Integrals of logarithmic and hypergeometric functions

Communications in Mathematics 24 (2016) 7–22 c 2016 The University of Ostrava Copyright

7

Integrals of logarithmic and hypergeometric functions Anthony Sofo

Abstract. Integrals of logarithmic and hypergeometric functions are intrinsically connected with Euler sums. In this paper we explore many relations and explicitly derive closed form representations of integrals of logarithmic, hypergeometric functions and the Lerch phi transcendent in terms of zeta functions and sums of alternating harmonic numbers.

1

Introduction and Preliminaries

Let N := {1, 2, 3, · · · } be the set of positive integers, N0 := N ∪ {0} , also let R and C denote, respectively, the sets of real and complex numbers. In this paper we develop identities, new families of closed form representations of alternating harmonic numbers and reciprocal binomial coefficients, including integral representations, of the form: Z 1 x ln2m−1 (x) 1, 2; F − x dx, (k, m ∈ N), (1) 2 1 2 + k; 1−x 0 ·, ·; z is the classical generalized hypergeometric function. Some ·; specific cases of similar integrals of the form (1) have been given by [11]. We also investigate integrals of the form

where 2 F1

Z 0

1

x ln2m−1 (x) Φ(−x, 1, 1 + r) dx 1−x

and show that these integrals may be expressed in terms of a linear rational combination of zeta functions and known constants. 2010 MSC: Primary 33C20, 05A10, 05A19; Secondary 11Y60, 11B83, 11M06. Key words: Logarithm function, Hypergeometric functions, Integral representation, Lerch transcendent function, Alternating harmonic numbers, Combinatorial series identities, Summation formulas, Partial fraction approach, Binomial coefficients. DOI: 10.1515/cm-2016-0002

- 10.1515/cm-2016-0002 Downloaded from De Gruyter Online at 09/08/2016 12:29:08AM via free access

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Anthony Sofo

P∞ zm Here Φ(z, t, a) = m=0 (m+a) t is the Lerch transcendent defined for |z| < 1 and 0 and satisfies the recurrence Φ(z, t, a) = zΦ(z, t, a + 1) + a−t . The Lerch transcendent generalizes the Hurwitz zeta function at z = 1, Φ(1, t, a) =

∞ X

1 (m + a)t m=0

and the Polylogarithm, or de Jonqui`ere’s function, when a = 1, Lit (z) :=

∞ X zm , t ∈ C when |z| < 1; 1 when |z| = 1. mt m=1

Moreover Z

1

 

ζ(1 + t), for p = 1

Lit (px) dx =  x

. (2−r − 1)ζ(1 + t), for p = −1 A generalized binomial coefficient µλ (λ, µ ∈ C) is defined, in terms of the familiar gamma function, by λ Γ(λ + 1) (λ, µ ∈ C), := Γ(µ + 1)Γ(λ − µ + 1) µ 0

which, in the special case when µ = n, n ∈ N0 , yields λ λ λ(λ − 1) · · · (λ − n + 1) (−1)n (−λ)n := 1 and := = 0 n n! n!

(n ∈ N),

where (λ)ν (λ, ν ∈ C) is the Pochhammer symbol. Let, for n ∈ N, Hn =

n X 1 r=1

r

= γ + ψ(n + 1),

(H0 := 0)

(2)

be the n-th harmonic number. Here γ denotes the Euler-Mascheroni constant and ψ(z) is the Psi (or Digamma) function defined by Z z d Γ0 (z) ψ(z) := {log Γ(z)} = or log Γ(z) = ψ(t) dt. dz Γ(z) 1 An unusual, but intriguing representation, of the n-th harmonic number, has recently been given by [8], as Z 1 x−

Hn = π 0

1 2

cos

(4n+1)πx − 2 sin πx 2

cos

πx 2

dx.


Integrals of logarithmic and hypergeometric functions

9

Choi [3] has also given the definition, in terms of log-sine functions Z π2 ln(sin x) sin x(cos x)2n−1 dx Hn = −4n 0

Z

π 2

= −4n

ln(cos x) cos x(sin x)2n−1 dx.

0 (m)

A generalized harmonic number Hn Hn(m) :=

n X r=1

1 rm

of order m is defined, as follows:

(m, n ∈ N)

and

(m)

H0

:= 0

(m ∈ N),

in terms of integral representations we have the result Z (−1)m 1 (ln x)m (1 − xn ) dx. Hn(m+1) = m! 1−x 0

(3)

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

dn dn+1 {ψ(z)} = {log Γ(z)} n dz dz n+1

(n ∈ N0 ),

by (−1)m (m) ψ (ρ + 1) m! (ρ ∈ R \ {−1, −2, −3, · · · } ; m ∈ N),

Hρ(m+1) = ζ(m + 1) +

(4)

where ζ(z) is the Riemann zeta function. Whenever we encounter harmonic num(m) bers of the form Hρ at admissible real values of ρ, they may be evaluated by means of this known relation (4). In the exceptional case of (4) when m = 0, we (1) may define Hρ by Hρ(1) = Hρ = γ + ψ(ρ + 1) ρ ∈ R \ {−1, −2, −3, . . . } . In the case of non integer values of the argument z = rq , we may write the gener(α+1)

alized harmonic numbers, Hz (α+1)

Hr q

, in terms of polygamma functions

(−1)α (α) r ψ ( + 1), α! q

= ζ(α + 1) +

r 6= {−1, −2, −3, . . . } , q

r where ζ(z) is the zeta function. The evaluation of the polygamma function ψ (α) ( ) a at rational values of the argument can be explicitly done via a formula as given by Kölbig [14], or Choi and Cvijovic [4] in terms of the Polylogarithmic or other special functions. Some specific values are listed in the books [18], [25] and [26]. Let us define the alternating zeta function ¯ := ζ(z)

∞ X (−1)n+1 = (1 − 21−z )ζ(z) z n n=1


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Anthony Sofo

¯ = ln 2, and also let with ζ(1) +− Sp,q :=

∞ (p) X (−1)n+1 Hn . nq n=1

In the case where p and q are both positive integers and p + q is an odd integer, Flajolet and Salvy [12] gave the identity: X p + i − 1 +− ¯ +2 ¯ 2Sp,q = 1 − (−1)p ζ(p)ζ(q) ζ(p + i)ζ(2k) p−1 i+2k=q (5) X q + j − 1 ¯ + q) − 2 ¯ + j)ζ(2k) ¯ + ζ(p (−1)j ζ(q , q−1 j+2k=p

¯ ¯ = ln 2, ζ(1) = 0 and ζ(0) = − 12 in accordance with the where ζ(0) = 21 , ζ(1) analytic continuation of the Riemann zeta function. Some results for sums of alternating harmonic numbers may be seen in the works [1], [2], [5], [6], [7], [9], [10], [12], [15], [16], [17], [20], [21], [19], [22], [23], [27], [28], [29] and [30] and references therein. The following lemma will be useful in the development of the main theorems. Lemma 1. Let r, p ∈ N. Then we have r X (−1)j 1 (p) (p) (p) = p H r + H r−1 − H2 r+1 −1 , [ 2 ] [2] [ 2 ] jp 2

(6)

j=1

where [x] is the integer part of x. The following identities hold. For 0 < t ≤ 1 ∞ X 1+t (−t)n+1 Hn t ln ( )=2 t n+1 n=1 2

and when t = 1, ln2 2 = 2

∞ 1 X (−1)n+1 Hn = ζ(2) − 2Li2 . n+1 2 n=1

Proof. The proof is given in the paper [24].

(7)

Lemma 2. The following identity holds: for m ∈ N, M (m, 0) =

∞ (2m) X (−1)n+1 Hn ¯ = mζ(2m + 1) − ζ(2m) ln 2 n n=1

−

m−1 X

(8)

¯ ¯ ζ(2j + 1)ζ(2m − 2j)

j=1

1 = (2m − 1)!

Z 0

1

ln2m−1 x ln(1 + x) − ln 2 dx. 1−x

(9)



11

Proof. From (5) we choose q = 1 and p = 2m, m ∈ N. Then (8) follows. Next from the definition (3), Z 1 2m−1 X ∞ ln x 1 (−1)n+1 (1 − xn ) M (m, 0) = − dx (2m − 1)! 0 1 − x n=1 n Z 1 2m−1 ln x 1 = ln(1 + x) − ln 2 dx. (2m − 1)! 0 1−x Lemma 3. Let m ∈ N. Then M (m, 1) =

∞ (2m) X (−1)n+1 Hn ¯ = ζ(2m + 1) − mζ(2m + 1) n + 1 n=1

¯ + ζ(2m) ln 2 +

m−1 X

¯ ¯ ζ(2j + 1)ζ(2m − 2j)

(10)

ln2m−1 x ln(1 + x) (ln 2 − ) dx . 1−x x

(11)

j=1

1 = (2m − 1)!

1

Z 0

Proof. From the left hand side of (10) and by a change of counter M (m, 1) =

(2m) ∞ X (−1)n Hn−1 ¯ = −M (m, 0) + ζ(2m + 1) , n n=1

substituting for M (m, 0) we obtain the right hand side of (10). The integral (11) is obtained by the same method as that used in Lemma 2. Lemma 4. Let r ≥ 2 be a positive integer, m ∈ N and define M (m, r) :=

∞ (2m) X (−1)n+1 Hn . n+r n=1

We have the recurrence relation (−1)r r M (m, r) + M (m, r − 1) = H − H r−1 − 1 + (−1) ) ln 2 r−1 [ 2 ] (r − 1)2m +

2m X j=2

(−1)j ¯ ζ(j) (r − 1)2m+1−j

(12)

with solution M (m, r) = (−1)r+1 M (m, 1) − (−1)r ln 2

r−1 X (−1)ρ+1 ρ=1

+ (−1)r

ρ2m

r−1 r−1 X 2m X X 1 (−1)ρ+1+j ¯ r (H − H − ln 2) + (−1) ζ(j), ρ ρ 2m [ ] 2 ρ ρ2m+1−j ρ=1 ρ=1 j=2

(13) with M (m, 0) and M (m, 1) given by (8) and (10) respectively. - 10.1515/cm-2016-0002 Downloaded from De Gruyter Online at 09/08/2016 12:29:08AM via free access

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Anthony Sofo

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

(2m) ∞ ∞ X X (−1)n Hn−1 1 (−1)n (2m) Hn − 2m = n+r−1 n+r−1 n n=1 n=1

=−

∞ ∞ (2m) X X (−1)n+1 Hn (−1)n+1 + n+r−1 n2m (n + r − 1) n=1 n=1

= −M (m, r − 1) +

−

∞ r−1 X X 1 1 (−1)n+r (−1)n+r − (r − 1)2m n=1 n (r − 1)2m n=1 n

∞ 2m ∞ X X (−1)n+1 X (−1)j (−1)n+1 1 + . 2m 2m+1−j (r − 1) n=1 n (r − 1) nj n=1 j=2

From Lemma 1 and using the known results, M (m, r) = −M (m, r − 1) −

(−1)r ln 2 + H r−1 − Hr−1 [ 2 ] (r − 1)2m

2m

X (−1)j ln 2 ¯ + − ζ(j). (r − 1)2m j=2 (r − 1)2m+1−j

(14)

From (14) we have the recurrence relation M (m, r) + M (m, r − 1) =

(−1)r (Hr−1 − H[ r−1 ] − (1 + (−1)r ) ln 2) 2 (r − 1)2m +

2m X j=2

(−1)j ¯ , ζ(j) (r − 1)2m+1−j

for r ≥ 2, with M (m, 0) given by (8) and M (m, 1) given by (10). The recurrence relation is solved by the subsequent reduction of the M (m, r), M (m, r − 1), M (m, r − 2), . . . , M (m, 1) terms, finally arriving at the relation (13).

It is of some interest to note that M (m, r) may be expanded in a slightly different way so that it gives rise to another unexpected harmonic series identity. This is pursued in the next lemma. Lemma 5. For a positive integer r ≥ 2, and m ∈ N we have the identity Υ(m, r) =

∞ X

(2m)

H2n (2n + r − 1)(2n + r) n=1 2m

= M (m, r) −

X 1 (−1)j H r−1 + ζ(j) , 2m j 2 2(r − 1) 2 (r − 1)2m+1−j j=2

(15)


13


with M (m, r) given by (13). For r = 1, Υ(m, 1) =

∞ X

(2m)

H2n ¯ = ζ(2m + 1) + (2−1−2m − m)ζ(2m + 1) 2n(2n + 1) n=1 ¯ + ζ(2m) ln 2 +

m−1 X

(16)

¯ ¯ ζ(2j + 1)ζ(2m − 2j) ,

j=1

and for r = 0 Υ(m, 0) =

∞ X

(2m)

H2n ¯ = mζ(2m + 1) − ζ(2m) ln 2 + ln 2 2n(2n − 1) n=1 −

m−1 X

¯ ¯ ζ(2j + 1)ζ(2m − 2j) −

j=1

2m X

(17) 2−j ζ(j) .

j=2

Proof. From M (m, r) =

∞ ∞ (2m) (2m) X X (−1)n+1 Hn H2n = n+r (2n + r − 1)(2n + r) n=1 n=1

−

∞ X

1 . (2n + r − 1)(2n)2m n=1

Hence Υ(m, r) :=

∞ (2m) X 1 H2n = M (m, r) + , (2n + r − 1)(2n + r) (2n + r − 1)(2n)2m n=1 n=1 ∞ X

then, Υ(m, r) = M (m, r) +

+

2m X j=2

∞ X (−1)j 1 j 2m+1−j 2 (r − 1) nj n=1

= M (m, r) − +

2m X j=2

∞ X 1 1 1 − 2m (r − 1) n=1 2n + r − 1 2n

1 H r−1 2 2(r − 1)2m

(−1)j ζ(j) . 2j (r − 1)2m+1−j

For r = 1 Υ(m, 1) =

∞ X

(2m)

H2n 1 = M (m, 1) + 2m+1 ζ(2m + 1) 2n(2n + 1) 2 n=1 - 10.1515/cm-2016-0002 Downloaded from De Gruyter Online at 09/08/2016 12:29:08AM via free access

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Anthony Sofo

and substituting for M (m, 1), we obtain (16). For r = 0 Υ(m, 0) =

∞ X

2m (2m) X H2n = M (m, 0) + ln 2 − 2−j ζ(j) 2n(2n − 1) n=1 j=2

and substituting for M (m, 0), we obtain (17).

We develop some integral identities for Lemma 4 and Lemma 5, namely: Lemma 6. For r ∈ N0 \ {0, 1}, Z 1 x ln2m−1 x 1 Φ(−x, 1, 1+r) dx = M (m, r)−ζ(2m)Φ(−1, 1, 1+r) (18) (2m − 1)! 0 1−x where M (m, r) is given by (13). P∞ (−1)n+1 Hn(2m) and by the use of the integral repreProof. From M (m, r) := n=1 n+r sentation (3) Z 1 2m−1 X ∞ 1 ln x (−1)n+1 (1 − xn ) M (m, r) = dx (2m − 1)! 0 1 − x n=1 n+r Z 1 2m−1 1 ln x = xΦ(−x, 1, 1 + r) − Φ(−1, 1, 1 + r) dx (2m − 1)! 0 1−x Z 1 x ln2m−1 x 1 Φ(−x, 1, 1 + r) dx + ζ(2m)Φ(−1, 1, 1 + r) = (2m − 1)! 0 1−x and by re-arrangement we obtain (18).

Remark 1. For the case r = 0, we have Z 1 2m−1 ln x ln(1 + x) 1 dx = M (m, 0) − ζ(2m) ln 2. (2m − 1)! 0 1−x This identity could also be obtained from (9), since it is known that Z 1 2m−1 ln x dx = (2m − 1)! Li2m (1) = (2m − 1)! ζ(2m). − 1 − x 0 For r = 1 we obtain 1 (2m − 1)!

Z 0

1

ln2m−1 x ln(1 + x) dx = −M (m, 1) − ζ(2m) ln 2, x(1 − x)

for m ≥ 2. An obvious result of the preceding two integrals yields the delightful identity Z 1 2m−1 1 ln x ln(1 + x) ¯ dx = −ζ(2m + 1) (19) (2m − 1)! 0 x ∞ 1 X 1 = −M (m, 0) − M (m, 1) = − (−1)n+1 Hn(2m) + n n+1 n=1 The Wolfram on-line integrator yields no solution to these general integrals. - 10.1515/cm-2016-0002 Downloaded from De Gruyter Online at 09/08/2016 12:29:08AM via free access


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Lemma 7. For r ∈ N0 \ {0, 1}, 1 (2m − 1)!

1

Z 0

x2 ln2m−1 x 2 + r 1 + r 2 2 − Φ x , 1, dx Φ x , 1, 2(1 − x) 2 2 1 = Υ(m, r) + ζ(2m) H r−1 − H r2 (20) 2 2

where Υ(m, r) is given by (15) and Φ(x2 , 1, 1+r 2 ) is the Lerch transcendent. For r=0 Z 1 2m−1 ln x 1 1 x tanh−1 x + ln(1 − x2 ) dx = Υ(m, 0) − ζ(2m) ln 2. (2m − 1)! 0 1−x 2 For r = 1 1 (2m − 1)!

Z

1

0

ln2m−1 x tanh−1 x 1 + ln(1 − x2 ) dx = −Υ(m, 1) − ζ(2m) ln 2. 1−x x 2

By addition we have the result Z 1 2m−1 1 ln x 1 (x + ) tanh−1 x + ln(1 − x2 ) dx (2m − 1)! 0 1−x x = Υ(m, 0) − Υ(m, 1) − 2ζ(2m) ln 2 2m 1 X 2−j ζ(j) = M (m, 0) − M (m, 1) − 2m+1 + ln 2 ζ(2m) − ln 2 − 2 j=2 and by further simplification Z 1 1 (1 + x) ln2m−1 x − tanh−1 x dx (2m − 1)! 0 x = Υ(m, 1) + Υ(m, 0) = ζ(2m + 1) + ln 2 −

2m+1 X

(21)

2−j ζ(j)

j=2

=

∞ X n=0

(2

n 2

1 . + 1)(n + 1)2m

(22)

It may also be shown that the alternating companion to (22) is, ∞ X n=0

(2

n 2

(−1)n 1 =− 2m (2m − 1)! + 1)(n + 1)

Z 0

1

(1 − x) ln2m−1 x tanh−1 x dx x

¯ = ζ(2m + 1) − ln 2 +

2m+1 X

2−j ζ(j),

j=2

where [z] is the integer part of z. - 10.1515/cm-2016-0002 Downloaded from De Gruyter Online at 09/08/2016 12:29:08AM via free access

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Anthony Sofo

Proof. Follow the same pattern as used in Lemma 6. The new sum (22) can be obtained from the Taylor series expansion of (21) in the following way: Z 1 1 (1 + x) ln2m−1 x − tanh−1 x dx (2m − 1)! 0 x Z 1 ∞ X 1 1 xn ln2m−1 x dx =− (2m − 1)! n=0 (2 n2 + 1) 0 =

=

∞ X n=0 ∞ X

(2

n 2

1 + 1)(n + 1)2m (2m)

2H2n . (2n − 1)(2n + 1) n=1

For m = 3, we have ∞ X

∞ (6) X 2H2n 1 n = (2n − 1)(2n + 1) + 1)(n + 1)6 (2 2 n=1 n=0

ζ(2) ζ(3) ζ(4) ζ(5) ζ(6) 127ζ(7) + + + + − − ln 2. 4 8 16 32 64 128 Mathematica cannot sum either of these two series. =

The next few theorems relate the main results of this investigation, namely the integral and 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 of order 2m and inverse binomial coefficients. First we indicate the closed form representation of X(m, k, p) =

∞ (2m) X (−1)n+1 Hn np n+k k n=1

(23)

for the two cases (k, p) = (k, 1), (k, 0) and k ∈ N. Theorem 1. Let k ∈ N. Then from (23) with p = 1 we have, X(m, k, 1) =

k ∞ (2m) X X (−1)n+1 Hn 1+r k = M (m, 0) − (−1) M (m, r) , r n( n+k k n=1 r=1

(24)

where M (m, r) is given by (13). Proof. Consider the expansion X(m, k, 1) =

=

∞ ∞ (2m) (2m) X X (−1)n+1 Hn (−1)n+1 k!Hn = n(n + 1)k n n+k k n=1 n=1 ∞ X

(−1)n+1 k!Hn(2m)

n=1

Ω n

+

k X Λr , n+r r=1


17


where

   

   

n+r (−1)r+1 Λr = lim = − k n→−r   Q k!  n  n + r

k , r

Ω=

1 . k!

(25)

r=1

We can now express X(m, k, 1) =

∞ X n=1

=

∞ (2m) X Hn n n=1

k X

1 (−1)r+1 k + n+r r n r=1 ∞ k (2m) X k X Hn − (−1)1+r . r n=1 n + r r=1

(−1)n+1 Hn(2m)

−

From (13) we have M (m, r), hence substituting into (26), (24) follows.

(26)

The other case of X(m, k, 0) 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 for p = 0, X(m, k, 0) =

∞ (2m) X (−1)n+1 Hn n+k k

n=1

=

k X

(−1)1+r r

r=1

k M (m, r) , r

(27)

and where M (m, r) is given by (13). Proof. Same procedure as that used in Theorem 1 can be used here.

The following integral identities can be exactly evaluated by using the alternating harmonic number sums in Theorems 1 and 2. The following integral identities are the main results of this paper. Theorem 3. Let m, k ∈ N. Then we have Z 1 1 x ln2m−1 x 1, 1; − x dx 2 F1 2 + k; (1 + k)(2m − 1)! 0 1−x ζ(2m) 1, 1; − 1 , (28) = X(m, k, 1) − 2 F1 2 + k; 1+k where X(m, k, 1) is given by (24), 1 (1 + k)(2m − 1)!

Z 0

1

x ln2m−1 x 2 F1 1−x

1, 2; − x dx 2 + k;

ζ(2m) = X(m, k, 0) − 2 F1 1+k

1, 2; − 1 , (29) 2 + k;

where X(m, k, 0) is given by (27). - 10.1515/cm-2016-0002 Downloaded from De Gruyter Online at 09/08/2016 12:29:08AM via free access

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Anthony Sofo

Proof. From the identity (3), we can write Z 1 2m−1 X ∞ ∞ (2m) X (−1)n+1 Hn 1 ln x (−1)n+1 (1 − xn ) = − dx (2m − 1)! 0 1 − x n=1 n n+k n n+k k k n=1  1, 1; Z 1 2m−1  x2 F1 2 + k; − x 1 ln x =  (1 + k)(2m − 1)! 0 1−x  1, 1; − 2 F1 −1 2 + k;

    dx . 

·, ·; · is the generalized hypergeometric function. Since ·;

where 2 F1 1 (2m − 1)!

Z 0

1

ln2m−1 x 2 F1 1−x

1, 1; 1, 1; − 1 dx = −2 F1 − 1 ζ(2m), 2 + k; 2 + k;

then by re-arrangement we obtain the integral identity (28). The identity 29 follows in a similar way. Remark 2. From Theorem 3 we detail the following two examples. From (28) with {m, k} = {3, 4}, we obtain 1 720

Z

Z

1

1 720

1

0

0

x ln5 x 2 F1 1−x

ζ(6) 1, 1; 1, 1; F − x dx = X(3, 4, 1) − − 1 ; 2 1 6; 6; 5

12 + 42x + 52x2 + 25x3 ln5 x 12(1 + x)4 ln(1 + x) − dx 1−x x4 x3 3090551 84353ζ(2) 15743ζ(4) 3739ζ(6) 16040 ln 2 + + + − = 279936 15552 128 192 729 63 ln 2ζ(6) 13867ζ(3) 21ζ(3)ζ(4) 1775ζ(5) − − − − 2 1728 4 192 15ζ(2)ζ(5) 2127ζ(7) + . − 2 64

From (29) with {m, k} = {3, 4}, we obtain x ln5 x ζ(6) 1, 2; 1, 2; F − x dx = X(3, 4, 0) − F − 1 2 1 2 1 6; 6; 5 0 1−x 40832 ln 2 492155 13373ζ(2) 2471ζ(4) − − − 43ζ(6) + + 63 ln 2ζ(6) =− 17496 972 108 729 2191ζ(3) 275ζ(5) 129ζ(7) + + 21ζ(3)ζ(4) + + 15ζ(2)ζ(5) − . 108 12 2

1 720

Z

1

The Wolfram on-line integrator, yields no solution to these general slow converging integrals. - 10.1515/cm-2016-0002 Downloaded from De Gruyter Online at 09/08/2016 12:29:08AM via free access

19


Remark 3. It appears that, for r ∈ N0 , m ∈ N ∞ (2m) X (−1)n+1 Hn K(m, r) = (n + r)2 n=1

may not have a closed form solution, in terms of some common special functions. Remarkably, however, the sum of two consecutive terms of K(m, r) does have a closed form solution, this result is pursued in the next Lemma. Lemma 8. For r ∈ N, m ∈ N K(m, r) + K(m, r + 1) =

∞ X

(−1)n+1 Hn(2m)

n=1

=

1 1 + (n + r)2 (n + r + 1)2

1 (2) (2) H r − H r−1 2 4r2m 2 2m(−1)r + 2m+1 1 − (−1)r ln 2 + H[ r ] − Hr 2 r 2m j X (−1) (2m + 1 − j) ¯ + ζ(j) r2m+2−j j=2

(30)

and for r = 0, K(m, 0) + K(m, 1) =

∞ X

(−1)n+1 Hn(2m)

n=1

1 1 + n2 (n + 1)2

¯ = ζ(2m + 2) =

1 (2m − 1)!

(31) Z

1

ln

2m−1

0

x Li2 (−x) dx . x

(32)

Proof. K(m, r) + K(m, r + 1) =

∞ X

(−1)n+1 Hn(2m)

n=1

1 1 + 2 2 (n + r) (n + r + 1)

and by a change of counter in the second sum, after simplification we obtain K(m, r) + K(m, r + 1) =

∞ X n=1

(−1)n+1 + r)2

n2m (n

∞ ∞ 1 X (−1)n+1 2m X (−1)n+1 = 2m + r n=1 (n + r)2 r2m+1 n=1 n + r 2m ∞ X (−1)j (2m + 1 − j) X (−1)n+1 r2m+2−j nj n=1 j=1 2m(−1)r 1 (2) 2m (2) = 2m H r − H r−1 + 2m+1 ln 2 + H[ r ] − Hr − 2m+1 ln 2 2 2 4r r r 2 2m j X (−1) (2m + 1 − j) ¯ + ζ(j) r2m+2−j j=2

+


20

Anthony Sofo

and upon simplification we obtain (30). For the case r = 0, (31) and (32) follow directly. Remark 4. The identity (32) with (31) is obtained using the definition (3). Also considering the Taylor expansion of (32) we notice 1 (2m − 1)!

Z 0

1

ln2m−1 x Li2 (−x) dx x Z ∞ X (−1)j+1 1 j 2m−1 1 x ln x dx = (2m − 1)! j=0 (j + 1)2 0 ∞ X

=

j=0

(33)

(−1)j ¯ = ζ(2m + 2) . (j + 1)2m+2

This gives the even zeta functions of alternating type and (19) gives the odd zeta functions of alternating type. Moreover Z 1 2m−1 1 ln x Li2 (−x) lim dx = 1 m→∞ (2m − 1)! 0 x From (30), for {m, r} = {3, 3}, K(3, 3) + K(3, 4) =

31ζ(6) 5ζ(5) 7ζ(4) ζ(3) 2ζ(2) 4 ln 2 91 − + − + − + . 288 72 216 81 729 729 26244

It is possible to obtain a nice generalization of (33) as, 1 (2m − 1)!

1

Z 0

ln2m−1 x Lip (−x) ¯ dx = ζ(2m + p) x

for p ∈ N, and is analogous to the identity (see [13]) Z 0

1

lnm x Lip (x) dx = (−1)m m! ζ(m + p + 1) x

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[29] T. C. Wu, S. T. Tu, H. M. Srivastava: Some combinatorial series identities associated with the digamma function and harmonic numbers. Appl. Math. Lett. 13 (3) (2000) 101–106. [30] D. Y. Zheng: Further summation formulae related to generalized harmonic numbers. J. Math. Anal. Appl. 335 (1) (2007) 692–706. Author’s address: Victoria University, P. O. Box 14428, Melbourne City, Victoria 8001, Australia a E-mail: anthony.sofo vu.edu.au Received: 14 September, 2015 Accepted for publication: 10 May, 2016 Communicated by: Attila Bérczes
