How do I evaluate the following integral $\int_{-\infty}^\infty\mathop{dx} \frac{x^n}{(x^2+a^2)^m}$?

Question

I am interested in the following integral $$\int_{-\infty}^\infty\mathop{dx} \frac{x^n}{(x^2+a^2)^m},$$ given that $m>n/2$ (this is just what I wrote so that the integral converges. If this is not exactly true, just assume that $m$ and $n$ are such that the integral converges.) It is known that $m$ and $n$ are rational numbers and $a$ is a real constant. Any body has any idea? I felt that contour integral might be a good bet. However, the problem is that for non integer $m$, there are branch points.

Answer 1

It looks very like a loop integral in physics, that we do it like this(I hope it wll help): when n is odd, it is 0, when n is even, then $$\int _{-\infty}^{\infty}dx\frac{x^n}{(x^2+a)^m}=\int _0^{\infty}d(x^2)\frac{(x^2)^{\frac{n-1}{2}}}{(x^2+a)^m}$$ then, let $y=\frac{a}{x^2+a}$, it is $$\int _0^1 a y^{-2}dy\frac{(\frac{a}{y}(1-y))^{\frac{n-1}{2}}}{(\frac{a}{y})^m}=a^{\frac{n-1}{2}+1-m}\int _0^1 dy^{m-2-\frac{n-1}{2}}(1-y)^{\frac{n-1}{2}}$$ it is a beta-function so is $$a^{\frac{n-1}{2}+1-m} \frac{\Gamma (m-1-\frac{n-1}{2}) \Gamma (\frac{n-1}{2}+1)}{\Gamma (m)}$$, I hope I don't make mistakes.

Answer 2

Assume $a\neq0$, $m-\frac{n+1}2>0 $ and $n>-1$. Then

$$ \int_{-\infty}^\infty \frac{x^n}{(x^2+a^2)^m}\mathop{dx}=\frac{(e^{i\pi n}+1)}2\:|a|^{n-2m-1}\:\frac{\Gamma\left(m-\frac{n+1}2\right)\Gamma\left(\frac{n+1}2\right)}{\Gamma\left(m\right)}. \tag1 $$

Proof. Using the Euler beta function $B$, you may write $$ \begin{align} &\int_{-\infty}^\infty \frac{x^n}{(x^2+a^2)^m}\mathop{dx}\\\\ &=\int_{-\infty}^0 \frac{x^n}{(x^2+a^2)^m}\mathop{dx}+\int_0^\infty \frac{x^n}{(x^2+a^2)^m}\mathop{dx}\\\\ &=\int_0^{\infty} \frac{(-x)^n}{(x^2+a^2)^m}\mathop{dx}+\int_0^\infty \frac{x^n}{(x^2+a^2)^m}\mathop{dx}\\\\ & =(e^{i\pi n}+1)\int_0^\infty \frac{x^n}{(x^2+a^2)^m}\mathop{dx}\\\\ &=(e^{i\pi n}+1)\:|a|^{n-2m-1}\int_0^\infty \frac{v^n}{(v^2+1)^m}\mathop{dv}\qquad (x=|a|\times v)\\\\ &=\frac{(e^{i\pi n}+1)}2\:|a|^{n-2m-1}\int_0^1 u^{\large m-\frac{n+1}2-1}(1-u)^{\large \frac{n+1}2-1}\mathop{du}\quad \left(u=\frac1{v^2+1}\right)\\\\ &=\frac{(e^{i\pi n}+1)}2\:|a|^{n-2m-1}B\left(m-\frac{n+1}2,\frac{n+1}2\right) \end{align} $$ giving $(1)$.