Solving a 2 variable integral with a delta function

Question

How can this integrals that include the dirac delta function be solved?

$$\int_{-\infty}^{+\infty}dq\int_{-\infty}^{+\infty}dp\cdot p^n\cdot \delta(p^2+q^2-E)$$

Answer 1

In polar coordinates,

$$\int_{\theta=0}^{2\pi}\int_{r=0}^\infty r^n\cos^n\theta\,\delta(r^2-E)\,r\,dr\,d\theta =\int_{\theta=0}^{2\pi}\cos^n\theta\,d\theta\int_{r=0}^\infty r^{n+1}\,\frac{\delta(r-\sqrt E)}{2\sqrt E}\,dr\\ =\frac{ E^{n/2}}2\int_{\theta=0}^{2\pi}\cos^n\theta\,d\theta .$$

We used the fact that $2\sqrt E\,\delta(r^2-E)=\delta(r-\sqrt E)$.

Answer 2

Note that the the integral is zero by symmetry if $n$ is odd. Assume from now on that $n\geq 0$ is even.

One idea is to use polar coordinates $$(p,q)~=~(r\cos\theta,r\sin\theta).\tag{1}$$ Then $$I~:=~ \iint_{\mathbb{R}^2}\! \mathrm{d}p~\mathrm{d}q~p^n~\delta(p^2+q^2-E) ~=~ \int_{\mathbb{R}_+}\! \mathrm{d}r~r^{n+1}\delta(r^2-E)\int_{[0,2\pi]}\! \mathrm{d}\theta~\cos^n\theta $$ $$~=~\ldots~=~\frac{1}{2}H(E) |E|^{n/2} 2\pi \frac{(n-1)!!}{n!!}.\tag{2}$$