How to solve gaussian integral for $x^2e^{-\frac{x^2}{w}}$?

Question

I am trying to find $\sigma=\displaystyle\sqrt{\int _{-\infty \:}^{\infty \:}x^2e^{-\frac{x^2}{w}}dx}$ for the function $f(x)=e^{-\frac{x^2}{w}}$. I have tried tabular integration by parts, but it quickly got messy and I stopped after the second integration $\sqrt{w}\frac{\sqrt{\pi }}{2}\text{erf}\left(\frac{x}{\sqrt{w}}\right)$. From some quick research no elementary function exists for the indefinite integral. So how would I find the definite integral in this case? I would be grateful for any help.

Answer 1

Starting with $\displaystyle \int_{-\infty}^{\infty}e^{- x^2}\,\mathrm{dx} = \sqrt{\pi}$, let $x \mapsto \sqrt{\lambda} x$ then $\displaystyle \int_{-\infty}^{\infty}e^{-\lambda x^2}\,\mathrm{dx} = \frac{\sqrt{\pi}}{\sqrt{\lambda}}$

Define $\displaystyle $ $\displaystyle f(\lambda) := \int_{-\infty}^{\infty}e^{-\lambda x^2}\,\mathrm{dx} = \frac{\sqrt{\pi}}{\sqrt{\lambda}} $ then $\displaystyle f'(\lambda)=-\int_{-\infty}^{\infty}x^2 e^{-\lambda x^2}\,\mathrm{dx} = -\frac{\sqrt{\pi}}{2\lambda^{3/2}} $

so that $$\int_{-\infty}^{\infty}x^2 e^{-\frac{1}{w} x^2}\,\mathrm{dx} = \frac{1}{2}w^{3/2}\sqrt{\pi}.$$

Answer 2

Hint: Try integration by parts with $u = x(-\frac{w}{2})$ and $v = e^{-\frac{x^2}{w}}$. Then use that $\int_{-\infty}^\infty e^{-\frac{x^2}{w}}dx = \sqrt{2\pi w}$.