Problem 11

juliantalbot 07 Dec 2006 21:45

Start from

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

Substitute $u(r)=-\alpha/ r^n,\;\;n>3$ giving

(2)\begin{align} B_2=\frac{2\pi n\alpha\beta}{3}\int_0^{\infty}r^{2-n}e^{-\beta\alpha/r^n} dr \end{align}

This suggests the gamma function. So let $t=\beta\alpha/ r^n$ or $r=(\beta\alpha/t)^{1/n}$

Substituting gives

\begin{align} B_2=\frac{2\pi}{3}(\beta\alpha)^{3/n}\int_0^{\infty}t^{-3/n}e^{-t} dt \end{align}

Using the definition of the gamma function finally gives

(4)\begin{align} B_2=\frac{2\pi}{3}(\beta\alpha)^{3/n}\Gamma(1-3/n) \end{align}