Calculate the convolution of $xe^{-x^2}$

Question

$h=xe^{-x^2}$. Calculate the convolution $(h\ast h)(x)$

$\int^{\infty}_{-\infty}ye^{-y^2}(x-y)e^{-(x-y)^2}\,dy$

The answer is $\frac{1}{4}\sqrt{\frac{\pi}{2}}(x^2-1)e^{-x^2/2}$

Could someone give me some hints?

I know how to solve $\int^{\infty}_{-\infty}e^{-y^2}e^{-(x-y)^2}\,dy=\int^{\infty}_{-\infty}e^{-2(y-\frac{1}{2}x)^2-X^2/2}\,dy=e^{-X^2/2}\frac{1}{\sqrt{2}}\int^{\infty}_{-\infty}e^{-u^2}\,du=\sqrt{\frac{\pi}{2}}e^{-x^2/2}$

Above is another question of my homework.

Answer 1

You have:

$\int^{\infty}_{-\infty}ye^{-y^2}(x-y)e^{-(x-y)^2}\,dy \underset{\text{adding and removing $\frac{x^2}{2}$ to the exponent of $e$}}{=} e^{-\frac{1}{2}x^2}\int^{\infty}_{-\infty}(yx-y^2)e^{-(\sqrt 2 y - \frac {x}{ \sqrt {2}})^2}dy \underset{\text{defining $z=\sqrt 2 y - \frac{x}{\sqrt 2}$}}{=} $

$=\sqrt 2 e^{-\frac{1}{2}x^2}\int^{\infty}_{-\infty}((\frac {z}{\sqrt 2}+\frac x 2)x-(\frac {z}{\sqrt 2}+\frac x 2)^2)e^{-z^2}dy=\sqrt 2 e^{-\frac{1}{2}x^2}[\frac {x^2}{ 4} \int^{\infty}_{-\infty}e^{-z^2}dz-\frac 1 2 \int^{\infty}_{-\infty} z^2 e^{-z^2}dz]$

that brings you to the result thanks to the fact that:

(i) $\int^{\infty}_{-\infty}e^{-z^2}dz=\sqrt{\pi}$

(ii) $\int^{\infty}_{-\infty}z^2e^{-z^2}dz=\sqrt{\frac{\pi}{2}}$