Double Integral Representation of Sums
Anthony Sofo
School of Computer Science and Mathematics Victoria University, PO Box 14428 Melbourne City, VIC 8001, Australia
Abstract. We consider sums involving the product binomial coefficient and polynomial terms and develop some double integral identities. In particular cases it is possible to express the sums in closed form, recover some known results and produce new identities.
Mathematics Subject Classification: 05A10, 11B65, 05A19
Keywords: Double Integrals, Binomial coefficients, Combinatorial identi- ties
1. Introduction
In this paper we will develop integral identities for sums of the form X∞
n=0
tn ¡n+m−1
n
¢ (an+j+ 1)k+1
µ an+j bn
¶.
We recover and extend some results published by Batir [1] and other authors.
There has recently been renewed interest in the study of series involving bi- nomial coefficients and a number of authors have obtained either closed form representation or integral representation for some particular cases of these se- ries. The interested reader is referred to [1, 2, 3, 4, 5, 6, 7, 8] and references therein. The following Lemma and Theorem are the main results presented in this paper.
2. the main results
The following Lemma deals with the integral representation of (k+α)1 j.
Lemma 1. Let k+α≥0, k, α∈R and j ≥0 then,
(2.1) 1
(k+α)j =
⎧⎨
⎩
1 (j−1)!
R∞ 0
yj−1e−y(k+α)dy, for j ≥1
1, for j = 0
.
Proof. The proof follows easily by the application of integration by parts. ¤ We now state the following theorem.
Theorem 1. Let a be a positive real number, |t| ≤ 1 , j ≥ 0, k ≥ 0, and m ≥1, then
(2.2) S(a, j, k, m, t) = X∞ n=0
tn ¡n+m−1
n
¢ (an+j+ 1)k+1
à an+j j
!
=
⎧⎪
⎪⎨
⎪⎪
⎩
1 (k−1)!
R∞ y=0
R1 x=0
(1−x)j yk−1 e−y(j+1)
(1−txae−ay)m dxdy, for k≥1 R1
0
(1−x)j
(1−txa)mdx, for k = 0
.
Proof. Consider X∞
n=0
tn ¡n+m−1
n
¢ (an+j+ 1)k+1
à an+j j
! = X∞ n=0
tn ¡n+m−1
n
¢ Γ(j + 1)Γ(an+ 1)
(an+j+ 1)k Γ(an+j+ 2)
= X∞ n=0
tn ¡n+m−1
n
¢
(an+j+ 1)k B(an+ 1, j+ 1) (2.3)
whereB(α, β) = R1 0
xα−1(1−x)β−1 dx,is the classical Beta function andΓ(α) = R∞
0
uα−1e−udu is the Gamma function. From (2.3) we have X∞
n=0
tn ¡n+m−1
n
¢
(an+j + 1)k B(an+ 1, j+ 1) = X∞ n=0
tn ¡n+m−1
n
¢ (an+j+ 1)k
Z1
0
xan(1−x)jdx
= X∞ n=0
tn ¡n+m−1
n
¢ (k−1)!
Z∞
0
yk−1e−y(j+1)e−anydy Z1
0
xan(1−x)jdx
upon utilising Lemma 1. Now changing the order of integration and summation we have,
1 (k−1)!
Z∞
0
Z1
0
(1−x)jyk−1e−y(j+1) X∞ n=0
µn+m−1 n
¶¡
txae−ay¢n
dxdy,
= 1
(k−1)!
Z∞
y=0
Z1
x=0
(1−x)j yk−1 e−y(j+1)
(1−txae−ay)m dxdy, for k≥1.
The case of k= 0 follows in a similar way so that
S(a, j,0, m, t) = X∞ n=0
tn ¡n+m−1
n
¢ (an+j+ 1)
à an+j j
!
= Z1
0
(1−x)j (1−txa)mdx.
¤ An alternative representation of (2.2) is given in the following corollary.
Corollary 1. Let the conditions of Theorem 1 hold then : X∞
n=0
tn ¡n+m−1
n
¢ (an+j+ 1)k+1
à an+j j
!
= amt k!
Z∞
y=0
Z1
x=0
(1−x)j xa−1 yk e−y(a+j+1)
(1−txae−ay)m+1 dxdy, for k ≥0.
(2.4)
Proof. From Theorem 1 X∞
n=0
tn ¡n+m−1
n
¢ (an+j+ 1)k+1
à an+j j
! =a X∞ n=0
tn ¡n+m−1
n
¢ Γ(j + 1)nΓ(an) (an+j + 1)k+1 Γ(an+j+ 1)
= a X∞ n=0
ntn ¡n+m−1
n
¢
(an+j+ 1)k+1 B(an, j+ 1)
= a X∞ n=0
ntn ¡n+m−1
n
¢ (an+j+ 1)k+1
Z1
0
xan−1(1−x)jdx
= a k!
Z∞
0
Z1
0
(1−x)jyke−y(j+1) x
X∞ n=0
n
µn+m−1 n
¶¡
txae−ay¢n
dxdy
= a k!
Z∞
0
Z1
0
(1−x)jyke−y(j+1)mtxae−ay
x(1−txae−ay)m+1 dxdy for k ≥0,
and rearranging the integrand we obtain (2.4). ¤
Remark 1. For a = 1, t = 1, j =p, p a positive integer, m =p+ 3, and k a positive integer
S(1, p, k, p+ 3,1) = X∞ n=0
¡n+p+2
n
¢ (n+p+ 1)k+1
à n+p p
!
= 1
(k−1)!
Z∞
y=0
Z1
x=0
(1−x)p yk−1 e−y(p+1) (1−xe−y)p+3 dxdy
= 1
(p+ 2) (p+ 1)
£ζ(k) +ζ(k−1)−Hp(k)−Hp(k−1)¤ , where ζ(z) is the Zeta function and Hp(k) =Pp
i=0 1
ik are the generalised Har- monic numbers of order k.
The next Theorem deals with a more general version of the above Theorem.
Theorem 2. Letaandbbe a positive real number with(a−b)≥0, j ≥0, k ≥ 0, t∈R and m≥1. For
¯¯
¯tbb(a−aab)a−b
¯¯
¯≤1 then
S(a, b, j, k, m, t) = X∞ n=0
tn ¡n+m−1
n
¢ (an+j + 1)k+1
à an+j bn
!
(2.5) =
⎧⎪
⎪⎨
⎪⎪
⎩
1 (k−1)!
R∞ y=0
R1 x=0
(1−x)j yk−1 e−y(j+1)
(1−txb(1−x)a−be−ay)mdxdy, for k ≥1 R1
0
(1−x)j
(1−txb(1−x)a−b)mdx, for k = 0 .
Proof. Consider
X∞ n=0
tn ¡n+m−1 n
¢ (an+j+ 1)k+1
à an+j bn
! = X∞ n=0
tn ¡n+m−1 n
¢ Γ((a−b)n+j+ 1)Γ(bn+ 1) (an+j+ 1)k Γ(an+j+ 2)
= X∞ n=0
tn ¡n+m−1
n
¢
(an+j+ 1)k B(bn+ 1,(a−b)n+j+ 1).
Now
[∞ n=0
tn n+m−1 n
B(bn+ 1,(a−b)n+j+ 1)
(an+j+ 1)k =
[∞ n=0
tn n+m−1 n
(an+j+ 1)k
]1
0
xbn(1−x)(a−b)n+jdx
= X∞ n=0
tn ¡n+m−1
n
¢ (k−1)!
Z∞
0
yk−1e−y(j+1)e−anydy Z1
0
xbn(1−x)(a−b)n+jdx upon utilising Lemma 1. By an allowable change of the order of integration and summation we have,
1 (k−1)!
Z∞
0
Z1
0
(1−x)jyk−1e−y(j+1) X∞ n=0
µn+m−1 n
¶ ³txb(1−x)(a−b)e−ay´n
dxdy,
= 1
(k−1)!
Z∞
y=0
Z1
x=0
(1−x)j yk−1 e−y(j+1)
³
1−txb(1−x)(a−b)e−ay´mdxdy, fork ≥1.
Similarly for k = 0,
S(a, b, j,0, m, t) = X∞ n=0
tn ¡n+m−1
n
¢ (an+j+ 1)
à an+j bn
!
= Z1
0
(1−x)j
³
1−txb(1−x)a−b´mdx.
¤
Remark 2. For the special case a= 2, b = 1, j = 0, k ≥1, m≥1 and |t|<4 we have that
S(2,1,0, k, m, t) = X∞ n=0
tn ¡n+m−1
n
¢ (2n+ 1)k+1
à 2n
n
!
= 1
(k−1)!
Z∞
y=0
Z1
x=0
yk−1 e−y
(1−tx(1−x)e−2y)mdxdy
= Z1
0
k+1Fk
" 1
2,12,12, . . . ,12, m
3
2,32,32, . . . , 32
¯¯
¯¯
¯
¡x−x2¢ t
# dx
= k+2Fk+1
" 1
2,12,12, . . . ,12, m
3
2,32,32, . . . , 32
¯¯
¯¯
¯ t 4
#
where the generalised hypergeometric representation pFq [·,·], is defined as
pFq
"
a1, a2, ..., ap
b1, b2, ..., bq
¯¯
¯¯
¯t
#
= X∞ n=0
(a1)n (a2)n...(ap)n tn (b1)n (b2)n...(bq)n n!
and (w)α = w(w+ 1) (w+ 2)...(w+α−1) = Γ(w+α)Γ(w) is Pochhammer’s sym- bol.
For k = 1, t= 1andm = 1we recover Batir’s [1] result X∞
n=0
1 (2n+ 1)2
µ 2n n
¶ = 8 3G−π
3 ln³ 2 +√
3´
= Z∞
y=0
Z1
x=0
e−y
(1−x(1−x)e−2y)dxdy.
where G is known as Catalan’s constant.
For k = 0, S(2,1,0,0, m, t) =
X∞ n=0
tn ¡n+m−1
n
¢ (2n+ 1)
µ 2n n
¶ = Z1
0
dx
(1−tx(1−x))m = 2F1
"
1, m
3 2
¯¯
¯¯
¯ t 4
# ,
in particular for m= 5andt =−54
X∞ n=0
¡−54
¢n ¡n+4
n
¢ (2n+ 1)
µ 2n n
¶ = Z1
0
¡ dx
1 +54x(1−x)¢5 = 2F1
"
1,5
3 2
¯¯
¯¯
¯− 5 16
#
= 2
583443 (
102879 + 2048√ 105 ln
Ã13 +√ 105 8
!) .
3. Conclusion
We have applied the method of integral representation for binomial sums, in some cases expressed them in closed form, and have extended some published results.
References
[1] Batir, N. Integral representations of some series involving ¡2k
k
¢−1
k−n and some related series.Appl. Math. Comp. 147(2004), 645-667.
[2] Batir, N. On the series P∞
k=1
¡3k
k
¢−1
k−nxk .Proceedings of the Indian Academy of Sci- ences: Mathematical Sciences 115(4) (2005), 371-381.
[3] Borwein, J. M. and Girgensohn, R. Evaluations of binomial series.Aequationes Mathe- maticae,70,(2005), 25-36.
[4] Sherman, T. Summation of Glaisher and Apéry-like series, available at http://math.edu.arizona.edu/~ura/001/sherman.travis/series.pdfs.
[5] Sofo, A. Computational techniques for the summation of series, Kluwer Acad- emic/Plenum Publishers, 2003.
[6] Sofo, A. General properties involving reciprocals of binomial coefficients, Journal of Integer Sequences,9(2006), Article 06.4.5.
[7] Sofo, A. Sums of reciprocals of triple binomial coefficients, Int. Journal of Math. and Math Sci., (2008), doi:10.1155/2008.
[8] Sofo, A. Some more identities involving rational sums,Applicable Analysis Discrete Math- ematics,2(2008), 56-66.
Received: June, 2008