# Convolution Of Exponential Function

• Jan 24th 2011, 09:41 AM
mathematicalbagpiper
Convolution Of Exponential Function
Let $\displaystyle f(x)=e^{-x^2}$. Compute $\displaystyle f*f$.

So I get:

$\displaystyle f*f=\int_{-\infty}^{\infty}e^{-y^2}e^{-(x-y)^2}dy$

$\displaystyle =\int_{-\infty}^{\infty}e^{-y^2}e^{-y^2+2xy-x^2}dy$

$\displaystyle =\int_{-\infty}^{\infty}e^{-2y^2+2xy-x^2}dy$

$\displaystyle =e^{-x^2}\int_{-\infty}^{\infty}e^{-2y^2+2xy}dy$

$\displaystyle =f(x)\int_{-\infty}^{\infty}e^{-2(y^2-xy)}dy$

$\displaystyle =f(x)\int_{-\infty}^{\infty}e^{-2(y-\frac{x}{2})^2+\frac{x^2}{2}}dy$

$\displaystyle =f(x)e^{\frac{x^2}{2}}\int_{-\infty}^{\infty}e^{-2(y-\frac{x}{2})^2}dy$

...and that's as far as I can get :-(
• Jan 24th 2011, 09:58 AM
Ackbeet
Looks good so far. Try the substitution $\displaystyle u=\sqrt{2}\,(y-x/2).$ Where does that lead you?
• Jan 25th 2011, 04:51 AM
mathematicalbagpiper
Leads me to a nice result using the improper integral of $\displaystyle e^{-u^2}du$, and the resulting following calculation.

Got it! Thanks! :-D
• Jan 25th 2011, 05:39 AM
Ackbeet
I ended up with $\displaystyle \sqrt{\frac{\pi}{2}}\cdot f(\frac{x}{\sqrt{2}})$ if I remember correctly.