Compute $e^{-x^2} * e^{-x^2}$

Question

How to compute the convolution of $e^{-x^2}$ with itself?

$$e^{-x^2} * e^{-x^2} = \int_{\mathbb R} e^{-(x-y)^2} e^{-y^2}dy = e^{-x^2}\int_{-\infty}^{\infty} e^{2xy - 2y^2} dy$$

I can't solve it. I tried integration by parts so neglecting $e^{-x^2}$:

$$2\int_{-\infty}^{\infty}(2y^2 - xy) e^{2xy - 2y^2}dy$$

please help. thanks.

Answer 1

Notice $$e^{2xy-2y^2}=e^{-2(y-\frac{x}{2})^2}e^{\frac{x^2}{2}}.$$ Hence $$e^{-x^2}\int_{-\infty}^{\infty} e^{2xy - 2y^2} dy = e^{-\frac{x^2}{2}}\int^{+\infty}_{-\infty}e^{-2(y-\frac{x}{2})^2} dy = e^{-\frac{x^2}{2}}\int^{+\infty}_{-\infty}e^{-2u^2} du,$$ the last equality is a substitution $u =y-\frac{x}{2}.$ Now use the famous Gaussian integral $$\int^{+\infty}_{-\infty}e^{-2u^2} du =\sqrt{\frac{\pi}{2}}$$ to conclude.

Answer 2

$$ \left( y - \frac{x}{2} \right)^2 = y^2 - xy + \frac{x^2}{4} $$ $$ \left( y - \frac{x}{2} \right)^2 - \frac{x^2}{4} = y^2 - xy $$ $$ 2 \left( y - \frac{x}{2} \right)^2 - \frac{x^2}{2} = 2 y^2 - 2 xy $$ $$ \frac{x^2}{2}-2 \left( y - \frac{x}{2} \right)^2 = 2 xy- 2 y^2 $$

Answer 3

use $2xy-2y^2 = -2(y-\frac{1}{2}x)^2 +\frac{1}{2}x^2$ $$\int_{-\infty}^{\infty} e^{2xy - 2y^2} dy = \int_{-\infty}^{\infty} e^{-2(y-\frac{1}{2}x)^2} e^{\frac{1}{2}x^2} dy = e^{\frac{1}{2}x^2} \int_{-\infty}^{\infty} e^{-2y^2}dy$$