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

thanks

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

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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 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
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@SF your last two comments make no sense. –  mrf May 4 '13 at 22:52
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That is a very well known integral. See for example mathworld.wolfram.com/GaussianIntegral.html –  mrf May 4 '13 at 22:59
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