Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I remember their being a special rule for this kind of function but I cant remember what it was. Anyone know how ?


share|cite|improve this question
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 $e^{-x^2/4} / \sqrt {2}$ – Will Jagy May 4 '13 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 '13 at 22:23
up vote 10 down vote accepted

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

share|cite|improve this answer
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 '13 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 '13 at 22:44
Furthermore why does e^-infinity - e^ infinity = square root(pi) ? – S F May 4 '13 at 22:49
@SF your last two comments make no sense. – mrf May 4 '13 at 22:52
That is a very well known integral. See for example – mrf May 4 '13 at 22:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.