# Fourier transform of $f(x) = e^{-x^2}$ [duplicate]

Possible Duplicate:
Fourier Transform of complicated product: $(1+x)^2 e^{-x^2/2}$

I calculate the Fourier Transform of $f(x)$ by $$\mathbb{F}(t) =\int_{-\infty}^{\infty}e^{-x^2} \cdot e^{-ixt}dx,$$ but my result is not equal to the Mathematica result. I tried to integrate by parts it, and next do an equal with the integral above.

-
 "but my result is not equale to the Mathematica result" - but you know that Mathematica's normalization convention might be different from the one you're accustomed to? – J. M. Nov 20 '11 at 11:04 Yes, but in my book and a few other sources is other (the same as Mathematica) result. – D.M. Nov 20 '11 at 11:05 In my answer to the thread linked to above I gave one way of computing that $$\mathbb{F}(t) = \sqrt{\pi} e^{-t^2/4}:$$ Derive an ordinary differential equation (differentiate under the integral sign, integrate by parts once) and you get $\mathbb{F'}(t) = -C t \mathbb{F}(t)$ and $\mathbb{F}(0) = 0$ and solve it . – t.b. Nov 20 '11 at 11:25 That should have been $\mathbb{F}(0) = \sqrt{\pi}$ of course. – t.b. Nov 20 '11 at 11:31

## marked as duplicate by t.b., joriki, Davide Giraudo, Jonas Teuwen, J. M.Nov 20 '11 at 15:11

The simplest way to do this is: $\int_{-\infty}^{\infty} e^{-(x^2+ixt)}dx=\int_{-\infty}^{\infty} e^{-(x+it/2)^2}e^{-t^2/4}dx=e^{-t^2/4} \sqrt{\pi}$ because $\int_{-\infty}^{\infty}e^{-(x+it/2)^2}dx=\int_{-\infty}^{\infty}e^{-(x+it/2)^2}d(x+it/2)=\int_{-\infty+it/2}^{\infty+it/2}e^{-z^2}dz=\sqrt{\pi}$.