What is the Fourier transform of $f(x)=e^{-x^2}$?

I remember there is a special rule for this kind of function, but I can't remember what it was.

Does anyone know?

• The way it is usually normalized, the transform of $e^{-x^2/2}$ is itself. If you drop the half as you wrote, you get $e^{-x^2/4} / \sqrt {2}$ May 4, 2013 at 22:14
• my textbook says we first have to calculate the derivative and solve it by making the derivative = -w/2f(w) , are you familiar with that method ?
– S F
May 4, 2013 at 22:23

Caveat: I'm using the normalization $\hat f(\omega) = \int_{-\infty}^\infty f(t)e^{-it\omega}\,dt$.

A cute way to to derive the Fourier transform of $f(t) = e^{-t^2}$ is the following trick: Since $$f'(t) = -2te^{-t^2} = -2tf(t),$$ taking the Fourier transfom of both sides will give us $$i\omega \hat f(\omega) = -2i\hat f'(\omega).$$

Solving this differential equation for $\hat f$ yields $$\hat f(\omega) = Ce^{-\omega^2/4}$$ and plugging in $\omega = 0$ finally gives $$C = \hat f(0) = \int_{-\infty}^\infty e^{-t^2}\,dt = \sqrt{\pi}.$$

I.e. $$\hat f(\omega) = \sqrt{\pi}e^{-\omega^2/4}.$$

• Thanks yes that seems familiar could you explain how you get to the step ................ taking the Fourier transfom of both sides will give us $$i\omega \hat f(\omega) = -2i\hat f'(\omega).$$
– S F
May 4, 2013 at 22:40
• Those should be familiar "rules" for Fourier transforms: The Fourier transform of $f'(t)$ is $i\omega \hat f(\omega)$ and the FT of $tf(t)$ is $-i\hat f'(\omega)$. If they are not familiar, they follow fairly easily from the definition of the Fourier transform.
– mrf
May 4, 2013 at 22:44
• @Leon You have $\hat f(\omega) = Ce^{-\omega^2/4}$ for some value of $C$. You can determine the constant from any value of $\hat f(\omega)$, but it seems like $\omega = 0$ gives the simplest computations, doesn't it?