How do I integrate $I = \int_0^\infty dr \: r e^{-ar^2}$, $a \in \mathbb{C}$?

Question

I know that the answer has to be $\frac{1}{2a}$, $Re(a) > 0$. I tried using partial integration and I am left with $I = -\frac{e^{-ar^2}}{2a} \bigg\rvert_0^\infty + \int_0^\infty dr \frac{e^{-ar^2}}{2ar}$, where the first term already gives me the desired result. Is my calculation up till now correct? How can I argue that the second term vanishes, even though the integral seems to be divergent?

Answer 1

Partial integration isn't the way to go here. Instead, use $u = r^2$, so $du = 2r\, dr$ and $\frac{1}{2} \, du = r \, dr$.

$$\begin{align}I &= \int_0^\infty r e^{-ar^2} \,dr\\ &=\frac{1}{2}\int_0^\infty e^{-au}\,du \\ &= \frac{1}{2} \left[-\frac{1}{a}e^{-au}\right]_0^\infty \\ &= \frac{1}{2a} \end{align}$$ where the final limit $u \to \infty$ is valid since $Re(a) > 0$.

Answer 2

Hint remark that the derivative of $e^{-ar^2}$ is $-2rae^{-ar^2}$.