Problem 10

JT 06 Dec 2006 21:54

Start from $B_2(T)=-\frac{1}{2}\int_0^{\infty}(e^{-\beta u(r)}-1)4\pi r^2dr$. Integrate by parts to get

(1)\begin{align} B_2(T)=-\frac{2\pi}{3}[r^3(e^{-\beta u(r)}-1)]_0^{\infty}-\frac{\beta}{6}\int_0^{\infty}r\frac{du(r)}{dr}e^{-\beta u(r)}4\pi r^2dr \end{align}

If $\lim_{r\rightarrow\infty}r^3(e^{-\beta u(r)}-1)=0$ then

(2)\begin{align} B_2(T)=-\frac{\beta}{6}\int_0^{\infty}r\frac{du(r)}{dr}e^{-\beta u(r)}4\pi r^2dr \end{align}