Integrals of Logarithmic and Hypergeometric Functions

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

Integrals of logarithmic and hypergeometric functions

Anthony Sofo

Abstract. Integrals of logarithmic and hypergeometric functions are intrin- sically 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; 2F1 − x dx, (k, m ∈ N), (1) 0 1 − x 2 + k;

·, ·; where F z is the classical generalized hypergeometric function. Some 2 1 ·; speciﬁc cases of similar integrals of the form (1) have been given by [11]. We also investigate integrals of the form

Z 1 x ln2m−1(x) Φ(−x, 1, 1 + r) dx 0 1 − x and show that these integrals may be expressed in terms of a linear rational com- bination 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, Summa- tion formulas, Partial fraction approach, Binomial coeﬃcients. DOI: 10.1515/cm-2016-0002

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P∞ zm Here Φ(z, t, a) = m=0 (m+a)t is the Lerch transcendent deﬁned for |z| < 1 and <(a) > 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,

∞ X 1 Φ(1, t, a) = (m + a)t m=0 and the Polylogarithm, or de Jonqui`ere’sfunction, when a = 1,

∞ X zm Li (z) := , t ∈ when |z| < 1; <(t) > 1 when |z| = 1. t mt C m=1

Moreover  ζ(1 + t), for p = 1 Z 1 Li (px)  t dx = . x 0  (2−r − 1)ζ(1 + t), for p = −1

λ A generalized binomial coeﬃcient µ (λ, µ ∈ C) is deﬁned, in terms of the familiar gamma function, by

λ Γ(λ + 1) := (λ, µ ∈ ), µ Γ(µ + 1)Γ(λ − µ + 1) C which, in the special case when µ = n, n ∈ N0, yields λ λ λ(λ − 1) ··· (λ − n + 1) (−1)n(−λ) := 1 and := = n (n ∈ ), 0 n n! n! N where (λ)ν (λ, ν ∈ C) is the Pochhammer symbol. Let, for n ∈ N,

n X 1 H = = γ + ψ(n + 1), (H := 0) (2) n r 0 r=1 be the n-th harmonic number. Here γ denotes the Euler-Mascheroni constant and ψ(z) is the Psi (or Digamma) function deﬁned by

d Γ0(z) Z 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 re- cently been given by [8], as

Z 1 (4n+1)πx πx 1 cos 2 − cos 2 Hn = π x − πx dx. 0 2 sin 2

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Choi [3] has also given the deﬁnition, in terms of log-sine functions

π Z 2 2n−1 Hn = −4n ln(sin x) sin x(cos x) dx 0 π Z 2 = −4n ln(cos x) cos x(sin x)2n−1 dx. 0 (m) A generalized harmonic number Hn of order m is defined, as follows: n X 1 (m) H(m) := (m, n ∈ ) and H := 0 (m ∈ ), n rm N 0 N r=1 in terms of integral representations we have the result m Z 1 m n (m+1) (−1) (ln x) (1 − x ) Hn = dx. (3) m! 0 1 − x In the case of non-integer values of n such as (for example) a value ρ ∈ R, the (m+1) generalized harmonic numbers Hρ may be defined, in terms of the Polygamma functions dn dn+1 ψ(n)(z) := {ψ(z)} = {log Γ(z)} (n ∈ ), dzn dzn+1 N0 by (−1)m H(m+1) = ζ(m + 1) + ψ(m)(ρ + 1) (4) ρ m! (ρ ∈ R \ {−1, −2, −3, ···} ; m ∈ N), 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 (1) Hρ = Hρ = γ + ψ(ρ + 1) ρ ∈ R \ {−1, −2, −3,... } . r In the case of non integer values of the argument z = q , we may write the gener- (α+1) alized harmonic numbers, Hz , in terms of polygamma functions α (α+1) (−1) (α) r r H r = ζ(α + 1) + ψ ( + 1), 6= {−1, −2, −3,... } , q α! q 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 ∞ X (−1)n+1 ζ¯(z) := = (1 − 21−z)ζ(z) nz n=1

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with ζ¯(1) = ln 2, and also let

∞ n+1 (p) X (−1) Hn S+− := . p,q 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 2S+− = 1 − (−1)pζ(p)ζ¯(q) + 2 ζ(p + i)ζ¯(2k) p,q p − 1 i+2k=q (5) X q + j − 1 + ζ¯(p + q) − 2 (−1)jζ¯(q + j)ζ¯(2k) , q − 1 j+2k=p

¯ 1 ¯ 1 where ζ(0) = 2 , ζ(1) = ln 2, ζ(1) = 0 and ζ(0) = − 2 in accordance with the 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 j X (−1) 1 (p) (p) (p) p = p H r + H r−1 − H r+1 , (6) j 2 [ 2 ] [ 2 ] 2[ 2 ]−1 j=1 where [x] is the integer part of x. The following identities hold. For 0 < t ≤ 1

∞ n+1 1 + t X (−t) Hn t ln2( ) = 2 t n + 1 n=1 and when t = 1,

∞ n+1 X (−1) Hn 1 ln2 2 = 2 = ζ(2) − 2Li . (7) n + 1 2 2 n=1

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

Lemma 2. The following identity holds: for m ∈ N,

∞ n+1 (2m) X (−1) Hn M(m, 0) = = mζ(2m + 1) − ζ¯(2m) ln 2 (8) n n=1 m−1 X − ζ¯(2j + 1)ζ¯(2m − 2j) j=1 1 Z 1 ln2m−1 x = ln(1 + x) − ln 2 dx. (9) (2m − 1)! 0 1 − x

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Proof. From (5) we choose q = 1 and p = 2m, m ∈ N. Then (8) follows. Next from the deﬁnition (3), ∞ 1 Z 1 ln2m−1 x X (−1)n+1(1 − xn) M(m, 0) = − dx (2m − 1)! 1 − x n 0 n=1 Z 1 2m−1 1 ln x = ln(1 + x) − ln 2 dx. (2m − 1)! 0 1 − x

Lemma 3. Let m ∈ N. Then ∞ n+1 (2m) X (−1) Hn M(m, 1) = = ζ¯(2m + 1) − mζ(2m + 1) n + 1 n=1 m−1 X + ζ¯(2m) ln 2 + ζ¯(2j + 1)ζ¯(2m − 2j) (10) j=1 1 Z 1 ln2m−1 x ln(1 + x) = (ln 2 − ) dx . (11) (2m − 1)! 0 1 − x x Proof. From the left hand side of (10) and by a change of counter

∞ (2m) X (−1)nH M(m, 1) = n−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 ∞ n+1 (2m) X (−1) Hn M(m, r) := . n + r n=1 We have the recurrence relation r (−1) r M(m, r) + M(m, r − 1) = Hr−1 − H r−1 − 1 + (−1) ) ln 2 (r − 1)2m [ 2 ] 2m (12) X (−1)j + ζ¯(j) (r − 1)2m+1−j j=2 with solution r−1 X (−1)ρ+1 M(m, r) = (−1)r+1M(m, 1) − (−1)r ln 2 ρ2m ρ=1 r−1 r−1 2m ρ+1+j r X 1 r X X (−1) ¯ + (−1) (Hρ − H ρ − ln 2) + (−1) ζ(j), ρ2m [ 2 ] ρ2m+1−j ρ=1 ρ=1 j=2 (13) with M(m, 0) and M(m, 1) given by (8) and (10) respectively.

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Proof. By a change of counter

∞ (2m) ∞ X (−1)nH X (−1)n 1 M(m, r) = n−1 = H(2m) − n + r − 1 n + r − 1 n n2m n=1 n=1 ∞ n+1 (2m) ∞ n+1 X (−1) Hn X (−1) = − + n + r − 1 n2m(n + r − 1) n=1 n=1 ∞ r−1 1 X (−1)n+r 1 X (−1)n+r = −M(m, r − 1) + − (r − 1)2m n (r − 1)2m n n=1 n=1 ∞ 2m ∞ 1 X (−1)n+1 X (−1)j X (−1)n+1 − + . (r − 1)2m n (r − 1)2m+1−j nj n=1 j=2 n=1

From Lemma 1 and using the known results, (−1)r M(m, r) = −M(m, r − 1) − ln 2 + H r−1 − Hr−1 (r − 1)2m [ 2 ] 2m (14) ln 2 X (−1)j − + ζ¯(j). (r − 1)2m (r − 1)2m+1−j j=2

From (14) we have the recurrence relation

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

∞ (2m) X H Υ(m, r) = 2n (2n + r − 1)(2n + r) n=1 2m 1 X (−1)j = M(m, r) − H r−1 + ζ(j) , (15) 2(r − 1)2m 2 2j(r − 1)2m+1−j j=2

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

∞ (2m) X H Υ(m, 1) = 2n = ζ¯(2m + 1) + (2−1−2m − m)ζ(2m + 1) 2n(2n + 1) n=1 m−1 (16) X + ζ¯(2m) ln 2 + ζ¯(2j + 1)ζ¯(2m − 2j) , j=1 and for r = 0

∞ (2m) X H Υ(m, 0) = 2n = mζ(2m + 1) − ζ¯(2m) ln 2 + ln 2 2n(2n − 1) n=1 m−1 2m (17) X X − ζ¯(2j + 1)ζ¯(2m − 2j) − 2−jζ(j) . j=1 j=2

Proof. From

∞ n+1 (2m) ∞ (2m) X (−1) Hn X H M(m, r) = = 2n n + r (2n + r − 1)(2n + r) n=1 n=1 ∞ X 1 − . (2n + r − 1)(2n)2m n=1 Hence

∞ (2m) ∞ X H X 1 Υ(m, r) := 2n = M(m, r) + , (2n + r − 1)(2n + r) (2n + r − 1)(2n)2m n=1 n=1 then,

∞ 1 X 1 1 Υ(m, r) = M(m, r) + − (r − 1)2m 2n + r − 1 2n n=1 2m ∞ X (−1)j X 1 + 2j(r − 1)2m+1−j nj j=2 n=1 1 = M(m, r) − H r−1 2(r − 1)2m 2 2m X (−1)j + ζ(j) . 2j(r − 1)2m+1−j j=2

For r = 1

∞ (2m) X H 1 Υ(m, 1) = 2n = M(m, 1) + ζ(2m + 1) 2n(2n + 1) 22m+1 n=1

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

∞ (2m) 2m X H X Υ(m, 0) = 2n = M(m, 0) + ln 2 − 2−jζ(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}, 1 Z 1 x ln2m−1 x 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).

n+1 (2m) P∞ (−1) Hn Proof. From M(m, r) := n=1 n+r and by the use of the integral representation (3) ∞ 1 Z 1 ln2m−1 x X (−1)n+1(1 − xn) M(m, r) = dx (2m − 1)! 1 − x n + r 0 n=1 1 Z 1 ln2m−1 x = xΦ(−x, 1, 1 + r) − Φ(−1, 1, 1 + r) dx (2m − 1)! 0 1 − x 1 Z 1 x ln2m−1 x = 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 1 Z 1 ln2m−1 x ln(1 + x) 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 ln2m−1 x − dx = (2m − 1)! Li2m(1) = (2m − 1)! ζ(2m). 0 1 − x For r = 1 we obtain 1 Z 1 ln2m−1 x ln(1 + x) dx = −M(m, 1) − ζ(2m) ln 2, (2m − 1)! 0 x(1 − x) for m ≥ 2. An obvious result of the preceding two integrals yields the delightful identity 1 Z 1 ln2m−1 x ln(1 + x) dx = −ζ¯(2m + 1) (19) (2m − 1)! 0 x ∞ X 1 1 = −M(m, 0) − M(m, 1) = − (−1)n+1H(2m) + n n n + 1 n=1 The Wolfram on-line integrator yields no solution to these general integrals.

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Lemma 7. For r ∈ N0 \{0, 1},

1 Z 1 x2 ln2m−1 x 1 + r 2 + r Φ x2, 1, − Φ x2, 1, dx (2m − 1)! 0 2(1 − x) 2 2 1 = Υ(m, r) + ζ(2m) H r−1 − H r (20) 2 2 2

2 1+r where Υ(m, r) is given by (15) and Φ(x , 1, 2 ) is the Lerch transcendent. For r = 0 1 Z 1 ln2m−1 x 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 Z 1 ln2m−1 xtanh−1 x 1 + ln(1 − x2) dx = −Υ(m, 1) − ζ(2m) ln 2. (2m − 1)! 0 1 − x x 2 By addition we have the result

1 Z 1 ln2m−1 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 = M(m, 0) − M(m, 1) − + ln 2 ζ(2m) − ln 2 − 2−jζ(j) 22m+1 j=2 and by further simpliﬁcation

1 Z 1 (1 + x) ln2m−1 x − tanh−1 x dx (21) (2m − 1)! 0 x = Υ(m, 1) + Υ(m, 0) 2m+1 X = ζ(2m + 1) + ln 2 − 2−jζ(j) j=2 ∞ X 1 = . (22) (2 n + 1)(n + 1)2m n=0 2 It may also be shown that the alternating companion to (22) is,

∞ X (−1)n 1 Z 1 (1 − x) ln2m−1 x = − tanh−1 x dx (2 n + 1)(n + 1)2m (2m − 1)! x n=0 2 0 2m+1 X = ζ¯(2m + 1) − ln 2 + 2−jζ(j), j=2 where [z] is the integer part of z.

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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: 1 Z 1 (1 + x) ln2m−1 x − tanh−1 x dx (2m − 1)! 0 x ∞ 1 X 1 Z 1 = − xn ln2m−1 x dx (2m − 1)! (2 n + 1) n=0 2 0 ∞ X 1 = (2 n + 1)(n + 1)2m n=0 2 ∞ (2m) X 2H = 2n . (2n − 1)(2n + 1) n=1 For m = 3, we have ∞ (6) ∞ X 2H X 1 2n = (2n − 1)(2n + 1) (2 n + 1)(n + 1)6 n=1 n=0 2 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

∞ n+1 (2m) X (−1) Hn X(m, k, p) = (23) pn+k n=1 n k 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, ∞ n+1 (2m) k X (−1) Hn X k X(m, k, 1) = = M(m, 0) − (−1)1+r M(m, r) , (24) n+k r n=1 n( k r=1 where M(m, r) is given by (13). Proof. Consider the expansion

∞ n+1 (2m) ∞ n+1 (2m) X (−1) Hn X (−1) k!Hn X(m, k, 1) = = n+k n(n + 1) n=1 n k n=1 k ∞ k X Ω X Λr = (−1)n+1k!H(2m) + , n n n + r n=1 r=1

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where      n + r  (−1)r+1 k 1 Λ = lim = − , Ω = . (25) r k n→−r  Q  k! r k! n n + r  r=1 We can now express

∞ k X X (−1)r+1 k 1 X(m, k, 1) = (−1)n+1H(2m) − + n n + r r n n=1 r=1 ∞ (2m) k ∞ (2m) X Hn X k X Hn = − (−1)1+r . (26) n r n + r n=1 r=1 n=1

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

∞ n+1 (2m) X (−1) Hn X(m, k, 0) = n+k n=1 k k X k = (−1)1+rr M(m, r) , (27) r r=1 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

1 Z 1 x ln2m−1 x 1, 1; 2F1 − x dx (1 + k)(2m − 1)! 0 1 − x 2 + k; ζ(2m) 1, 1; = X(m, k, 1) − F − 1 , (28) 1 + k 2 1 2 + k; where X(m, k, 1) is given by (24),

1 Z 1 x ln2m−1 x 1, 2; 2F1 − x dx (1 + k)(2m − 1)! 0 1 − x 2 + k; ζ(2m) 1, 2; = X(m, k, 0) − F − 1 , (29) 1 + k 2 1 2 + k; where X(m, k, 0) is given by (27).

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Proof. From the identity (3), we can write

∞ n+1 (2m) Z 1 2m−1 ∞ n+1 n X (−1) Hn 1 ln x X (−1) (1 − x ) = − dx n+k (2m − 1)! 1 − x n+k n=1 n k 0 n=1 n k  1, 1;  x2F1 − x 1 Z 1 ln2m−1 x  2 + k;    =   dx . (1 + k)(2m − 1)! 0 1 − x  1, 1;  − F − 1 2 1 2 + k;

·, ·; where F · is the generalized hypergeometric function. Since 2 1 ·;

1 Z 1 ln2m−1 x 1, 1; 1, 1; 2F1 − 1 dx = −2F1 − 1 ζ(2m), (2m − 1)! 0 1 − x 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 Z 1 x ln5 x 1, 1; ζ(6) 1, 1; 2F1 − x dx = X(3, 4, 1) − 2F1 − 1 ; 720 0 1 − x 6; 5 6;

1 Z 1 ln5 x 12(1 + x)4 12 + 42x + 52x2 + 25x3 4 ln(1 + x) − 3 dx 720 0 1 − x x x 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

1 Z 1 x ln5 x 1, 2; ζ(6) 1, 2; 2F1 − x dx = X(3, 4, 0) − 2F1 − 1 720 0 1 − x 6; 5 6; 492155 13373ζ(2) 2471ζ(4) 40832 ln 2 = − − − − 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 The Wolfram on-line integrator, yields no solution to these general slow converging integrals.

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Remark 3. It appears that, for r ∈ N0, m ∈ N ∞ n+1 (2m) X (−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 ∞ X 1 1 K(m, r) + K(m, r + 1) = (−1)n+1H(2m) + n (n + r)2 (n + r + 1)2 n=1 1 (2) (2) = 2m H r − H r−1 4r 2 2 r 2m(−1) r + 1 − (−1) ln 2 + H r − Hr r2m+1 [ 2 ] 2m X (−1)j(2m + 1 − j) + ζ¯(j) (30) r2m+2−j j=2 and for r = 0, ∞ X 1 1 K(m, 0) + K(m, 1) = (−1)n+1H(2m) + n n2 (n + 1)2 n=1 = ζ¯(2m + 2) (31) 1 Z 1 ln2m−1 x Li (−x) = 2 dx . (32) (2m − 1)! 0 x Proof. ∞ X 1 1 K(m, r) + K(m, r + 1) = (−1)n+1H(2m) + n (n + r)2 (n + r + 1)2 n=1 and by a change of counter in the second sum, after simpliﬁcation we obtain ∞ X (−1)n+1 K(m, r) + K(m, r + 1) = n2m(n + r)2 n=1 ∞ ∞ 1 X (−1)n+1 2m X (−1)n+1 = + r2m (n + r)2 r2m+1 n + r n=1 n=1 2m ∞ X (−1)j(2m + 1 − j) X (−1)n+1 + r2m+2−j nj j=1 n=1 r 1 (2) (2) 2m(−1) 2m = 2m H r − H r−1 + 2m+1 ln 2 + H r − Hr − 2m+1 ln 2 4r 2 2 r [ 2 ] r 2m X (−1)j(2m + 1 − j) + ζ¯(j) r2m+2−j j=2

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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 Z 1 ln2m−1 x Li (−x) 2 dx (33) (2m − 1)! 0 x ∞ 1 X (−1)j+1 Z 1 = xj ln2m−1 x dx (2m − 1)! (j + 1)2 j=0 0 ∞ X (−1)j = = ζ¯(2m + 2) . (j + 1)2m+2 j=0

This gives the even zeta functions of alternating type and (19) gives the odd zeta functions of alternating type. Moreover

1 Z 1 ln2m−1 x Li (−x) lim 2 dx = 1 m→∞ (2m − 1)! 0 x From (30), for {m, r} = {3, 3},

31ζ(6) 5ζ(5) 7ζ(4) ζ(3) 2ζ(2) 4 ln 2 91 K(3, 3) + K(3, 4) = − + − + − + . 288 72 216 81 729 729 26244 It is possible to obtain a nice generalization of (33) as,

1 Z 1 ln2m−1 x Li (−x) p dx = ζ¯(2m + p) (2m − 1)! 0 x for p ∈ N, and is analogous to the identity (see [13])

Z 1 lnm x Li (x) p dx = (−1)mm! ζ(m + p + 1) 0 x References

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Author’s address: Victoria University, P. O. Box 14428, Melbourne City, Victoria 8001, Australia E-mail: anthony.sofo a vu.edu.au

Received: 14 September, 2015 Accepted for publication: 10 May, 2016 Communicated by: Attila Bérczes

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