Mathematical Communications 13(2008), 303-309 303

Integral identities for sums

Anthony Sofo∗

Abstract. We consider some finite binomial sums involving the derivatives of the binomial coefficient and develop some integral iden- tities. In particular cases it is possible to express the sums in closed form.

Key words: integral representations, binomial coefficients, identi- ties, harmonic numbers

AMS subject classifications: 05A10, 11B65, 05A19

Received June 11, 2008 Accepted October 7, 2008

1. Introduction

The Beta function 1 β−1 Γ(α)Γ(β) B (α, β)= uα−1 (1 − u) du = for α, β > 0 Γ(α + β) 0 and the Gamma function ∞ Γ(α)= uα−1e−udu 0 are useful tools that are employed in the summation of series. In this paper we will develop integral identities for sums of the form

p p dq nktn Q a, j n dj q ( ) n=1 where an j −1 Q a, j + ( )= j

(q) dq th is the binomial coefficient and let Q (a, j)= dj q (Q(a, j)) be the q derivative of the reciprocal binomial coefficient. There has recently been renewed interest in the ∗School of Computer Science and Mathematics, Victoria University, PO Box 14428, Melbourne City, VIC 8001, Australia, e-mail: [email protected] 304 A. Sofo study of series involving binomial coefficients and a number of authors have obtained either closed form representation or integral representation for some of these series. The interested reader is referred to [1, 2, 3, 4, 5, 9, 10, 11, 12] and references therein. The following Lemma and Theorem are the main results presented in this paper.

2. Integral identity

The following Lemma deals with the derivatives of binomial coefficients. First we define N = {0, 1, 2, 3, ...} and note the polygamma functions ψ(k) (z) , k ∈ N are defined by dk+1 dk Γ (z) ψ(k) (z):= log Γ (z)= dzk+1 dzk Γ(z) 1 k [log(t)] tz−1 = − dt, 1 − t 0 and ψ(0) (z)=ψ (z) , denotes the digamma function, defined by d Γ (z) ψ (z)= log Γ (z)= . dz Γ(z) We also recall [7] the series representation for ψ (z) ∞ ψ z 1 − 1 − γ, ( )= r r z r=0 +1 + n 1 where γ = lim r − log (n) = −ψ (1) ≈ 0.5772156649015328606065 is the n→∞ r=1 Euler-Mascheroni constant. ∞ ζ q, b 1 ( )= k b q k=0 ( + ) is the generalised Zeta function where any term with k + b = 0 is excluded. We also define the generalised Harmonic number in power k as n H(k) 1 n = jk j=1 and for convenience we put for i =1, 2, 3, 4, ... diP di an P (i) a, j 1 ( )= dj i = dj i r j (1) r=1 + an i 1 =(−1) i! i+1 r=1 (r + j) =(−1)ii![ζ (i +1,j+1)− ζ (i +1,j+1+an)] . Integral identities for sums 305

Lemma 1. a j ≥ ,n Let be a positive integer with 0 a positive integer and an j −1 Q a, j Q(0) a, j + . ( )= ( )= j Then,   −Q(a, j)P (a, j), where dQ  an Q(1) a, j P (a, j)=P (0)(a, j)= 1 for j>0 , ( )= dj =  r+j (2)  r=1 = −Q(a, j)[ψ (j +1+an) − ψ(j +1)] and for λ ≥ 2 dλQ Q(λ) a, j ( )= dj λ (3) λ−1 λ − − 1 Q(ρ) a, j P (λ−1−ρ) a, j . = ρ ( ) ( ) ρ=0

Proof. Let −1 an + j Γ(an +1)Γ(j +1) Γ(an +1) Q(a, j)= = = , (4) j Γ(an + j +1) an (r + j) r=1 taking logs of both sides and differentiating with respect to j we obtain the result (2). Now from (2) and for λ ≥ 2 dλQ dλ−1 λ−1 λ − Q(λ) a, j Q(λ) a, j −QP − 1 Q(ρ)P (λ−1−ρ) ( )= dj λ = ( )=dj λ−1 ( )= ρ ρ=0 and P (λ−1−ρ) (a, j)isgivenby(1). ✷ We list the following −1 an (1) an + j 1 Q (a, j)=− , j r + j  r=1  2 an j −1 an an Q(2) a, j +  1 1  ( )= j r j + 2 r=1 + r=1 (r + j)   an 3 an −1 1 1 an j  r+j +2 (r+j)3  (3) +  r=1 r=1  Q (a, j)=−  an an  j 1 1 +3 (r+j)2 r+j r=1 r=1 and   an an 2 1 1  6 (r+j)2 r+j   r=1 r=1  −1  an an an 2  (4) an + j   Q (a, j)=  1 1 1  . (5) j  +8 (r+j)3 r+j +3 (r+j)2   r=1 r=1 r=1   an 4 an  1 1 + r+j +6 (r+j)4 r=1 r=1 306 A. Sofo

In the special case when a =1andj =0wemaywrite (1) (1) Q (1, 0) = −Hn , (6a)  2 (2) (1) (2) Q (1, 0) = Hn + Hn , (6b)  3 (3) (1) (1) (2) (3) Q (1, 0) = Hn +3Hn Hn +2Hn (6c) and  4  2  2 (4) (1) (1) (2) (1) (3) (2) (4) Q (1, 0) = Hn +6 Hn Hn +8Hn Hn +3 Hn +6Hn , (7)

(k) where Hn are the generalised Harmonic numbers. We now state the following theorem. Theorem 1. Let a be a positive real number, t ∈ R , p =1, 2, 3, ..., k = 1, 2, 3, ..., q =1, 2, 3... and j ≥ 0, then p p V a, j, k, q, t nktn Q(q) a, j p ( )= n ( )(8) n=1   k p k 1 q tr r xar − x j−1 txa p−r − x q−1 dx = r ! r (1 ) (1+ ) (log(1 )) r=1 0   k p k 1 j tr r xar − x j−1 txa p−r − x q dx + r ! r (1 ) (1+ ) (log(1 )) r=1 0 where Q(q)(a, j) is defined by (3), and   r k r 1 − µ r − µ k r = r ( 1) µ ( ) ! µ=0 areStirlingnumbersofthesecondkind. Proof. Consider n p p t p n p tn Γ(j)Γ(an +1) = j an j n an j n=0 + n=0 Γ( + +1) j p p j tn B an ,j = n ( +1 ) n=0 p p 1 j tn xan − x j−1 dx = n (1 ) n=0 0 where B (·, ·) is the classical Beta function and Γ (·) is the Gamma function. By an allowable change of sum and integral we have n p p t 1 n j−1 p = j (1 − x) (1+txa) dx. an j n=0 + 0 j Integral identities for sums 307

d Now apply consecutively k- times the operator t dt (·), so that n p p nt 1 n j−1 p−1 = jpt xa (1 − x) (1+txa) dx, an j n=1 + 0 j

2 n p p n t 1 n j−1 p−1 = jpt xa (1 − x) (1+txa) dx an j n=1 + 0 j 1 j−1 p−2 +jp(p − 1) t2 x2a (1 − x) (1+txa) dx, ...... 0 ...... p k n   p n t k 1 n p k j−1 p−r = j tr r! xar (1 − x) (1+txa) dx. an j r r n=1 + r=1 0 j

(q) dq Applying the operator Q (a, j)= dj q (Q(a, j)) as defined in Lemma 1 we have that p p nk tn Q(q) a, j n ( ) n=1   k p k 1 q tr r xar − x j−1 txa p−r − x q−1 dx = r ! r (1 ) (1+ ) (log (1 )) r=1 0   k p k 1 j tr r xar − x j−1 txa p−r − x q dx + r ! r (1 ) (1+ ) (log (1 )) r=1 0   k ✷ where r are Stirling numbers of the second kind. Some interesting results follow for particular parameter values, which involve Harmonic numbers. Consider the following corollary. Corollary 1. Let a =1,t= −1.Then

p p dq nk − n Q ,j ( 1) n dj q ( (1 )) n=1     k p k r r s − p q − q − r r − s = !( 1) ( 1) r ! r ( 1) s q+1 (9) r=1 s=0 (p + j − s) 308 A. Sofo

Proof. From Theorem 1

p p dq nk − n Q ,j ( 1) n dj q ( (1 )) n=1   k p k 1 q − r r xr − x p+j−r−1 − x q−1 dx = ( 1) r ! r (1 ) (log (1 )) r=1 0   k p k 1 j − r r xr − x p+j−r−1 − x q dx, + ( 1) r ! r (1 ) (log (1 )) r=1 0 evaluating the integrals we have   k p k r r q − q−1 − r r − s 1 = !( 1) ( 1) r ! r ( 1) s p j − r s q r=1 s=0 ( + + )   k p k r r jq − q − r r − s 1 , + !( 1) ( 1) r ! r ( 1) s q+1 r=1 s=0 (p + j − r + s) collecting like terms and renaming the counter s we obtain (9). ✷ Remark 1. For q =4,j=0, k =3and using (7)    4  2 p (1) (1) (2) (1) (3) Hn Hn Hn Hn Hn 3 n p  +6 +8  n (−1)   2  n (2) (4) n=1 +3 Hn +6Hn   −1296 + 9504p − 30235p2 + 54654p3 − 61280p4  +44016p5 − 20255p6 + 5770p7 − 926p8 +64p9    . =24 4 4 4  p (p − 1) (p − 2) (p − 3)

Moreover,    4  2 ∞ p (1) (1) (2) (1) (3) Hn Hn Hn Hn Hn 3 n p  +6 +8  n (−1)   2  n (2) (4) p=4 n=1 +3 Hn +6Hn 973 =24ζ (4) + . 2

3. Conclusion

We have applied the method of integral representation for binomial sums that also involve Harmonic numbers and in some cases expressed them in closed form. Integral identities for sums 309

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