integral of $x^2e^{-x^2}~dx$ from $-\infty$ to $+\infty$

Solution 1:

I'm surprised no one has given this answer. We have of course by $u$-substitution $$\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}.$$ Take a derivative of each side with respect to $\alpha$ to get $$-\int_{-\infty}^\infty x^2 e^{-\alpha x^2} dx = -\frac{1}{2} \frac{\sqrt{\pi}}{\alpha^{3/2}}.$$ Substitute $\alpha = 1$ and cancel negative signs.

EDIT: I see that Felix Marin does essentially the same thing, but I think this is a better explanation.

Solution 2:

$$\begin{align} \int_{-\infty}^\infty x^2e^{-x^2}\text dx&=2\int_0^\infty x^2e^{-x^2}\text dx\\ &=\int_0^\infty \sqrt ue^{-u}\text du\\ &=\Gamma\left(\frac32\right)\\ &=\frac12\Gamma\left(\frac12\right)\\ &=\frac12\int_0^\infty\frac{e^{-u}}{\sqrt u}\text du\\ &=\int_0^\infty e^{-x^2}\text dx\\ &=\frac{\sqrt \pi}2 \end{align}$$