I would like to work out the Fourier transform of the Gaussian function

$$f(x) = \exp \left(-n^2(x-m)^2 \right)$$

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

$$\mbox{FT}(f)(k)=\int_{-\infty}^\infty f(x)e^{-ikx}\,\mathrm d x$$

as my definition of a Fourier transform. Many thanks.

  • 5
    $\begingroup$ 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.) $\endgroup$
    – snar
    Jan 4, 2013 at 22:05
  • $\begingroup$ cse.yorku.ca/~kosta/CompVis_Notes/… $\endgroup$
    – vito
    Dec 7, 2015 at 20:25
  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/381597 $\endgroup$
    – Watson
    Mar 11, 2017 at 15:07

5 Answers 5


First we consider the case $m=0$ and $n=1$, i.e. $f(x) := \exp(-x^2)$ and $$\hat{f}(k) := \int_{\mathbb{R}} f(x) \cdot e^{-\imath \, k x} \, dx = \int_{\mathbb{R}} \exp \left(-x^2 \right) \cdot e^{-\imath \, k \cdot x} \, dx.$$ Differentiating with respect to $k$ yields $$\frac{d}{dk} \hat{f}(k) = \int_{\mathbb{R}} e^{-x^2} \cdot (-\imath \, x) \cdot e^{-\imath \, k x} \, dx = \frac{1}{2} \imath \int_{\mathbb{R}} \left( \frac{d}{dx} e^{-x^2} \right) \cdot e^{-\imath \, k x} \, dx.$$

Applying the integration by parts formula, we obtain

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

The unique solution to this ordinary differential equation is given by

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

Since $c=\hat{f}(0) = \int_{\mathbb{R}} f(x) \, dx$, it follows that $c = \sqrt{\pi}$. Moreover, applying 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.

  • $\begingroup$ Thank you, this is starting to make a lot of sense now!! $\endgroup$
    – user55225
    Jan 4, 2013 at 22:47
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    $\begingroup$ Could you elaborate why $x \exp(-x^2)$ is in $L^1$. We need this in order to interchange integration and differentiation in the second paragraph. $\endgroup$
    – el_tenedor
    Sep 18, 2016 at 13:35
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    $\begingroup$ @el_tenedor $exp(-x^2)$ is decaying faster than any polynomial, in particular, we can choose $c>0$ such that $\exp(-x^2) \leq \frac{c}{|x|^4}$ for all $|x| \geq 1$. Using that $x \exp(-x^2)$ is bounded on $[-1,1]$, this gives the integrability. $\endgroup$
    – saz
    Sep 18, 2016 at 13:38
  • $\begingroup$ thanks. This does the trick. $\endgroup$
    – el_tenedor
    Sep 18, 2016 at 13:43

$$ \begin{align} \int_{-\infty}^\infty e^{-x^2}\,e^{-ix\xi}\,\mathrm{d}x &=e^{-\xi^2/4}\int_{-\infty}^\infty e^{-(x+i\xi/2)^2}\,\mathrm{d}x\\ &=e^{-\xi^2/4}\int_{-\infty+i\xi/2}^{\infty+i\xi/2}e^{-x^2}\mathrm{d}x\\ &=e^{-\xi^2/4}\int_{-\infty}^\infty e^{-x^2}\mathrm{d}x\\ &=\sqrt{\pi}\,e^{-\xi^2/4}\tag{1} \end{align} $$ 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 $$ \begin{align} \int_{-\infty}^\infty e^{-n^2(x-m)^2}\,e^{-ix\xi}\,\mathrm{d}x &=\int_{-\infty}^\infty e^{-n^2x^2}\,e^{-i(x+m)\xi}\,\mathrm{d}x\\ &=e^{-im\xi}\int_{-\infty}^\infty e^{-n^2x^2}\,e^{-ix\xi}\,\mathrm{d}x\\ &=\frac{e^{-im\xi}}{n}\int_{-\infty}^\infty e^{-x^2}\,e^{-ix\xi/n}\,\mathrm{d}x\\ &=\frac{e^{-im\xi}}{n}\sqrt{\pi}\,e^{-\xi^2/(4n^2)}\tag{2} \end{align} $$

  • $\begingroup$ How exactly is the third equation justified by contour integration? $\endgroup$
    – Brian Bi
    Feb 22, 2014 at 22:38
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    $\begingroup$ @BrianBi: The difference of the integrals is the limit of the integral over the contour $$[-R,R]\cup\color{#C00000}{R+i\xi/2[0,1]}\cup [R,-R]+i\xi/2\cup\color{#C00000}{-R+i\xi/2[1,0]}$$ as $R\to\infty$ because the integral over the red pieces vanishes. Since the there are no singularities inside the contour, the integral around the contour is $0$. $\endgroup$
    – robjohn
    Feb 23, 2014 at 0:02
  • $\begingroup$ Oh, that makes sense. Sorry for being so dim lol $\endgroup$
    – Brian Bi
    Feb 23, 2014 at 0:58
  • $\begingroup$ would the downvoter care to comment? $\endgroup$
    – robjohn
    Jun 18, 2014 at 23:10
  • 1
    $\begingroup$ @LebronJames: unless you can show that they cancel, I don't see any immediate reason that they would. The usual way is to use pretty simple estimates as $R\to\infty$. $\endgroup$
    – robjohn
    Nov 11, 2015 at 6:52

While saz has already answered the question, I just wanted to add that this can be seen as one of the simplest examples of the Uncertainty Principle found in quantum mechanics, and generalizes to something called Hardy's uncertainty principle. In the QM context, momentum and position are each other's Fourier duals, and as you just discovered, a Gaussian function that's well-localized in one space cannot be well-localized in the other.

  • 1
    $\begingroup$ In this answer, I use this result to show that the Heisenberg Uncertainty Principle is sharp. $\endgroup$
    – robjohn
    Jan 5, 2013 at 20:24

This answer is basically an adaptation on robjohn's proof that instead of using contour integration relies on the identity theorem of complex analysis. For each $\xi \in \mathbb{R}$, we have that: \begin{equation*} \int_{-\infty}^\infty e^{-x^2}\,e^{-\xi x}\,\mathrm{d}x =e^{\xi^2/4}\int_{-\infty}^\infty e^{-\big(x+\frac{\xi}{2}\big)^2}\,\mathrm{d}x =e^{\xi^2/4}\int_{-\infty}^\infty e^{-x^2}\mathrm{d}x =\sqrt{\pi}\,e^{\xi^2/4}\;. \end{equation*} Now, we see that the two functions $f:\mathbb{C}\to\mathbb{C}, z\mapsto \int_{-\infty}^\infty e^{-x^2}\,e^{-xz}\,\mathrm{d}x$ and $g:\mathbb{C}\to\mathbb{C}, z\mapsto \sqrt{\pi}\,e^{z^2/4}$ are well-defined, holomorphic, and coincides on $\mathbb{R}$. We can then apply the identity theorem, obtaining that $f=g$. In particular, for each $\xi \in \mathbb{R}$, we have: \begin{equation*} \int_{-\infty}^\infty e^{-x^2}\,e^{-i\xi x}\,\mathrm{d}x=f(i\xi)=g(i\xi)=\sqrt{\pi}\,e^{(i\xi)^2/4}=\sqrt{\pi}\,e^{-\xi^2/4}\;. \end{equation*}

Then you can work out the rest of the proof exactly as in saz or robjohn's answer.


The following is instead a variant (that you may find in Folland's book on Real Analysis) of saz's proof, that does not rely on ODE uniqueness theorem, but instead on the following corollary of the Fundamental theorem of calculus: if $\varphi: \mathbb{R} \to \mathbb{R}$ is a continuously differentiable function such that $\varphi' = 0$, then $\varphi$ is constant.

With the same notations as in saz's answer and with the same argument, for each $\xi \in \mathbb{R}$, we have $$ \frac{\operatorname{d}}{\operatorname{d}\xi}\hat{f}(\xi)=-\frac{1}{2}\xi\hat{f}(\xi)\;. $$ In particular, since $\hat{f} \in C^0$, we have that $\hat{f}$ is $C^1$ (actually $C^\infty)$. Now: $$ \frac{\operatorname{d}}{\operatorname{d}\xi}\Big(\hat{f}(\xi)e^{\xi^2/{4}} \Big) = -\frac{1}{2}\xi\hat{f}(\xi)e^{\xi^2/4}+\hat{f}(\xi)e^{\xi^2/4}\frac{2\xi}{4} =0.$$ Then, the function $\varphi:\mathbb{R} \to \mathbb{R}, \xi \mapsto \hat{f}(\xi)e^{\xi^2/4}$ is $C^1$ and such that $\varphi'=0$. It follows that $\varphi$ is constant. To calculate this constant, we can compute $\varphi(0) = \hat{f}(0) = \int_{\mathbb{R}}e^{-x^2}\operatorname{d}x = \sqrt{\pi}$. Then, for each $\xi \in \mathbb{R}$ we have $$\hat{f}(\xi) = \varphi(\xi)e^{-\xi^2/4}=\varphi(0)e^{-\xi^2/4}=\sqrt{\pi}e^{-\xi^2/4} \;.$$ Then you can work out the rest of the proof exactly as in saz or robjohn's answer.


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