1. ## Challenge

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

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

Good luck...

2. Originally Posted by Mathstud28
We have some good integrators here...well...let see this one done. Prove

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

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

-Dan

3. ## First step

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

--Kevin C.

4. 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

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

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

5. ## Next steps

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

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

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

A little more to follow.

--Kevin C.

6. ## I'm not getting the same result

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

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

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

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

--Kevin C.