I know this is a difficult integral to compute, and I know that the answer is: $\int_{-\infty}^{\infty} \frac{dx}{(ax^2 + b)^k} = \frac{\sqrt\pi \Gamma(k-\frac{1}{2})}{\Gamma(k)}\frac{1}{\sqrt a b^{k-\frac{1}{2}}}$.
However, I don't know how to work it out.
I would really appreciate any hints.
If anyone is able to work the answer, would it also be possible to compute it for different bounds for the integral? (e.g. once a suitable change in variable x is performed so that one integrates between $\frac{-\pi}{2}$ and $\frac{\pi}{2}$, could I integrate just between, let's say $\frac{-\pi}{2}$ and $\frac{-\pi}{4}$?).
Thanks.
Let $x=\left(\frac ba\right)^{1/2}u$. Then $$\int_{-\infty}^{\infty}\frac{dx}{(ax^2+b)^k}=\frac1{\sqrt ab^{k-\frac12}}\int_{-\infty}^{\infty}\frac{du}{(u^2+1)^k}=\frac2{\sqrt ab^{k-\frac12}}\int_0^{\infty}\frac{du}{(u^2+1)^k}$$ Now let $t=\frac1{u^2+1}$. Then $$\int_{-\infty}^{\infty}\frac{dx}{(ax^2+b)^k}=\frac1{\sqrt ab^{k-\frac12}}\int_0^1t^{k-\frac32}(1-t)^{-\frac12}\,dt=\frac1{\sqrt ab^{k-\frac12}}\text{B}\left(k-\frac12,\frac12\right)$$ The key is to recognize the beta function in the answer and work towards that.