# Definite Integration

• Jul 24th 2007, 08:53 AM
dekar
Definite Integration
Hi,

(1+(x^m))^n

limits of integration are from 0 to 1. Integration is to be done with respect to x and m and n are positive real numbers.

Is it to do with the incomplete beta function, but the upper limit is coming out to be -1, which should lie in the interval 0 to 1.

Thanks in anticipation.

Dekar
• Jul 24th 2007, 09:15 AM
galactus
It appears to be hypergeometric. I ran it through Maple and it gave me.

$\int{(1+x^{m})^{n}}dx=xhypergeom\left([\frac{1}{m},-n],[1+\frac{1}{m}],-x^{m}\right)$.
• Jul 24th 2007, 09:19 AM
dekar
Thanks Galactus,

Thanks

Dekar
• Jul 24th 2007, 09:41 AM
topsquark
Quote:

Originally Posted by dekar
Thanks Galactus,

Thanks

Dekar

See here.

-Dan
• Jul 24th 2007, 09:41 AM
ThePerfectHacker
Quote:

Originally Posted by dekar
Hi,

(1+(x^m))^n

limits of integration are from 0 to 1. Integration is to be done with respect to x and m and n are positive real numbers.

I think the following works for $n,m>0$.

$\int_0^1 (1+x^m)^n dx$
Let $t=x^m \implies t' = mx^{m-1}$
Thus,
$\frac{1}{m}\int_0^1 (1+t)^n t^{-1 +\frac{1}{m}}dt= \frac{1}{m}\mbox{B}\left( n+1, \frac{1}{m} \right)$
• Jul 24th 2007, 09:57 AM
dekar
Hi,

It won't work because it is (1+t), to have it to get reduced to the beta function we need (1-t),

Thats why I guess, It may go to Incomplete Beta Function if we substitute say p=-t at this stage, but the upper limit then becomes -1.

Dekar
• Jul 24th 2007, 10:00 AM
ThePerfectHacker
Quote:

Originally Posted by dekar
Hi,

It won't work because it is (1+t), to have it to get reduced to the beta function we need (1-t),

Thats why I guess, It may go to Incomplete Beta Function if we substitute say p=-t at this stage, but the upper limit then becomes -1.

Dekar