How can we find the fourier transform of $e^\frac{-x^2}{2}$ where -$\infty $ < x < $\infty $. I tried applying the standard formulae but ended up in un defined form..

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    $\begingroup$ Why don't you show some work, which would help others identify the point where you are making an error. By the way, the title of your question is not what you're asking. You want to know the Fourier transform of $e^{-x^2/2}$ but the title of the question is asking for the Fourier transform of something else (at the time I write this). $\endgroup$
    – KCd
    Jan 17 '14 at 18:22
  • $\begingroup$ i started with $\int_{-\infty}^{\infty}{e^{isx}e^{\frac{-x^2}{2}}dx}$ after integrating, i got stuck at $\{\frac{e^{isx-\frac{x^2}{2}}}{is-x}\}_{-\infty}^{\infty}$ $\endgroup$
    – mahes
    Jan 17 '14 at 18:44
  • $\begingroup$ What do you get if you differentiating the Fourier transform with respect to $s$? $\endgroup$ Jan 17 '14 at 19:16
  • $\begingroup$ @user2332665 : The direct way (doing that integral) is tricky. The standard trick is to define $g(\omega)= \mathcal{F}(\exp(-x^2/2))(\omega)$ and show $g$ satisfies the IVP $g(0)=1$, $g'(\omega)=-\omega g(\omega)$. $\endgroup$ Jan 18 '14 at 3:29
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    $\begingroup$ You're essentially showing that the Fourier transform of a Gaussian is a Gaussian. With a little more work, you show essentially the lower bound achieved in Heisenberg's uncertainty principle. $\endgroup$
    – Batman
    Mar 1 '14 at 3:51

[Expanding on the comments by T.A.E. and Stefan Smith]. One of interesting features of $f(x)=e^{-x^2/2}$ is that differentiation has the same effect as multiplication by the independent variable (up to minus sign): $$f'(x) = -xf(x) \tag{1}$$ Recall that under the Fourier transform, differentiation becomes multiplication by independent variable, and vice versa (with multiplicative constants dependent on your convention for Fourier transform). E.g., with the convention $\hat f(\xi) = \int_{-\infty}^{\infty} e^{-i\xi x}f(x)\,dx$, identity (1) transforms into $$ i \xi \hat f(\xi) = -i \frac{d}{d\xi}\hat f(\xi) \tag{2}$$ Thus, $\hat f$ solves the same first-order ODE as $f$ itself. This implies $\hat f$ is a constant multiple of $e^{-\xi^2/2}$, and you can find that multiple from $\hat f(0)=\int f(x)\,dx=\sqrt{2\pi}$.


The way I do it is completing squares for the integral.

$$ \int_{-\infty}^{\infty} e^{ix\omega}e^{\frac{-x^2}{2}} \,dx =\int_{-\infty}^{\infty} e^{-\frac{1}{2}(x+i\omega)^2}e^{-\frac{\omega^2}{2}} \,dx \\=\sqrt{2\pi}e^{-\frac{\omega^2}{2}} $$ Where I used the fact that the gaussian integral $$ \int_{-\infty}^{\infty} e^{\frac{-(x+b)^2}{a}} \,dx = \sqrt{a\pi}$$ even when b is complex. When b is real, is easy to prove this with a change of variables because the limits doesn't change. For the imaginary case, this is proven using complex integration on the complex plane as you can see in this answer.


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