Inverse Fourier transform of Gaussian

I need to calculate the Inverse Fourier Transform of this Gaussian function:

$\frac{1}{\sqrt{2\pi}} exp(\frac{-k^2 \sigma^2}{2})$

where $\sigma > 0$, namely I have to calculate the following integral: $\int{\frac{1}{\sqrt{2\pi}} \exp(\frac{-k^2 \sigma^2}{2})} \exp(ikx) dk$.

How can I do that? If I am not wrong (in which case please correct me) the result should be $\frac{1}{\sqrt{2\pi}\sigma}exp(\frac{-x^2}{2\sigma^2})$...but I don't know how to get there!

Solution 1:

The Gaussian $e^{-\alpha x^{2}}$ satisfies the differential equation $$ \frac{df}{dx} = -2\alpha x f,\;\;\; f(0)=1. $$ The Fourier transform turns differentiation into multiplication, and multiplication into differentiation. So you get back the same differential equation with different constants. For example, if $f(x)=e^{-\alpha x^{2}}$, then \begin{align} \frac{d}{dk}\hat{f}(k) & =\frac{d}{dk}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}e^{-ikx}dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}e^{-ikx}(-ix)dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(-2\alpha xe^{-\alpha x^{2}})(\frac{i}{2\alpha}e^{-ikx})dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(\frac{d}{dx}e^{-\alpha x^{2}})(\frac{i}{2\alpha}e^{-ikx})dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}\left(-\frac{d}{dx}\frac{i}{2\alpha}e^{-ikx}\right)dx \\ & = -\frac{k}{2\alpha}\hat{f}(k). \end{align} And, $$ \hat{f}(0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}dx $$ Therefore, $$ \hat{f}(k) = \left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}dx\right]e^{-k^{2}/4\alpha} $$ Determining the multiplier constant is normally done with a trick using polar coordinates: \begin{align} \left[\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}dx\right]^{2} & = \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}e^{-\alpha y^{2}}dxdy \\ & = \frac{1}{2\pi}\int_{0}^{2\pi}\int_{0}^{\infty}e^{-\alpha r^{2}}rdrd\theta \\ & = \int_{0}^{\infty}e^{-\alpha r^{2}}rdr \\ & = -\frac{1}{2\alpha}\int_{0}^{\infty}\frac{d}{dr}e^{-\alpha r^{2}}dr \\ & = \frac{1}{2\alpha} \end{align} Finally, $$ \hat{f}(k) = \frac{1}{\sqrt{2\alpha}}e^{-k^{2}/4\alpha}. $$