Originally Posted by

**gosualite** Try integration by parts-- twice.

Let $\displaystyle u = r^2$, so $\displaystyle du = 2r\,dr$, $\displaystyle dv = \exp(-2r/a)$, and $\displaystyle v = -\frac{a}{2}\exp(-2r/a)$.

Your problem changes to:

$\displaystyle \lim_{b \to \infty}\left[-\frac{ar^2}{2}\exp(-2r/a)\right]_0^b + \lim_{b \to \infty} \int_0^b ar\exp(-2r/a)\,dr$

Now we have to assume a > 0 for the limit in the left term to exist. If we do this, we can compute it to 0.

Use integration by parts on the remaining integral. I computed it to $\displaystyle a^3/4$ as you said the answer should be.