# Evaluation of integrals with hypergeometric and logarithmic functions

Anthony Sofo
From the journal Open Mathematics

# Abstract

We provide an explicit analytical representation for a number of logarithmic integrals in terms of the Lerch transcendent function and other special functions. The integrals in question will be associated with both alternating harmonic numbers and harmonic numbers with positive terms. A few examples of integrals will be given an identity in terms of some special functions including the Riemann zeta function. In general none of these integrals can be solved by any currently available mathematical package.

MSC 2010: 05A10; 05A19; 33C20; 11B65; 11B83; 11M06

## 1 Introduction and Preliminaries

In this paper we will develop explicit analytical representations, identities, new families of integral representations, of the form:

01x2k1lnx3F212,2,23+p2,4+p2x2kdx(1)

for (k, p) being the set of positive integers and where 3F2,,,z is the classical generalized hypergeometric function. We also provide analytical solutions for integrals of the form

01x2k1ln(1x)Φx2k,1,1+r2dx,

where the Lerch transcendent function Φ is defined as the analytic continuation of the series

Φ(z,t,a)=m=0zm(m+a)t,

which converges for any real number a > 0 if z and t are any complex numbers with either |z| < 1 or |z| = 1 and ℜ (t) > 1. It is known that the Lerch transcendent extends by analytic continuation to a function Φ (z, t, a) which is defined for all complex t, z ∈ ℂ − [1, ∞) and a > 0, which can be represented, [3], by the integral formula

Φ(z,t,a)=1Γ(t)0xt1e(t1)xexzdx=1Γ(t)01xa1ln(1x)1xzdx

for ℜ (t) > 0. For a fuller account of the Lerch function see the excellent papers, [6], [7] and [8]. The Lerch transcendent generalizes the Hurwitz zeta function at z = 1,

ζ(t,a)=Φ(1,t,a)=m=01(m+a)t

and the Polylogarithm, or de Jonquière’s function, when a = 1,

Lit(z):=m=1zmmt=Φ(z,t,1),tCwhen |z|<1;(t)>1.

Moreover,

01Lit(px)xdx=ζ(1+t),for p=1(2t1)ζ(1+t),for p=1.

Let ℝ and ℂ denote, respectively, the sets of real and complex numbers and let ℕ := {1, 2, 3, …} be the set of positive integers, and ℕ0:= ℕ ∪ {0} . A generalized binomial coefficient λμ (λ, μ ∈ ℂ) is defined, in terms of the familiar gamma function, by

λμ:=Γ(λ+1)Γ(μ+1)Γ(λμ+1),(λ,μC),

which, in the special case when μ = n, n ∈ ℕ0, yields

λ0:=1andλn:=λλ1λn+1n!=1nλnn!(nN),

where (λ)ν (λ, ν ∈ ℂ) is the Pochhammer symbol. Let

Hn=r=1n1r=γ+ψn+1,H0:=0(2)

be the nth harmonic number. Here, as usual, γ denotes the Euler-Mascheroni constant and ψ (z) is the Psi (or Digamma) function defined by

ψ(z):=ddz{logΓ(z)}=Γ(z)Γ(z)orlogΓ(z)=1zψ(t)dt.

A generalized harmonic number Hn(m) of order mis defined, for positive integers n and m, as follows:

Hn(m):=r=1n1rm,m,nN andH0(m):=0(mN).

In the case of non-integer values of n such as (for example) a value ρ ∈ ℝ, the generalized harmonic numbers Hρm+1 may be defined, in terms of the Polygamma functions

ψ(n)(z):=dndzn{ψ(z)}=dn+1dzn+1{logΓ(z)}(nN0),

by

Hρm+1=ζm+1+1mm!ψmρ+1ρR1,2,3,;mN,(3)

where ζ (z) is the Riemann zeta function. Whenever we encounter harmonic numbers of the form Hρm at admissible real values of ρ, they may be evaluated by means of this known relation (3). In the exceptional case of (3) when m = 0, we may define Hρ1 by

Hρ1=Hρ=γ+ψρ+1(ρR1,2,3,).

We assume (as above) that

H0m=0(mN).

In the case of non integer values of the argument z=rq we may write the generalized harmonic numbers, Hzα+1, in terms of polygamma functions

Hrqα+1=ζα+1+1αα!ψαrq+1,rq1,2,3,...,

where ζ (z) is the zeta function. When we encounter harmonic numbers at possible rational values of the argument, of the form Hrqα they maybe evaluated by an available relation in terms of the polygamma function ψ(α)(z) or, for rational arguments z=rq, and we also define

Hrq1=γ+ψrq+1, and H0α=0.

The evaluation of the polygamma function ψαra at rational values of the argument can be explicitly done via a formula as given by Kölbig [4], or Choi and Cvijovic [1] in terms of the Polylogarithmic or other special functions. Polygamma functions at negative rational values of the argument can also be explicitly evaluated, for example

H34=π23ln2,H342=8G5ζ2,H163=23π390ζ3.

Some specific values are listed in the books [13] and [14]. Some results for sums of harmonic numbers may be seen in the works of [2], [15] and references therein.

The following lemma will be useful in the development of the main theorems.

## Lemma 1.1

Let k be a positive integer. Then:

Xk,0=n11n+1Hknn=k01xk1ln1x1+xkdx(4)
=12kζ2ln22+12j=1k11kj3F21,1,12,2jk12(5)
=1+k24kζ212j=0k1ln22sin2j+1π2k(6)

## Proof

Consider, for t ∈ [−1, 1) and j ∈ ℝ+ ∪ {0}

n1tnnkn+jj=kn1tnBj+1,kn

where the beta function

Bs,x=Bx,s=ΓsΓxΓs+x=01ts11tx1dt

for ℜ (s) > 0 and ℜ (x) > 0. We have

kn1tnBj+1,kn=k011xjxn1txkndx=kt011xjxk11txkdx,

now

limj0ddjn1tnnkn+jj=ktlimj0ddj011xjxk11txkdx

and with t = −1 we obtain the result (4). To prove (5), we note, from the properties of the polygamma function with multiple argument, that

ψ(n)(kz)=δn,0lnk+1kn+1j=0k1ψ(n)(z+jk),

where δn,0 is the Kronecker delta. By the use of the digamma function in terms of harmonic numbers, we have

Hkn=lnk+1kHn+1kj=1k1Hnjk,

where Hnrp may be thought of as shifted harmonic numbers. Summing over the integers

n=11n+1Hknn=n=11n+1nlnk+1kHn+1kj=1k1Hnjk=ln2lnk+1kn=11n+1Hnn+1kj=1k1n=11n+1Hnjkn=ln2lnk+12kζ2ln22+1kj=1k1ln2Hjk+k2kj3F21,1,12,2jk12,

the first sum is obtained from [11] and the second sum is deduced from [9].Since j=1k1Hjk=k ln kthen (5) follows. The closed form representation (6) can be evaluated by contour integration, the details are in [5. □

## Lemma 1.2

Let k be a positive integer. Then:

n11n+1Hknn+1=Xk,1=k01ln1xx11+xkln1+xkxkdx(7)
=Xk,0+12kζ2Hk1ln2+12j=1k11jHj2kHk+j2k,(8)

where X(k, 0) is given in(4).

## Proof

The proof of (7) is concluded in the same manner as used in Lemma 1.1. Consider

Xk,1=n11n+1Hknn+1

and by a change of summation index

Xk,1=n11n+1Hknn+1=n11nHknkn=n11n+1Hknn+n11n+1kn2+j=1k1n11n+1nknj=Xk,0+12kζ2Hk1ln2+12j=1k11jHj2kHk+j2k.

The integral identity following (7) is obtained by the Beta method as described in Lemma 1.1 and therefore the details will not be outlined. It is of some interest to note that from (4) and (7)

k01ln1xx1xk1+xkln1+xkxkdx=n11n+12n+1Hknnn+1=12kζ2Hk1ln2+12j=1k11jHj2kHk+j2k.

□

## Lemma 1.3

Let k and r be positive integers. Then:

n=11n+1Hknn+r=Xk,r=k1+r01xk1ln1x2F12,1+r2+rxkdx(9)

and

Xk,r=1r+1Xk,1+1r+1kHr12Hr1ln2+1r+12ks=1r11ssHs12Hs2+1r+12s=1r11sHs12Hs2Hks+k1Hks+1r+12j=1k1s=1r11sks+jHj2kHk+j2k,(10)

with X(k, 1) given by(8).

## Proof

By a change of summation index

Xk,r:=n=11n+1Hknn+r=n=11nHknkn+r1=n=11n+1Hknn+r1+n=11n+1knn+r1+j=1k1n=11n+1knjn+r1=Xk,r1+1kr1ln2+12Hr22Hr12
+j=1k11j+kr1n=11n+1kknj1n+r1=Xk,r1+1kr1ln2+12Hr22Hr12+j=1k11j+kr1Φ1,1,1jk+12Hr22Hr12.

Since the Lerch transcendent

Φ1,1,1jk=12ψ2kj2kψkj2k=12Hj2kHj+k2k,

so that

Xk,r=Xk,r1+ln2kr1+12kr1Hr22Hr12+12Hr22Hr12Hkr1Hkrk+12j=1k11kr1+jHj2kHk+j2k.(11)

From (11) we have the recurrence relation

Xk,r+Xk,r1=ln2kr1+12kr1Hr22Hr12+12Hr22Hr12Hkr1Hkrk+12j=1k11kr1+jHj2kHk+j2k

for r ≥ 2, and with X(k, 1) given by (8). The recurrence relation is solved by the subsequent reduction of the

Xk,r,Xk,r1,Xk,r2,....,Xk,1

terms, finally arriving at the relation (10). The integral identity (9) is obtained by the Beta method as described in Lemma 1.1 and details will not be outlined. □

A slightly different re-arrangement of the terms in X(k, r) leads to the following Lemma.

## Lemma 1.4

Let k and r ≥ 2 be a positive integers. Then:

Yk,r=k201x2k1ln1xr1Φ(x2k,1,r+12)Φ(x2k,1,r+22)dx(12)
=Xk,r+12Hr12Hkr1Hkrk112j=1k11j+r1kHj2k=n=1H2kn2n+r12n+r(13)

with X(k, r) given by(10).

## Proof

By expansion,

Xk,r=n=11n+1Hknn+r=n=1H2kn2n+r12n+rj=0k1n=112n+r12nkj,

by re arrangement

Yk,r=n=1H2kn2n+r12n+r=Xk,r+j=0k1Hr12Hj2k2j+r1k=Xk,r+12Hr12Hkr1Hkrk112j=1k11j+r1kHj2k.

The integral (12) is obtained by considering for t ∈ [−1, 1) and j ∈ ℝ+∪ {0}

n1tn2n+r12n+r2kn+jj=2kn1ntnBj+1,2kn2n+r12n+r2kn1ntnBj+1,2kn2n+r12n+r=2k011xjxn1ntx2kn2n+r12n+rdx.

Now differentiating with respect to jand replacing the limit as japproaches zero, with t = −1, we obtain the result (12). Two special cases, furnish the following. For r = 0,

k201xk1ln1xln1+xk1xkdx=Xk,0Hkln2+12j=1k11jkHj2k.(14)

For r = 1,

k01ln1xx112xkln1+xk1xkdx=Xk,1+14kζ212j=1k11jHj2k,

from which we deduce the integral identity,

k201ln1xxk+1ln1+xk1xkdx=12j=1k11jHj2kXk,11+4k24kζ2,

and for k = 4,

01ln1xx5ln1+x41x4dx=π481822516ζ2116ln221924ln2+ln2+126+14ln21.

□

The next few theorems relate the main results of this investigation, namely the closed form representation of integrals of the type (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 and inverse binomial coefficients. First we indicate the closed form representation of

n=11n+1Hknnqn+pp(15)

for q = 0, 1, and k, p ≥ 1 are positive integers.

## Theorem 2.1

Let k ≥ 1 be real positive integer, then from(15)with q = 0 and p be real positive integer:

kp+101xk1ln1x2F12,2p+2xkdx=Λk,p=n=11n+1Hknn+pp(16)
=r=1p11+rrprXk,r,(17)

where X(k, r) is given by(10).

## Proof

Consider the expansion

Λk,p=n=11n+1Hknn+pp=n=11n+1p!Hknn+1p=n=11n+1p!Hknr=1kλrn+r

where

λr=limnrn+rr=1pn+r=11+rrp!pr.(18)

We can now express

Λk,p=n=11n+1p!Hknr=1pλrn+r=r=1p11+rrprn=11n+1Hknn+r.(19)

From (10) we have X(k, r), hence substituting into (19), (17) follows. The integral (16) is evaluated as in Lemma 1.4. □

The other case of q = 1 can be evaluated in a similar fashion. We list the result in the next Theorem.

## Theorem 2.2

Under the assumptions of Theorem 2.1, with q = 1, we have,

kp+101xk1ln1x2F11,2p+2xkdx=Mk,p=n=11n+1Hknnn+pp=r=0p1rprXk,r.(20)

and where X(k, r) is given by(10).

## Proof

The proof of (20) follows using the same technique as used in Theorem 2.1 and also using (18). □

It is possible to gain some further integral identities from Theorems 2.1 and 2.2 regarding the representation of a sequence of alternating shifted harmonic numbers as follows.

## Theorem 2.3

For p ∈ ℕ ∪ {0} and k ∈ ℕ:

1kp+1r=1k101ln1xxrkk2F11,2p+2xr2F11,1p+2xdx=kMk,pklnkp+12F11,1p+21Sp(21)
=r=1k1n=11n+1Hnrknn+pp,(22)

where M(k, p) is given by(20)and

Sp=12ζ2+2p11ln22+m=1pmpm2Hm1Hm12ln2+Hm1Hm2Hmj=1m11jjHmj2Hmj+jHjj+1,(23)

where [x] is the integer part of x.

## Proof

From the properties of harmonic numbers,

Hkn=lnk+1kr=0k1Hnrk,r=1k1n=11n+1Hnrknn+pp=kn=11n+1Hknnn+ppklnkn=11n+1nn+ppn=11n+1Hnnn+pp=kMk,pklnkp+12F11,1p+21Sp,

the details for the calculation of (23) may be seen in [11]. The integral representation (21) is obtained in the same manner as in Lemma 1.4.

For the simple case of p = 0, we have

r=1k101ln1xx1+rkrkln1+xx1+xdx=kXk,0kln2lnk12ζ2+12ln22,

and when k = 6,

r=1501ln1xx1+r6r6ln1+xx1+xdx=354ζ210ln226ln2ln3+12ln3+1ln31.

It is also possible to represent, individually, some results of shifted harmonic numbers of (22), see for example, [9] and [10]. □

The following integral identities can be exactly evaluated by using the alternating harmonic number sums in Theorems 2.1 and 2.2.

## Theorem 2.4

Let k and p be real positive integers, then:

4kp+1p+201x2k1ln1x3F21,32,2p+32,p+42x2kdx=Ωk,p=n=1H2knn2n+pp=2pr=1p11+rrprXk,r(24)
+2pp+1j=0k112kj4F31,1,32,2kj2kp+22,p+32,4kj2k1,(25)

where X (k, r) is given by(10).

## Proof

From

Λk,p=n=11n+1Hknn+pp=n=1H2n1k2n1+ppn=1H2kn2n+pp=n=1pH2kn2n2n+ppj=1k1n=112nkj2n1+ppn=112nk2n1+pp.

Re-arranging

Ωk,p=n=1H2knn2n+pp=2pΛk,p+2pp+1j=0k112kj4F31,1,32,2kj2kp+22,p+32,4kj2k1

and (25) follows. The integral (24) is evaluated as in Lemma 1.4. □

## Theorem 2.5

Let k and p be real positive integers, then:

4kp+1p+201x2k1ln1x3F212,1,2p+32,p+42x2kdx=Ξk,p=n=1H2knn2n12n+pp=2p+1r=0p1rprXk,r+2p+12j=0k112kj4F312,1,1,2kj2kp+22,p+32,4kj2k1,

where X (k, r) is given by (10).

## Proof

Follows the same pattern as used in Theorem 2.4. □

## Theorem 2.6

Let k and p be real positive integers, then:

4kp+1p+201x2k1ln1x3F212,2,2p+32,p+42x2kdx=Υk,p=n=1H2kn2n12n+pp=2pΛk,p+2p+1Mk,p+2p+12j=0k112kj4F312,1,1,2kj2kp+22,p+32,4kj2k1+2pp+1j=0k112kj4F31,1,32,2kj2kp+22,p+32,4kj2k1,

where Λ (k, p) is given by (17) and M (k, p) by (20).

## Proof

Follows the same pattern as used in Theorem 2.4. □

A number of special cases follow in the next Corollary.

## Corollary 2.7

Some examples of integrals are given below. For p = 0, Theorem 2.5 reduces to (14).

For p = 1, from Theorem 2.6 we have

k011+x2kxk+1ln1xln1+xk1xkdx=3+4k22kζ2+2kln2+4Hk1+j=1k11kjHj2kHj2k,

for k = 3, we have

3011+x6x4ln1xln1+x31x3dx=132ζ2+536π+23ln2,

this integral is highly oscillatory near the origin of x. From Theorem 2.5, with k = 6.

01x121x7ln1xln1+x61x6dx=7372ζ2+12ln22+3940ln343ln1+3235+ln31+11543+116ln2,

this integral is highly oscillatory near the origin of x.

## Conclusion 2.8

We have established a number of integral identities in closed form in terms of special functions. A number of oscillatory integrals are also given in closed form. The integral identities established in this paper complement and extend the results in the paper [12]. Some particular identities obtained are

1801x+1xln2xln1+x1xdx=103ζ2143ζ3+2627π+329G5627ln2+10427+π312.12011x41xx52ln2xln1+x21x2dx=162714213π+163225ζ2+13322π3.

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