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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.

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    $\begingroup$ 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 '13 at 22:05
  • $\begingroup$ cse.yorku.ca/~kosta/CompVis_Notes/… $\endgroup$ – vito Dec 7 '15 at 20:25
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    $\begingroup$ Related: math.stackexchange.com/questions/381597 $\endgroup$ – Watson Mar 11 '17 at 15:07
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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.

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  • $\begingroup$ Thank you, this is starting to make a lot of sense now!! $\endgroup$ – user55225 Jan 4 '13 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 '16 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 '16 at 13:38
  • $\begingroup$ thanks. This does the trick. $\endgroup$ – el_tenedor Sep 18 '16 at 13:43
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$$ \begin{align} \int_{-\infty}^\infty e^{-x^2}\,e^{-ix\xi}\,\mathrm{d}x &=e^{-\xi^2/4}\int_{-\infty}^\infty e^{-(x+i\xi/2)^2}\,\mathrm{d}x\\ &=e^{-\xi^2/4}\int_{-\infty+i\xi/2}^{\infty+i\xi/2}e^{-x^2}\mathrm{d}x\\ &=e^{-\xi^2/4}\int_{-\infty}^\infty e^{-x^2}\mathrm{d}x\\ &=\sqrt{\pi}\,e^{-\xi^2/4}\tag{1} \end{align} $$ 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 $$ \begin{align} \int_{-\infty}^\infty e^{-n^2(x-m)^2}\,e^{-ix\xi}\,\mathrm{d}x &=\int_{-\infty}^\infty e^{-n^2x^2}\,e^{-i(x+m)\xi}\,\mathrm{d}x\\ &=e^{-im\xi}\int_{-\infty}^\infty e^{-n^2x^2}\,e^{-ix\xi}\,\mathrm{d}x\\ &=\frac{e^{-im\xi}}{n}\int_{-\infty}^\infty e^{-x^2}\,e^{-ix\xi/n}\,\mathrm{d}x\\ &=\frac{e^{-im\xi}}{n}\sqrt{\pi}\,e^{-\xi^2/(4n^2)}\tag{2} \end{align} $$

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  • $\begingroup$ How exactly is the third equation justified by contour integration? $\endgroup$ – Brian Bi Feb 22 '14 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 '14 at 0:02
  • $\begingroup$ Oh, that makes sense. Sorry for being so dim lol $\endgroup$ – Brian Bi Feb 23 '14 at 0:58
  • $\begingroup$ would the downvoter care to comment? $\endgroup$ – robjohn♦ Jun 18 '14 at 23:10
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    $\begingroup$ @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 '15 at 6:52
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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.

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    $\begingroup$ In this answer, I use this result to show that the Heisenberg Uncertainty Principle is sharp. $\endgroup$ – robjohn♦ Jan 5 '13 at 20:24

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