# How to calculate the Fourier transform of a Gaussian function.

I would like to work out the Fourier transform for the Gaussian function: $f(x)=\exp(-n^2(x-m)^2)$. It seems likely that I will need to use differentiation and the shift rule at some point, but I can't seem to get the calculation to work. Does anyone have any advice?

By the way, I am using $FT(f)(k)=\int_{-\infty}^\infty f(x)e^{-ikx}\,dx$ as my definition of a Fourier transform.

Many thanks.

-
You can easily google this if you want the answer, since the Fourier transform of the Gaussian has a special property. Do you know what $\int_{-\infty}^\infty e^{-x^2} dx$ is? (Hint: write $\left(\int_{-\infty}^\infty e^{-x^2} dx\right)^2$ as an iterated integral, use polar coordinates. Then to calculate the Fourier transform, complete the square and change variables.) –  snarski Jan 4 at 22:05

First we consider $f(x) := \exp(-x^2)$. Let $$\hat{f}(k) := \int_{\mathbb{R}} f(x) \cdot e^{-\imath \, k x} \, dx = \int_{\mathbb{R}} \exp \left(-x^2 \right) \cdot e^{-\imath \, k \cdot x} \, dx$$ By differentiating we obtain $$\frac{d}{dk} \hat{f}(k) = \int_{\mathbb{R}} e^{-x^2} \cdot (-\imath \, x) \cdot e^{-\imath \, k x} \, dx = \frac{1}{2} \imath \int_{\mathbb{R}} \left( \frac{d}{dx} e^{-x^2} \right) \cdot e^{-\imath \, k x} \, dx$$

Using partial integration we get

$$\frac{d}{dk} \hat{f}(k) = - \frac{1}{2} k \cdot \int_{\mathbb{R}} e^{-x^2} \cdot e^{-\imath \, k \, x} \, dx =- \frac{1}{2} k \cdot \hat{f}(k)$$

Hence

$$\hat{f}(k) =c \cdot \exp \left(- \frac{k^2}{4} \right)$$

Since $c=\hat{f}(0) = \int_{\mathbb{R}} f(x) \, dx$ one can determine the constant $c$ (correct result: $c=\sqrt{\pi}$). Moreover, by using the following well-known formulas

\begin{align} \widehat{f(x+m)}(k) &= e^{\imath \, k \cdot m} \hat{f}(k) \\ \widehat{f(\alpha \cdot x)}(k) &= \frac{1}{\alpha} \cdot \hat{f} \left( \frac{k}{\alpha} \right) \qquad (\alpha>0) \end{align}

one can calculate the fourier transform of $f(x) = \exp \left(-n^2 \cdot (x-m)^2 \right)$ by some straight-forward computations.

-
Thank you, this is starting to make a lot of sense now!! –  user55225 Jan 4 at 22:47
\begin{align} \int_{-\infty}^\infty e^{-x^2}\,e^{-ix\xi}\,\mathrm{d}x &=e^{-\xi^2/4}\int_{-\infty}^\infty e^{-(x+i\xi/2)^2}\,\mathrm{d}x\\ &=e^{-\xi^2/4}\int_{-\infty+i\xi/2}^{\infty+i\xi/2}e^{-x^2}\mathrm{d}x\\ &=e^{-\xi^2/4}\int_{-\infty}^\infty e^{-x^2}\mathrm{d}x\\ &=\sqrt{\pi}\,e^{-\xi^2/4}\tag{1} \end{align} The third equation is justified by contour integration since $e^{-x^2}=O\left(e^{-\mathrm{Re}(x)^2}\right)$ as $|\mathrm{Re}(x)|\to\infty$ for bounded $|\mathrm{Im}(x)|$.
Now, simple manipulation of $(1)$ yields \begin{align} \int_{-\infty}^\infty e^{-n^2(x-m)^2}\,e^{-ix\xi}\,\mathrm{d}x &=\int_{-\infty}^\infty e^{-n^2x^2}\,e^{-i(x+m)\xi}\,\mathrm{d}x\\ &=e^{-im\xi}\int_{-\infty}^\infty e^{-n^2x^2}\,e^{-ix\xi}\,\mathrm{d}x\\ &=\frac{e^{-im\xi}}{n}\int_{-\infty}^\infty e^{-x^2}\,e^{-ix\xi/n}\,\mathrm{d}x\\ &=\frac{e^{-im\xi}}{n}\sqrt{\pi}\,e^{-\xi^2/(4n^2)}\tag{2} \end{align}