# Definite integral formula from wikipedia

I need to solve a certain definite integral, and several places (for example Wikipedia) I've come across the following formula:

$\displaystyle\int_0^{\infty}\frac{x^mdx}{(x^n+a^n)^r}=\frac{(-1)^{r-1}\pi\ a^{m+1-nr}\ \Gamma\big((m+1)/n\big)}{n\sin((m+1)\pi/n)(r-1)!\ \Gamma\big((m+1)/n-r+1\big)}$

with the sole condition $0<m+1<nr$. However this is seems to me undefined for $(m+1)/n=3,5,7,\ldots$ because of the sine, which are exactly the cases I need. Since this is not contained in the conditions, am I mistaken? And if not how would you go about solving the integral?

As an example mathematica gives me the answer $\frac{\Gamma(A-3)}{\Gamma(A)}$ when I use $m=5,\ a=1,\ n=2,\ r=A$, which is one of the cases I need.

You are not mistaken - rather, Wikipedia does not give the answer in its simplest form. It is actually $$\int_{0}^{\infty}\frac{x^mdx}{(x^n+a^n)^r}=a^{m+1-nr}\frac{\Gamma\left(\frac{m+1}{n}\right)\Gamma\left(r-\frac{m+1}{n}\right)}{n\,\Gamma(r)}.\tag{1}$$ The formula you have seen before is obtained from this one using gamma function reflection relation $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}$ with $z=r-\frac{m+1}{n}$. However, (1) is ok for your parameter values.
You can consider this expression valid as a limit. Let's denote the RHS of your expression $I$. Then for e.g. $r=10,m=3,n=0.8,a=1$ you have: $$\int_0^\infty \frac{x^mdx}{(x^n+a^n)^r}=\lim_{m\to3}I=\frac{1}{504}$$, although substituting $m=3$ directly wouldn't work.
• No, the point is that as $m\to3$, $\sin((m+1)\pi/n)\to0$. Thus you have a $0$ in denominator. But at the same time, $\Gamma$ function in that same denominator becomes infinite. So the limit is needed to solve the $0\cdot\infty$ indeterminate, which appears to be finite. – Ruslan May 9 '13 at 17:18