An integral equality with respect to Gamma function

Question

How to prove that $$\int_0^{\infty}\biggl(\frac{r}{1+r^2}\biggr)^q\frac{1}{r^{1+\alpha}}dr=\frac{\Gamma(\frac{q+\alpha}{2})\Gamma(\frac{q-\alpha}{2})}{2\Gamma(q)}\quad?$$

Answer 1

Using $t=\frac {r^2} {1+r^2}$, and thus $r=\sqrt{\frac{t}{1-t}}$ $$\begin{split} \int_0^{\infty}\biggl(\frac{r}{1+r^2}\biggr)^q\frac{1}{r^{1+\alpha}}dr &= \int_0^1 t^q t^{-\frac {q+1+\alpha}{2}}(1-t)^{\frac{q+1+\alpha}{2}}\frac 1 2 \frac{1}{\sqrt{t}(1-t)^{\frac 3 2}}dt\\ &=\frac 1 2\int_0^1t^{\frac{q-\alpha}{2}-1}(1-t)^{\frac {q+\alpha}2-1}dt\\ &=\frac 1 2B\left(\frac{q-\alpha}2,\frac{q+\alpha}2\right)\\ \end{split}$$ where $B$ is the Beta function. To conclude, use the known identity (whose proof can be found following that link): $$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$