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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?

Rodrigo de Azevedo
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S F
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    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 04 '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 04 '13 at 22:23

1 Answers1

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

mrf
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  • 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 04 '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 04 '13 at 22:44
  • Furthermore why does e^-infinity - e^ infinity = square root(pi) ? – S F May 04 '13 at 22:49
  • the Fourier transform of t , you mean integral limits +infitniy,-infinity te^-iwt – S F May 04 '13 at 22:51
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    @SF your last two comments make no sense. – mrf May 04 '13 at 22:52
  • my question was after plugging in w=0 , why does the answer = root pi ? – S F May 04 '13 at 22:56
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    That is a very well known integral. See for example http://mathworld.wolfram.com/GaussianIntegral.html – mrf May 04 '13 at 22:59
  • I have no come across Gaussian integral before, perhaps I am able to assume it, and not expected to work it out .............. could you explain how the Fourier transform of f′(t) is iωf^(ω) – S F May 05 '13 at 12:17
  • If you can't find it in your textbook (which you should be able to), try partial integration. – mrf May 05 '13 at 14:47
  • @mrf when FT the right side of the first equation, $-2 tf(t)$, using that FT of $tf(t)$ is $-iF'(ω)$ shouldn't that yield $+2 iF'(ω)$? I just don't get it where the minus in $−2 i F′(ω)$ comes from. Thanks. – Daniel K. May 25 '21 at 15:37
  • @mrf I think I found it. According to your definition $$\hat f(\omega) = \int_{-\infty}^\infty f(t)e^{-it\omega},dt.$$ Differentiation yields $$i\frac{d}{d\omega}\hat f(\omega) = i \int f(t) \frac{d}{d\omega} e^{-it\omega},dt = \int tf(t) e^{-it\omega},dt$$ In other words, the FT of $tf(t)$ is +$i\hat f'(\omega)$. There is no minus in this relationship when using your definition of FT. So your brilliant answer is correct just the comment below contains a superflous minus when stating that relationship in my opinion. – Daniel K. May 28 '21 at 09:32
  • Why actually plugging in $0$ for $\omega$ in order to determine $C$ ? Why that initial condition? – Leon Jun 19 '21 at 18:49
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    @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? – mrf Jun 20 '21 at 14:06