Vol.12 (24), No.1, (2016), 17–32
POLYLOGARITHMIC CONNECTIONS WITH EULER SUMS
ANTHONY SOFO
Abstract. Polylogarithmic functions are intrinsically connected with sums of harmonic numbers. In this paper we explore many relations and explicitly derive closed form representations of integrals of polylogarith-mic functions and the Lerch phi transcendent in terms of zeta functions and sums of alternating harmonic numbers.
1. Introduction and preliminaries
In this paper we will develop identities, new families of closed form rep-resentations of alternating harmonic numbers and reciprocal binomial coef-ficients, including integral representations, of the form:
Z 1 0
xln2(x) 1−x 3F2
1,1,2 2 +k,2 +k
−x
dx (1.1)
fork ≥1 and where 3F2
·,·,· ·,·
z
is the classical generalized
hypergeo-metric function, also for integrals of the form
Z 1 0
xln2(x)
1−x Φ (−x,2,1 +r) dx
where Φ (z, t, a) =P∞ m=0 z
m
(m+a)t is the Lerch transcendent defined for|z|<1
and R(a)>0 and satisfies the recurrence
Φ (z, t, a) =z Φ (z, t, a+ 1) +a−t.
2010Mathematics Subject Classification. Primary: 05A10, 05A19, 33C20; Secondary: 11B65, 11B83, 11M06.
Key words and phrases. Polylogarithm function, integral representation, Lerch tran-scendent function, alternating harmonic numbers, combinatorial series identities, summa-tion formulas, partial fracsumma-tion approach, binomial coefficients.
The Lerch transcendent generalizes the Hurwitz zeta function at z= 1,
Φ (1, t, a) = ∞
X m=0
1 (m+a)t
and the Polylogarithm, or de Jonqui`ere’s function, when a= 1,
Lit(z) := ∞
X m=1
zm
mt, t∈Cwhen |z|<1; R(t)>1 when |z|= 1.
Moreover
Z 1 0
Lit(px) x dx=
(
ζ(1 +t), forp= 1 (2−r−1)ζ(1 +t), forp=−1.
We also obtain identities for integrals of the typeR01 ln 2x Li
2(−x)+x−x 2 4
x3(1−x) dx. LetRand Cdenote, respectively the sets of real and complex numbers and
let N:= {1,2,3, . . .} be the set of positive integers, andN0 :=N∪ {0}. A
generalized binomial coefficient λµ
(λ, µ ∈ C) is defined, in terms of the familiar gamma function, by
λ µ
:= Γ (λ+ 1)
Γ (µ+ 1) Γ (λ−µ+ 1), (λ, µ∈C),
which, in the special case whenµ=n, n∈N0,yields
λ
0
:= 1 and
λ n
:= λ(λ−1)· · ·(λ−n+ 1)
n! =
(−1)n(−λ)n
n! (n∈N),
where (λ)ν (λ, ν∈C) is the Pochhammer symbol. Let
Hn= n X r=1
1
r =γ+ψ(n+ 1), (H0:= 0) (1.2)
be thenthharmonic number. Here, as usual,γdenotes the Euler-Mascheroni constant and ψ(z) is the Psi (or Digamma) function defined by
ψ(z) := d
dz{log Γ(z)}=
Γ0(z)
Γ(z) or log Γ(z) =
Z z 1
ψ(t)dt.
A generalized harmonic number Hn(m) of order m is defined, for positive
integers nand m, as follows:
Hn(m):=
n X r=1
1
rm , (m, n∈N) and H (m)
in terms of integral representations we have the result
Hn(m+1) = (−1)
m
m!
Z 1 0
(lnx)m (1−xn)
1−x dx. (1.3)
In the case ofnon-integer values ofnsuch as (for example) a valueρ∈R, the generalized harmonic numbers Hρ(m+1) may be defined, in terms of the
Polygamma functions
ψ(n)(z) := d
n
dzn{ψ(z)}= dn+1
dzn+1{log Γ(z)} (n∈N0),
by
Hρ(m+1)=ζ(m+ 1) + (−1)
m
m! ψ
(m)(ρ+ 1) (1.4)
(ρ∈R\ {−1,−2,−3, . . .}; m∈N),
whereζ(z) is theRiemann zeta function. Whenever we encounter harmonic numbers of the formHρ(m)at admissible real values ofρ, they may be
evalu-ated by means of this known relation (1.4). In the exceptional case of (1.4) when m= 0, we may defineHρ(1) by
Hρ(1)=Hρ=γ+ψ(ρ+ 1) (ρ∈R\ {−1,−2,−3, . . .}).
We assume (as above) that
H0(m)= 0 (m∈N).
In the case of non integer values of the argument z= rq, we may write the
generalized harmonic numbers,Hz(α+1), in terms of polygamma functions H(α+1)r
q
=ζ(α+ 1) +(−1)
α
α! ψ
(α)
r q + 1
, r
q 6={−1,−2,−3, ...},
where ζ(z) is the zeta function. When we encounter harmonic numbers at possible rational values of the argument, of the formH(α)r
q they maybe
eval-uated by an available relation in terms of the polygamma functionψ(α)(z) or, for rational argumentsz= rq,and we also define
H(1)r
q =γ+ψ
r q + 1
, and H0(α)= 0.
Some specific values are listed in the books[16], [20] and [21]. Let us define the alternating zeta function
−
ζ(z) = ∞
X n=1
(−1)n+1
nz = 1−2 1−z
ζ(z)
with −
ζ(1) = ln 2,
Sp,q+−= ∞
X n=1
(−1)n+1 Hn(p) nq .
For first order powers (p= 1),of harmonic numbers, Sitaramachandra Rao [12] gave, for 1 +q an odd integer,
2S1,q+−= (1 +q) −
ζ(1 +q)−ζ(1 +q)−2
q
2−1
X j=1
−
ζ(2j) ζ(1 +q−2j).
For the positive terms, [1] gave
∞
X n=1
Hn(p)
nq = ζ(p)ζ(q) +
(−1)p (p−1)!
Z 1 0
lnp−1(t)Liq(t)
1−t dt
= ζ(p+q)− (−1)
q
(q−1)!
Z 1 0
lnq−1(t)Lip(t)
1−t dt, by symmetry.
Some results for sums of alternating harmonic numbers may be seen in the works of [2], [4], [5], [6], [7], [8], [10], [11], [13], [14], [15], [17], [18], [22], [23], [24] and [25] and references therein.
The following lemma will be useful in the development of the main theo-rems.
Lemma 1. Let r be a positive integer and p∈N.Then:
r X j=1
(−1)j
jp =
1 2p
H(p)
[r
2]
+H(p)
[r−1 2 ]
−H(p) 2[r+1
2 ]−1
(1.5)
where [x] is the integer part of x.
Proof. The proof is given in the paper [19].
Lemma 2. The following identities hold. For 0< t≤1
tln2
1 +t t
= 2 ∞
X n=1
(−t)n+1Hn
and when t= 1,
ln22 = 2 ∞
X n=1
(−1)n+1Hn
n+ 1 =ζ(2)−2Li2
1 2
=:S1.
tln
1 +t t
= ∞
X n=1
(−t)n+1
n , hence
ln 2 = ∞
X n=1
(−1)n+1
n =
∞
X n=1
1
n2n =
1 2
∞
X n=1
Hn
2n. (1.7)
Also
M(0) := ∞
X n=1
(−1)n+1Hn(3)
n =
19
16ζ(4)− 3
4ln 2ζ(3), (1.8)
M(1) := ∞
X n=1
(−1)n+1Hn(3) n+ 1 =
3
4ln 2ζ(3)− 5 16ζ(4)
and
X(0) := ∞
X n=1
(−1)n+1Hn(3) n2 =
3
4ζ(2)ζ(3)− 21
32ζ(5), (1.9)
X(1) := ∞
X n=1
(−1)n+1Hn(3)
(n+ 1)2 = 51
32ζ(5)− 3
4ζ(2)ζ(3).
Proof. Firstly (1.6) and (1.7) are standard known results. Next from the
definition (1.3),
M(0) = ∞
X n=1
(−1)n+1Hn(3)
n =
1 2
Z 1 0
ln2x
1−x
∞
X n=1
(−1)n+1(1−xn)
n dx
= 1 2
Z 1 0
ln2x
1−x(ln 2−ln (1 +x))dx
= 19 16ζ(4)−
3
4ln 2ζ(3). Here we have used the integral result
Z 1 0
ln2x ln (1 +x) 1−x dx=
7
2ln 2ζ(3)− 19
8 ζ(4).
M(1) = ∞
X n=1
(−1)n+1Hn(3) n+ 1 =
3
follows by a change of counter, also by the integral expression we deduce
Z 1 0
ln2x Li2(−x)
x(1−x) dx= 1
2ζ(2)ζ(3)− 51
16ζ(5), (1.10)
and hence, (1.8) follows. In a similar fashion
X(0) = ∞
X n=1
(−1)n+1Hn(3) n2 =
1 2
Z 1 0
ln2x
1−x
∞
X n=1
(−1)n+1(1−xn)
n2 dx
= 1 2
Z 1 0
ln2x
1−x
ζ(2)
2 −Li2(−x)
dx
= 1
2ζ(2)ζ(3)− 1 2
Z 1 0
ln2x Li2(−x) 1−x dx=
3
4ζ(2)ζ(3)− 21 32ζ(5).
X(1) = ∞
X n=1
(−1)n+1Hn(3)
(n+ 1)2 = 51
32ζ(5)− 3
4ζ(2)ζ(3)
follows by a change of counter, or by the integral expression (1.10).
Lemma 3. Let r≥2 be a positive integer, defining
M(r) := ∞
X n=1
(−1)n+1Hn(3) n+r
then
M(r) = (−1)r+1M(1) + 3 (−1)
r
4
Hr−1−H
[r−1 2 ]
ζ(3)
+(−1)
r+1
2
H(2)
2[r
2]−1
−1
4
H(2)
[r−1 2 ]
+H(2)
[r−2 2 ]
ζ(2)
+ (−1)r
Hr(3)−1+H(3)
2[r2]−1
− 1
8
H(3)
[r−1 2 ]
+H(3)
[r−2 2 ]
ln 2
+ (−1)r+1
r−1 X j=1
1
j3
Hj−H[j
2]
, (1.11)
Proof. By a change of counter
M(r) : = ∞
X n=1
(−1)n+1Hn(3) n+r =
∞
X n=1
(−1)nHn(3)−1 n+r−1
= ∞
X n=1
(−1)n
n+r−1
Hn(3)− 1
n3
=−
∞
X n=1
(−1)n+1Hn(3) n+r−1 +
∞
X n=1
(−1)n+1
n3(n+r−1)
=−M(r−1) + 1
r−1 ∞
X n=1
(−1)n+1 n3 −
1
(r−1)2 ∞
X n=1
(−1)n+1 n2
+ 1
(r−1)3 ∞
X n=1
(−1)n+1
n −
1
(r−1)3 ∞
X n=1
(−1)n+1
n+r−1.
From Lemma 1 and using the known results,
M(r) =−M(r−1) + 3ζ(3) 4 (r−1)−
ζ(2) 2 (r−1)2 +
ln 2 (r−1)3
− (−1)
r
(r−1)3
ln 2 +Hr−1−H
[r−1 2 ]
. (1.12)
From (1.12) we have the recurrence relation
M(r) +M(r−1) = 3ζ(3) 4 (r−1)−
ζ(2) 2 (r−1)2 +
1 + (−1)r (r−1)3
ln 2
− (−1)
r
(r−1)3
H
[r−1 2 ]
−Hr−1
forr ≥2, and withM(0) andM(1) given by (1.8). The recurrence relation is solved by the subsequent reduction of the
M(r), M(r−1), M(r−2), . . . , M(1)
terms, finally arriving at the relation (1.11).
Lemma 4. For a positive integer r≥2, we have the identity
X(r) := ∞
X n=1
(−1)n+1Hn(3)
then
X(r) = (−1)r+1X(1) +3 (−1)
r+1 4 1 4 H(2)
[r−1 2 ]
+H(2)
[r−2 2 ]
−H(2)
2[r2]−1
ζ(3)
+ (−1)r
1 8
H(3)
[r−1 2 ]
+H(3)
[r−2 2 ]
−H(3)
2[r
2]−1
ζ(2)
+ 3 (−1)r
Hr(4)−1− 1
16
H(4)
[r−1 2 ]
+H(4)
[r−2 2 ]
+H(4)
2[r
2]−1
ln 2
+ (−1)r
r−1 X j=1 3 j4
H[j
2]
−Hj+ 1
4j3
H(2)
[j
2]
−H(2)
[j+1 2 ]−
1 2
,
(1.13)
with X(0) and X(1) given by (1.9).
Proof. By expansion, and a change of counter
X(r) := ∞
X n=1
(−1)n+1Hn(3)
(n+r)2 = ∞
X n=1
(−1)nHn(3)−n13
(n+r−1)2 , then,
X(r) =−
∞
X n=1
(−1)n+1Hn(3)
(n+r−1)2 + ∞
X n=1
(−1)n+1
n3(n+r−1)2
=−X(r−1) + ∞
X n=1
(−1)n+1
1 (r−1)2n3 −
2 (r−1)3n2 +
3 (r−1)4n
− 1
(r−1)3(n+r−1)2 −
3 (r−1)4(n+r−1)
X(r) =−X(r−1) + 3ζ(3) 4 (r−1)2 −
ζ(2)
(r−1)3 +
3 ln 2
(r−1)4 + (−1)
r
4 (r−1)3
H(2)
[r−1 2 ]
−H(2)
[r
2]− 1 2
+ 3 (−1)
r
4 (r−1)4
ln 2 +H[r−1 2 ]
−Hr−1
.
Hence we obtain the recurrence relation
X(r) +X(r−1) = 3ζ(3) 4 (r−1)2 −
ζ(2) (r−1)3 +
3 (1 + (−1)r) ln 2 (r−1)4 + 3 (−1)
r
4 (r−1)4
H[r−1
2 ]
−Hr−1
+ (−1)
r
4 (r−1)3
H(2)
[r−1 2 ]
−H(2)
[r
2]− 1 2
, (1.14)
for r ≥ 2, and with X(0) and (1) given by (1.9). The recurrence relation (1.14) is solved by the subsequent reduction of the
terms, finally arriving at the relation (1.13).
We know develop some integral identities for Lemma 3 and Lemma 4, namely:
Lemma 5. Forr ∈N, Z 1
0
xln2(x) 1−x 2F1
1,1 +r
2 +r
−x
dx
= (1 +r)
Hr
2 −Hr −1 2
ζ(3)−2 (1 +r)M(r) (1.15)
where M(r) is given by (1.11).
Proof. FromM(r) := P∞
n=1
(−1)n+1Hn(3)
n+r and by the use of the integral
rep-resentation (1.3)
M(r) = 1 2
Z 1 0
ln2(x) 1−x
∞
X n=1
(−1)n+1(1−xn)
n+r dx
= 1 2
Z 1 0
ln2(x) 1−x
Hr
2 −Hr−12
2 −
x 2F1
1,1 +r
2 +r
−x
1 +r
dx
= ζ(3) 2
Hr
2 −Hr−12
− 1
2 (1 +r)
Z 1 0
xln2(x) 1−x 2
F1
1,1 +r
2 +r
−x
dx
and by re-arrangement we obtain (1.15).
Remark 1. For the caser= 1,we have
Z 1 0
ln2x ln (1 +x)
x(1−x) dx= 7
2ln 2ζ(3)− 5 8ζ(4)
which is a modification of the integral used in Lemma 2. Forr= 3 we obtain the very slow converging integral
Z 1 0
ln2x 2x−x2−2 ln (1 +x) x3(1−x) dx
= 5
4ζ(4)−7 ln 2ζ(3) + 7 2ζ(3)−
3
2ζ(2) + 8 ln 2− 1 4. The Wolfram on-line integrator yields no solution to this integral.
Lemma 6. Forr ∈N, Z 1
0
xln2(x)
1−x Φ (−x,2,1 +r) dx= ζ(3)
2
H(2)r
2
−H(2)r−1 2
whereX(r)is given by (1.13) andΦ (−x,2,1 +r)is the Lerch transcendent.
Proof. FromX(r) :=P∞
n=1
(−1)n+1Hn(3)
(n+r)2 and by the use of the integral
rep-resentation (1.3)
X(r) = 1 2
Z 1 0
ln2(x) 1−x
∞
X n=1
(−1)n+1(1−xn) (n+r)2 dx
= 1 2
Z 1 0
ln2(x) 1−x
H(2)r
2
−H(2)r−1 2
4 −x Φ (−x,2,1 +r)
dx
= ζ(3) 4
H(2)r
2
−H(2)r−1 2
−1
2
Z 1 0
xln2(x)
1−x Φ (−x,2,1 +r) dx
and by re-arrangement we obtain (1.16).
Remark 2. For the caser= 1,we have
Z 1 0
ln2x Li2(−x)
x(1−x) dx= 1
2ζ(2)ζ(3)− 51
16ζ(5) (1.17)
which is a modification of the integral used in Lemma 2. The Wolfram on-line integrator yields no solution to this integral.
The next few theorems relate the main results of this investigation, namely the integral and closed form representation of integrals of the type (1.1).
2. Integral and closed form identities
In this section we investigate integral identities in terms of closed form representations of infinite series of harmonic numbers of order three and in-verse binomial coefficients. First we indicate the closed form representation of
V (k, p, q) = ∞
X n=1
(−1)n+1 Hn(3) np n+k
k
q (2.1)
for six cases (k, p, q) = (k,0,1),(k,1,1),(k,2,1),(k,0,2),(k,1,2),(k,2,2),
and k≥1 is a positive integer.
Theorem 1. Let k ≥ 1 be real positive integer, then from (2.1) with p =
0,1,2 and q= 1 we have:
V (k,0,1) = ∞
X n=1
(−1)n+1 Hn(3) n+k
k
= k X r=1
(−1)1+rr
k r
V (k,1,1) = ∞
X n=1
(−1)n+1 Hn(3) n n+kk
= 19
16ζ(4)− 3
4ln (2)ζ(3)−
k X r=1
(−1)1+r
k r
M(r), (2.3)
and
V (k,2,1) = ∞
X n=1
(−1)n+1 Hn(3) n2 n+k
k
=
3ζ(2)ζ(3)
4 −
21ζ(5) 32
−Hk
19ζ(4) 16 −
3 ln 2ζ(3) 4
−
k X r=1
(−1)r krM(r)
r (2.4)
where M(r) is given by (1.11).
Proof. Consider the expansion
V (k,0,1) = ∞
X n=1
(−1)n+1 Hn(3) n+k
k
=
∞
X n=1
(−1)n+1k!Hn(3)
(n+ 1)k
= ∞
X n=1
(−1)n+1k!Hn(3) k X r=1
Θ (r)
n+r
where
Θ (r) = lim
n→−r (
n+r k Q r=1
(n+r)
)
= (−1)
1+r r k!
k r
. (2.5)
We can now express
V (k,0,1) = ∞
X n=1
(−1)n+1k!Hn(3) k X r=1
Θ (r)
n+r
=
k X r=1
(−1)1+r r
k r
k X r=1
Hn(3)
n+r. (2.6)
From (1.11) we haveM(r),hence substituting into (2.6), (2.2) follows. The identities (2.3) and (2.4) follow in a similar fashion.
Theorem 2. Under the assumptions of Theorem 1, we have,
V (k,0,2) = ∞
X n=1
(−1)n+1 Hn(3) n+k
k 2
=
k X r=1
r2
k r
2
(M(r) + 2 (Hr−1−Hk−r)X(r)), (2.7)
V (k,1,2) = ∞
X n=1
(−1)n+1 Hn(3) n n+kk 2
= k 2
2k k
19ζ(4) 16 −
3 ln 2ζ(3) 4
−
k X r=1 r
k r
2 M(r)
+ 2
k X r=1
k r
2
(Hr−1−Hk−r)
19ζ(4) 16 −
3 ln 2ζ(3)
4 −M(r)−rX(r)
(2.8)
and
V (k,2,2) = ∞
X n=1
(−1)n+1 Hn(3) n2 n+k
K 2
= k 2
2k k
3
4ζ(2)ζ(3)− 21 32ζ(5)
−
2k k
−1 19ζ(4)
16 −
3 ln 2ζ(3) 4
+
k X r=1
k r
2 M(r)
+ 2
k X r=1
k r
2
(Hr−1−Hk−r)
3
4ζ(2)ζ(3)− 2 r
19ζ(4)
16 −
3 ln 2ζ(3) 4
−2132ζ(5) + 2rM(r) +X(r)
(2.9)
where M(r) is given by (1.11) andX(r) is given by (1.13).
Proof. Consider the expansion
V (k,0,2) = ∞
X n=1
(−1)n+1 Hn(3) n+k
k
2 =
∞
X n=1
(−1)n+1(k!)2 Hn(3)
(n+ 1)1+k2 =
∞
X n=1
(−1)n+1(k!)2 Hn(3) k X r=1
Ω (r)
n+r +
∆ (r)
(n+r)2
where
Ω (r) = lim
n→−r (
(n+r)2
k Q r=1
(n+r)2
)
=
r k!
k r
2
and
∆ (r) = lim
n→−r d dn
(
(n+r)2
k Q r=1
(n+r)2
)
= 2
r k!
k r
2
(Hr−1−Hk−r).
We can now express
V (k,0,2) = ∞
X n=1
(−1)n+1(k!)2 Hn(3) k X r=1
Ω (r)
n+r +
∆ (r)
(n+r)2
and by substitution (2.7) follows. Similarly (2.8) and (2.9) follow using the
same technique.
The following integral identities can be exactly evaluated by using the alternating harmonic number sums in Theorems1 and 2. The following six integral identities are the main results of this paper.
Theorem 3. Let k be a positive integer, then we have:
Z 1 0
xln2(x) 2F1
1,2 2 +k
−x
1−x dx
= 2ζ(3) 2F1
1,2 2 +k
−1
−2 (1 +k)V (k,0,1) (2.10)
where V (k,0,1) is given by (2.2),
Z 1 0
xln2(x) 2F1
1,1 2 +k
−x
1−x dx
= 2ζ(3) 2F1
1,1 2 +k
−1
−2 (1 +k)V (k,1,1), (2.11)
where V (k,1,1) is given by (2.3),
Z 1 0
xln2(x) 3F2
1,1,1 2,2 +k
−x
1−x dx
= 2ζ(3)3F2
1,1,1 2,2 +k
−1
where V (k,2,1) is given by (2.4),
Z 1 0
xln2(x) 3F2
1,2,2 2 +k,2 +k
−x
1−x dx
= 2ζ(3) 3F2
1,2,2 2 +k,2 +k
−1
−2 (1 +k)2V (k,0,2) (2.13)
where V (k,0,2) is given by (2.7),
Z 1 0
xln2(x) 3F2
1,1,2 2 +k,2 +k
−x
1−x dx
= 2ζ(3) 3F2
1,1,2 2 +k,2 +k
−1
−2 (1 +k)2V (k,1,2) (2.14)
where V (k,1,2) is given by (2.8),
Z 1 0
xln2(x) 3F2
1,1,1 2 +k,2 +k
−x
1−x dx
= 2ζ(3) 3F2
1,1,1 2 +k,2 +k
−1
−2 (1 +k)2V (k,2,2) (2.15)
where V (k,2,2) is given by (2.9).
Proof. From the identity (1.3), we can write
∞
X n=1
(−1)n+1 Hn(3) n+k k = 1 2 Z 1 0
ln2x
1−x
∞
X n=1
(−1)n+1(1−xn)
n+k k
dx
= 1
2 (1 +k)
Z 1 0
ln2x
1−x
2F1
1,2 2 +k
−1
−x2F1
1,2 2 +k
−x dx. where2F1 ·,· · ·
is the generalized hypergeometric function. Since
1 2 (1 +k)
Z 1 0
ln2x
1−x 2F1
1,2 2 +k
−1
dx= ζ(3) 2 (1 +k) 2F1
1,2 2 +k
−1 ,
Remark 3. From Lemma 6 and Theorem 3 both the Lerch transcendent and the hypergeometric function introduce the polylogarithmic function
Li2(−x), from which we are able to evaluate their integrals. The follow-ing examples are given. From (1.16) with r = 1, we obtain (1.17). For
r= 2,
Z 1 0
ln2x (Li2(−x) +x) x2(1−x) dx=
51
16ζ(5)− 1
2ζ(2)ζ(3)−3ζ(3)
+ζ(2)−12 ln 2 + 8.
Forr = 3,
Z 1 0
ln2x Li2(−x) +x−x
2
4
x3(1−x) dx=
51
16ζ(5)− 1
2ζ(2)ζ(3) + 3 8ζ(3)
+5
8ζ(2)−12 ln 2 + 67
8 , and for r= 4
Z 1 0
ln2x
Li2(−x) +x−x42 +x93
x4(1−x) dx=−
51 16ζ(5) +
1
2ζ(2)ζ(3)
+217 72 ζ(3)−
143 216ζ(2) +
328 27 ln 2−
16525 1944 . The Wolfram on-line integrator, yields no solution to these slow converging integrals.
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(Received: March 22, 2015) Victoria University (Revised: June 22, 2015) P. O. Box 14428
Melbourne City Victoria 8001 Australia
