Mar 13, 2015 - Abstract. This purpose of this paper is to note an interesting identity derived from an integral in Gradshteyn and Ryzhik using techniques from ...
An Interesting Identity Brett Pansano Northwest Arkansas Community College March 13, 2015 Abstract This purpose of this paper is to note an interesting identity derived from an integral in Gradshteyn and Ryzhik using techniques from George Boros’(deceased) Ph.D thesis. The idenity equates a sum to a product by evaluating an integral in two di¤erent ways. A more general form of the idenity is left for further investigation.
1
Introduction
Theorem 1 Let n be non-negative integer. Then n X
k
( 1)
k=0
n 1 k nk + n
= nn n!
1
Proof. We will evaluate the integral Z 1 (1 0
n Y
k=0
1 nk + n
1
:
n
x)
dx
1 n
x
in two di¤erent ways. We start with n
(1
x) =
n X
k
( 1)
k=0
n k x ; k 1
which follows directly from the binomial theorem. Dividing both sides by x n yields n n X 1 (1 x) k n xk n ; = ( 1) 1 k xn k=0 so that Z 1 (1 0
n
x) x
1 n
dx =
n X
k=0
k
( 1)
n k
Z
1
xk
0
1 n
dx =
n X
k=0
1
k
( 1)
n n k nk + n
1
:
R1
But the integral
(1 x)n 1
0
xn
Z
dx can be evaluated using the fact that
1
xm
1
(1
n 1
x)
dx = B (m; n)
0
where B (m; n) is the beta function. We get Z 1 n (1 x) 1 dx = B n + 1; 1 1 n n x 0 since
= ::: = n so that (since
n
1 n
1 n
n
1 n
1 n
;
1 n
1
1 n
n
n
1
2
1 n
1 n
n
1 n
2
(n + 1) = n!) Z
1
=
n
(1
x) x
0
=
1 n
n+1
n+1
;
(m) (n) ; (m + n)
B(m; n) = where is the gamma function. We next note that 1 1 = n+1 n+2 n n 1 = n+1 n
1 n
(n + 1) 1 n + 2 n1
=
1 n
dx n!
n+1
1 n
n! n+1 Y
k
n n1 n+1 Y
= 1 n
n
1
(nk
1 n
n
1
1 n
1) :
k=1
k=1
Thus, n X
k
( 1)
k=0
n n k nk + n
1
=
n Y nn+1 n! 1 n = n n! n+1 nk + n Y k=0 (nk 1)
1
:
k=1
so that, n X
k=0
k
( 1)
n 1 k nk + n
1
=
nn n! n+1 Y
(nk
k=1
2
= nn n! 1)
n Y
k=0
1 nk + n
1
:
Corollary 2 Let a be a real number. Then 1.
n X
k
( 1)
k=0
2.
n X
k
( 1)
k=0
3.
n X
k
( 1)
k=0
n 1 k nk + an
n 1 k ank + n
1
= nn n!
k=0
n
1
n 1 k ank + an
= (an) n!
1 nk + an
n Y
k=0
n
1
n Y
= (an) n!
n Y
k=0
1
1 ank + n
1
1 ank + n
1
Proof. This is clear. The following identity was con…rmed via Mathematica 3 Y d 1 n4 da nk+an 1 : (an 1)2 (n+an 1)2 (2n+an 1)2 (3n+an 1)2 by noting that k=0
d 1 da nk + an
1
1
=
(nk + an
2 n:
1)
The following might be interesting to investigate: Conjecture 3 Let n be a natural number. Suppose f and g are di¤ erentiable functions. Then n X
n
k
( 1)
k=0
Y n 1 1 = nn n! : k f (n) k + g (n) f (n) k + g (n) k=0
References [1] I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, Academic Press, INC., 1980 [2] G. Boros and V. Moll, Irresistible Integrals, Symbolics, Analysis, and Experiments in the Evaluation of Integrals, Cambridge University Press, 2004
