Definite Integration

**dekar** · July 24th 2007, 07:53 AM

Hi,

Could somebody please help me in getting a closed form expression for integrating the following expression:

(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

**galactus** · July 24th 2007, 08:15 AM

It appears to be hypergeometric. I ran it through Maple and it gave me.

$\int{(1+x^{m})^{n}}dx=xhypergeom\left([\frac{1}{m},-n],[1+\frac{1}{m}],-x^{m}\right)$ .

**dekar** · July 24th 2007, 08:19 AM

Thanks Galactus,

Can you please help me in getting the expansion of this function?

Thanks

Dekar

**topsquark** · July 24th 2007, 08:41 AM

Originally Posted by dekar

Thanks Galactus,

Can you please help me in getting the expansion of this function?

Thanks

Dekar

See here.

-Dan

**ThePerfectHacker** · July 24th 2007, 08:41 AM

Originally Posted by dekar

Hi,

Could somebody please help me in getting a closed form expression for integrating the following expression:

(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,
$\frac{1}{m}\int_0^1 (1+t)^n t^{-1 +\frac{1}{m}}dt= \frac{1}{m}\mbox{B}\left( n+1, \frac{1}{m} \right)$

**dekar** · July 24th 2007, 08:57 AM

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

**ThePerfectHacker** · July 24th 2007, 09:00 AM

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

Yes, I made a mistake.
You can see more here.

**dekar** · July 24th 2007, 09:06 AM

So isn't there a better (a less messy) way to arrive at the closed form solution other than that hypergeometric?

I need it a program friendly way to code this.

Any help is highly appriciated.

Thanks

Dekar

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