Polylogarithmic connections with Euler sums

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 fraction approach, binomial coefficients.

Copyright c 2016 by ANUBIH.

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

z^m

m^t, t∈Cwhen |z|<1; R(t)>1 when |z|= 1.

Moreover Z 1

0

Li_t(px) x dx=

(ζ(1 +t), forp= 1 (2^−r−1)ζ(1 +t), forp=−1. We also obtain identities for integrals of the typeR1

0 ln²x

Li2(−x)+x−^x2

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 ^λ_µ

(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

H_n=

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:

H_n^(m):=

n

X

r=1

1

r^m , (m, n∈N) and H₀^(m):= 0 (m∈N),

in terms of integral representations we have the result H_n^(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 H_ρ^(m+1) may be defined, in terms of the Polygamma functions

ψ⁽ⁿ⁾(z) := dⁿ

dzⁿ{ψ(z)}= dⁿ⁺¹

dzⁿ⁺¹{log Γ(z)} (n∈N0), by

H_ρ^(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 defineH_ρ⁽¹⁾ by

H_ρ⁽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= ^r_q, 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= ^r_q,and we also define

H⁽¹⁾r

q =γ+ψ

r q + 1

, and H₀^(α)= 0.

The evaluation of the polygamma functionψ^{(α) r}_a

at rational values of the argument can be explicitly done via a formula as given by K¨olbig [9], or Choi and Cvijovic [3] in terms of the Polylogarithmic or other special functions.

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

n^z = 1−2^1−z ζ(z)

with

−

ζ(1) = ln 2,

S_p,q⁺⁻=

∞

X

n=1

(−1)ⁿ⁺¹ Hn^(p)

n^q .

For first order powers (p= 1),of harmonic numbers, Sitaramachandra Rao [12] gave, for 1 +q an odd integer,

2S_1,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)

n^q = ζ(p)ζ(q) + (−1)^p (p−1)!

Z 1 0

ln^p−1(t)L_iq(t) 1−t dt

= ζ(p+q)− (−1)^q (q−1)!

Z 1 0

ln^q−1(t)L_ip(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 j^p = 1

2^p

H^(p)

[^r2]+H^(p) [^r−12 ]

−H^(p)

2[^r+12 ]⁻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⁺¹H_n

n+ 1 (1.6)

and when t= 1, ln²2 = 2

∞

X

n=1

(−1)ⁿ⁺¹H_n

n+ 1 =ζ(2)−2L_i2 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⁺¹H_n⁽³⁾ n+ 1 = 3

4ln 2ζ(3)− 5 16ζ(4)

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−12 ]

ζ(3) +(−1)^r+1

2

H⁽²⁾

2[^r₂]⁻¹ −1 4

H⁽²⁾

[^r−12 ]+H⁽²⁾ [^r−22 ]

ζ(2) + (−1)^r

H_r−1⁽³⁾ +H⁽³⁾

2[^r₂]⁻¹− 1 8

H⁽³⁾

[^r−12 ]+H⁽³⁾ [^r−22 ]

ln 2 + (−1)^r+1

r−1

X

j=1

1 j³

H_j−H[₂^j]

, (1.11)

with M(0) andM(1) given by (1.8).

Proof. By a change of counter M(r) : =

∞

X

n=1

(−1)ⁿ⁺¹Hn⁽³⁾

n+r =

∞

X

n=1

(−1)ⁿH_n−1⁽³⁾ n+r−1

=

∞

X

n=1

(−1)ⁿ n+r−1

H_n⁽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−12 ]

. (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−12 ]−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⁺¹H_n⁽³⁾ (n+r)²

then

X(r) = (−1)^r+1X(1) +3 (−1)^r+1 4

1 4

H⁽²⁾

[^r−12 ]+H⁽²⁾ [^r−22 ]

−H⁽²⁾

2[^r₂]⁻¹

ζ(3) + (−1)^r

1 8

H⁽³⁾

[^r−12 ]+H⁽³⁾ [^r−22 ]

−H⁽³⁾

2[^r₂]⁻¹

ζ(2) + 3 (−1)^r

H_r−1⁽⁴⁾ − 1 16

H⁽⁴⁾

[^r−12 ]+H⁽⁴⁾ [^r−22 ]

+H⁽⁴⁾

2[^r₂]⁻¹

ln 2 + (−1)^r

r−1

X

j=1

3 j⁴

H[^j₂]−Hj

+ 1

4j³

H⁽²⁾

[^j2]−H⁽²⁾ [^j+12 ]⁻¹2

,

(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)ⁿ

Hn⁽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−1₂ ]−H⁽²⁾ [^r₂]⁻¹₂

+ 3 (−1)^r 4 (r−1)⁴

ln 2 +H[^r−12 ]−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−1₂ ]−Hr−1

+ (−1)^r 4 (r−1)³

H⁽²⁾

[^r−12 ]−H⁽²⁾ [^r2]⁻¹2

, (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

X(r), X(r−1), X(r−1), . . . , X(1)

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)

H^r

2 −H^r−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







 H^r

2 −H^r−1

2

2 −

x ₂F₁

1,1 +r 2 +r

−x

1 +r







 dx

= ζ(3) 2

H^r

2 −H^r−1

2

− 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

−2X(r) (1.16)

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⁾

n^{p 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 H_n⁽³⁾

n+k k

=

k

X

r=1

(−1)^1+rr k

r

M(r). (2.2)

V (k,1,1) =

∞

X

n=1

(−1)ⁿ⁺¹ Hn⁽³⁾

n ^n+k_k

= 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⁾

n^{2 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 k}_r M(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!H_n⁽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!H_n⁽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.

The other case of V (k, p,2), can be evaluated in a similar fashion. We list the result in the next theorem.

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+k_k 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 H_n⁽³⁾ n^{2 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) + ²_rM(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 H_n⁽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 H_n⁽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

−2 (1 +k)V (k,2,1) (2.12)

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

, then by re-arrangement we obtain the integral identity (2.10). The identities (2.11), (2.12), (2.13), (2.14) and (2.15) follow in a similar way.

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−^x₄²

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

anthony.sofo@vu.edu.au