∞ xn
Li (x) = , ℜ(s) ≥ 1,−1 ≤ x < 1,
s ns
n=1
X
with Li (x) = −ln(1−x).
1
Proof. Motivated by [2,7], for real −1≤ x,y < 1 and integers s and t, we consider the function
∞ xnym ∞ xnym(m−n+n+a)
T (x,y)= =
s,t (n+a)s(m+a)t(m−n) (n+a)s(m+a)t+1(m−n)
m,n=1 m,n=1
X X
m6=n m6=n
∞ xnym ∞ xnym
= +
(n+a)s(m+a)t+1 (n+a)s−1(m+a)t+1(m−n)
m,n=1 m,n=1
X X
m6=n m6=n
∞ xn ∞ ym yn
= − +T (x,y)
(n+a)s ( (m+a)t+1 (n+a)t+1) s−1,t+1
n=1 m=1
X X
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= Li (a,x)Li (a,y)−Li (a,xy)+T (x,y).
s t+1 s+t+1 s−1,t+1
Telescoping this gives
s
T (x,y) = T (x,y)−sLi (a,xy)+ Li (a,x)Li (a,y), s ∈N .
s,t 0,s+t s+t+1 j s+t+1−j 0
j=1
X
With t = 0, this becomes
s
T (x,y) = T (x,y)−sLi (a,xy)+ Li (a,x)Li (a,y), s∈ N. (2.18)
s,0 0,s s+1 j s+1−j
j=1
X
But for any integers s and t, there holds
∞ xnym
T (x,y) =
t,s (n+a)t(m+a)s(m−n)
m,n=1
X
m6=n
∞ ymxn
= −
(m+a)s(n+a)t(n−m)
m,n=1
X
m6=n
= −T (y,x). (2.19)
s,t
From the definition of T (x,y), we can deduce that
s,t
∞ xnym ∞ xnyn ∞ ym−n ∞ xn n−1 ym
T (x,y) = = −
s,0 (n+a)s(m−n) (n+a)s m−n (n+a)s n−m
m,n=1 n=1 m=n+1 n=1 m=1
X X X X X
m6=n
∞ xn n−1yn−j
= Li (a,xy)Li (y)− . (2.20)
s 1 (n+a)s j
n=1 j=1
X X
Substituting (2.19), (2.20) into (2.18) yields the desired result. (cid:3)
Setting x = y in (2.17), by simple calculation, we obtain the result
∞ xn n−1xn−j s
= Li a,x2 +Li a,x2 Li (x)−Li (a,x)Li (a,x)
(n+a)s j 2 s+1 s 1 s 1
n=1 j=1
X X (cid:0) (cid:1) (cid:0) (cid:1)
s−1
1
− Li (a,x)Li (a,x), (2.21)
j s+1−j
2
j=2
X
where in (2.21), we now require s ∈ N\{1} because the terms j = 1 and j = s were separated,
and assumed to be distinct. Taking x = −1 in (2.21), we get
∞ s−2
L (1) 1 s
n = ζ¯(s−j,a+1)ζ¯(j +1,a+1)− ζ(s+1,a+1)
(n+a)s 2 2
n=1 j=1
X X
∞ (−1)n−1
+ζ(s,a+1)ln2+ζ¯(s,a+1)ζ¯(1,a+1)+ . (2.22)
n(n+a)s
n=1
X
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