Challenge

**Mathstud28** · June 15th 2008, 06:54 PM

We have some good integrators here...well...let see this one done. Prove

$\int_0^{\infty}\frac{x^m}{(a^n+x^n)^p}dx=\frac{(-1)^{p-1}\cdot\pi\cdot{a^{m+1-n\cdot{p}}}\Gamma\left[\frac{m+1}{n}\right]}{n\sin\left[\frac{(m+1)\pi}{n}\right](p-1)!\Gamma\left[\frac{m+1}{n-p+1}\right]}$

Good luck...

**topsquark** · June 15th 2008, 06:58 PM

Originally Posted by Mathstud28

We have some good integrators here...well...let see this one done. Prove

$\int_0^{\infty}\frac{x^m}{(a^n+x^n)^p}dx=\frac{(-1)^{p-1}\cdot\pi\cdot{a^{m+1-n\cdot{p}}}\Gamma\left[\frac{m+1}{n}\right]}{n\sin\left[\frac{(m+1)\pi}{n}\right](p-1)!\Gamma\left[\frac{m+1}{n-p+1}\right]}$

Good luck...

From the way your solution is written I am presuming m, n, and p are all positive integers?

-Dan

**TwistedOne151** · June 15th 2008, 07:32 PM

Well, right away I see you can make the substitution $x=au$ . Then we have
$\int_0^{\infty}\frac{x^m}{(a^n+x^n)^p}\,dx=\int_0^ {\infty}\frac{a^mu^m}{(a^n+a^nu^n)^p}a\,du$
$=\int_0^{\infty}\frac{a^{m+1}u^m}{a^{np}(1+u^n)^p} \,du$
$=a^{m+1-np}\int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du$
Which takes care of the term containing a. Proving that
$\int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du=\frac{(-1)^{p-1}\pi\Gamma\left(\frac{m+1}{n}\right)}{n\sin\left( \frac{(m+1)\pi}{n}\right)(p-1)!\Gamma\left(\frac{m+1}{n-p+1}\right)}$
will require a bit more thought. I'm thinking a beta function integral may be involved at some point.

--Kevin C.

**Mathstud28** · June 15th 2008, 08:43 PM

Originally Posted by topsquark

From the way your solution is written I am presuming m, n, and p are all positive integers?

-Dan

No integers just

$0<m+1<np$ Although you may be right due to the $(p-1)!$

It should also be noted that I do not have the solution to this, that is why I am so curious.

**TwistedOne151** · June 16th 2008, 07:52 AM

Next, I see trying to find
$\int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du$
by the substitution $x=1+u^n$ . Thus gives $dx=nu^{n-1}$ , and thus we see:
$\int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du=\frac{1}{ n}\int_1^{\infty}\frac{(x-1)^{\frac{m-n+1}{n}}}{x^p}\,du$ .

Next, try the substitution $u=\frac{1}{x}$ . This gives:
$\frac{1}{n}\int_1^{\infty}\frac{(x-1)^{\frac{m-n+1}{n}}}{x^p}\,du=\frac{1}{n}\int_1^0u^p\left(\fr ac{1}{u}-1\right)^{\frac{m-n+1}{n}}\left(-\frac{du}{u^2}\right)$ .
$=\frac{1}{n}\int_0^1u^{\frac{np-m-n-1}{n}}(1-u)^{\frac{m-n+1}{n}}\,du$ .
This is the Eulerian integral of the first kind, and so we have:
$\int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du=\frac{1}{ n}\int_0^1u^{\frac{np-m-n-1}{n}}(1-u)^{\frac{m-n+1}{n}}\,du$
$=\frac{1}{n}B(\frac{np-m-n-1}{n}+1,\frac{m-n+1}{n}+1)$
$=\frac{1}{n}B(\frac{np-m-1}{n},\frac{m+1}{n})$
Where $B(x,y)$ is the Beta function.

Now, $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$
so
$\int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du=\frac{1}{ n}\frac{\Gamma(\frac{np-m-1}{n})\Gamma(\frac{m+1}{n})}{\Gamma(\frac{np-m-1}{n}+\frac{m+1}{n})}$
$=\frac{1}{n}\frac{\Gamma(\frac{np-m-1}{n})\Gamma(\frac{m+1}{n})}{\Gamma(p)}$
$=\frac{1}{n(p-1)!}\Gamma(\frac{np-m-1}{n})\Gamma(\frac{m+1}{n})$

A little more to follow.

--Kevin C.

**TwistedOne151** · June 16th 2008, 08:13 AM

I'm getting a different result on the last steps:

The gamma function has a relationship known as Euler's reflection formula:
$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin{\pi{z}}}$ .
So:
$\Gamma(z)=\frac{\pi}{\sin(\pi{z})\Gamma(1-z)}$
and so:
$\Gamma(\frac{np-m-1}{n})=\frac{\pi}{\sin(\pi\frac{np-m-1}{n})\Gamma(1-\frac{np-m-1}{n})}$
$=\frac{\pi}{\sin\left(\pi{p}-\pi\frac{m+1}{n}\right)\Gamma(\frac{n+m+1-np}{n})}$
$=\frac{(-1)^p\pi}{\sin\left(-\frac{(m+1)\pi}{n}\right)\Gamma(\frac{n+m+1-np}{n})}$
$=\frac{(-1)^{p-1}\pi}{\sin\left(\frac{(m+1)\pi}{n}\right)\Gamma(\ frac{n+m+1-np}{n})}$

This combines with the earlier results to give:
$\int_0^{\infty}\frac{x^m}{(a^n+x^n)^p}\,dx=\frac{(-1)^{p-1}\pi{a}^{m+1-np}\Gamma\left(\frac{m+1}{n}\right)}{n\sin\left(\f rac{(m+1)\pi}{n}\right)(p-1)!\Gamma\left({\color{red}\frac{n+m+1-np}{n}}\right)}$ .

I'm not sure why I obtained a different value for the arguement of that gamma.

--Kevin C.

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