Problem with an integral in statistical mechanics I have problems with this integral:
There are N identical particles contained in a circle with radius R with Hamiltonian $H=\frac{p^2}{2m} -Aq^2$, A is constant.
Now the integral i want to calculate is:
\begin{align}
\int d^2qd^2pe^{-\beta H} \delta (H-\epsilon) &= 2\pi\int_0^Rrdr\frac{d}{d\epsilon}\int d^2p\theta(\epsilon-\frac{p^2}{2m}+Ar^2) \\
&=4\pi^2m\int_0^Rrdr\frac{d}{d\epsilon}(\epsilon+Ar^2)\theta(\epsilon+Ar^2) \\
&=2\pi^2m\int_0^{R^2}dt\theta(\epsilon+At)
\end{align}
Can you explain this integral step by step to me? Thank you
 A: I view the problem like this: let $x=r\cos\psi$, $y=r\sin\psi$, $p_x=p\cos\phi$, and $p_y=p\sin\phi$. Then
$$\begin{align}I&=\int\int d^2\vec q\int\int d^2\vec pe^{-\beta H}\delta(H-\epsilon)\\
&=\int_0^Rr\,dr\int_0^{2\pi}d\psi\int_0^{\infty}p\,dp\int_0^{2\pi}d\phi e^{-\beta H}\delta(H-\epsilon)\\
&=4\pi^2\int_0^Rr\,dr\operatorname{\theta}(Ar^2+\epsilon)\int_0^{\infty}p\,dpe^{-\beta H}\delta\left(\frac{p^2}{2m}-Ar^2-\epsilon\right)\end{align}$$
Because the angular variables didn't enter into the integrand or limits and the Dirac $\delta$ function requires $0\le\frac{p^2}{2m}=Ar^2+\epsilon$. Then recall
that
$$\int_0^{\infty}g(p)\delta(f(p))dp=\int_{f(0)}^{\lim_{p\rightarrow\infty}f(p)}g(f^{-1}(u))\delta(u)\frac{du}{f^{\prime}(f^{-1}(u))}=\frac{g(f^{-1}(0))}{f^{\prime}(f^{-1}(0))}$$
Since $f(p)=\frac{p^2}{2m}-Ar^2-\epsilon=0$ we have $p=\sqrt{2m(Ar^2+\epsilon)}$, $g(p)=pe^{-\beta H}=\sqrt{2m(Ar^2+\epsilon)}e^{-\beta\epsilon}$, and $f^{\prime}(p)=p/m=\frac1m\sqrt{2m(Ar^2+\epsilon)}$, we have
$$\int_0^{\infty}g(p)\delta(f(p))dp=me^{-\beta\epsilon}$$
So we have
$$I=4\pi^2me^{-\beta\epsilon}\int_0^Rr\,dr\operatorname{\theta}(Ar^2+\epsilon)$$
