$$\int^\infty_0f(r) dr = \int^\infty_0 \frac{Ar}{1+Cr^\alpha} e^{-Br^2} dr$$

If I let $u = Br^2$, then I get

$$ = \frac{A}{2B} \int^\infty_0\frac{\exp(-u)}{1+(u/B)^{2/\alpha}} du$$

But I am stuck while proceeding further. Any idea?

- Thread starter sjaffry
- Start date