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!
 A: 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,
$$
      \hat{f}(k) = \frac{1}{\sqrt{2\alpha}}e^{-k^{2}/4\alpha}.
$$
