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

Does anyone know?

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    $\begingroup$ 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, 2013 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, 2013 at 22:23

1 Answer 1

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

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  • $\begingroup$ 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, 2013 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, 2013 at 22:44
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    $\begingroup$ @SF your last two comments make no sense. $\endgroup$
    – mrf
    May 4, 2013 at 22:52
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    $\begingroup$ That is a very well known integral. See for example mathworld.wolfram.com/GaussianIntegral.html $\endgroup$
    – mrf
    May 4, 2013 at 22:59
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    $\begingroup$ @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? $\endgroup$
    – mrf
    Jun 20, 2021 at 14:06

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