13
$\begingroup$

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

Does anyone know?

$\endgroup$
2
  • $\begingroup$ 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 $e^{-x^2/4} / \sqrt {2}$ $\endgroup$ – Will Jagy May 4 '13 at 22:14
  • $\begingroup$ 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 ? $\endgroup$ – S F May 4 '13 at 22:23
29
$\begingroup$

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

$\endgroup$
9
  • $\begingroup$ Thanks yes that seems familiar could you explain how you get to the step ................ taking the Fourier transfom of both sides will give us $$i\omega \hat f(\omega) = -2i\hat f'(\omega).$$ $\endgroup$ – S F May 4 '13 at 22:40
  • $\begingroup$ 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. $\endgroup$ – mrf May 4 '13 at 22:44
  • $\begingroup$ Furthermore why does e^-infinity - e^ infinity = square root(pi) ? $\endgroup$ – S F May 4 '13 at 22:49
  • 3
    $\begingroup$ @SF your last two comments make no sense. $\endgroup$ – mrf May 4 '13 at 22:52
  • 2
    $\begingroup$ That is a very well known integral. See for example mathworld.wolfram.com/GaussianIntegral.html $\endgroup$ – mrf May 4 '13 at 22:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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