# Convolution of Gaussian Function with itself

Trying to solve this question:

Let $$f(x)=e^{-x^2}$$ be a Gaussian. Compute explicitly $$(f*f)(x)$$.

Using the definition of the convolution, and given the fact that the convolution of 2 Gaussians is another Gaussian, I got \begin{align*} (f*f)(x) &= \int_{-\infty}^{\infty}f(x-y)f(y)\,dy\\ &=\int_{-\infty}^{\infty}e^{-(x-y)^2}e^{-y^2}\,dy \end{align*} but I'm not sure how to proceed from here.

Any tips would be appreciated!

• Where did you use the fact that convolution of two Gaussians is a Gaussian? – Gae. S. Oct 7 '19 at 21:50
• don't think i used it in the questions, but it is a well known property – nickoba Oct 7 '19 at 21:55
• You could rewrite the integrand so you can complete the square involving $y$ – Henry Oct 7 '19 at 21:55
• math.stackexchange.com/questions/1745174/… – cmk Oct 7 '19 at 22:07
• Next step: $\displaystyle\int_{-\infty}^\infty e^{-x^2+2xy-2y^2}\,\mathrm{d}y$ – robjohn Oct 8 '19 at 7:13

First, complete the square to get $$-a((y+b)^2+cx^2)$$, then you could take $$e^{-acx^2}$$ beyond the sign of the integral since integration goes over $$y$$ and change the integration variable to $$(y+b)$$. Finally, use the well-known formula for the Gaussian integral. As an answer, I've got $$\sqrt{\frac{\pi}{2}}\cdot e^{-\frac{x^2}{2}}$$