How to do this integral

Question

Prove that $$\int \frac{d^{n}q}{(2\pi)^{n} }\frac{q^{2a}}{(q^{2}+D)^{b}}=D^{-(b-a-n/2)}\frac{\Gamma (b-a-n/2)\Gamma (a+n/2)}{(4\pi )^{n/2}\Gamma (n)\Gamma (n/2)}$$

The angular part is easy to do as the integrand is spherically symmetric so the result is $V_{d-1}\times A$ where $V_{d-1}$ is the volume of the Euclidean n-ball of unit radius. It remains to compute the radial part but I don't know how it's done. This integral arises in the calculation of loop corrections in the propagators of scalar fields

Answer 1

With the spherical coordinate change followed by the substitution $r = \sqrt{D}\tan\theta$, \begin{align*} &\frac{1}{(2\pi)^{n}} \int_{\Bbb{R}^{n}} \frac{r^{2a}}{(r^2 + D)^{b}} \, dx \\ &=\frac{|S^{n-1}|}{(2\pi)^{n}} \int_{0}^{\infty} \frac{r^{2a}}{(r^2 + D)^{b}} \, r^{n-1}dr \\ &=\frac{2\pi^{n/2}}{(2\pi)^{n}\Gamma(n/2)} D^{a-b+n/2} \int_{0}^{\frac{\pi}{2}} \frac{\tan^{2a+n-1}\theta}{\sec^{2b}\theta} \sec^{2}\theta \, d\theta \\ &=\frac{1}{(4\pi)^{n/2}\Gamma(n/2)} D^{a-b+n/2} \cdot 2 \int_{0}^{\frac{\pi}{2}} \sin^{2a+n-1}\theta \cos^{2b-2a-n-1}\theta \, d\theta \\ &=\frac{1}{(4\pi)^{n/2}\Gamma(n/2)} D^{a-b+n/2} \beta\left(a+\frac{n}{2},b-a-\frac{n}{2}\right) \\ &=\frac{1}{(4\pi)^{n/2}\Gamma(n/2)} D^{a-b+n/2} \cdot \frac{\Gamma\left(a+\frac{n}{2}\right)\Gamma\left(b-a-\frac{n}{2}\right)}{\Gamma(b)}. \end{align*} Here we used the beta function identity $$ 2\int_{0}^{\frac{\pi}{2}} \sin^{2z-1}\theta \cos^{2w-1}\theta \, d\theta = \beta(z, w) = \frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)}. $$