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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.

Rodrigo de Azevedo
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    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.) – snar Jan 04 '13 at 22:05
  • http://www.cse.yorku.ca/~kosta/CompVis_Notes/fourier_transform_Gaussian.pdf – vito Dec 07 '15 at 20:25
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    Related: http://math.stackexchange.com/questions/381597/ – Watson Mar 11 '17 at 15:07

5 Answers5

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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.

saz
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  • Thank you, this is starting to make a lot of sense now!! –  Jan 04 '13 at 22:47
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    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. – el_tenedor Sep 18 '16 at 13:35
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    @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. – saz Sep 18 '16 at 13:38
  • thanks. This does the trick. – 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 $$ \begin{align} \int_{-\infty}^\infty e^{-n^2(x-m)^2}\,e^{-ix\xi}\,\mathrm{d}x &=\int_{-\infty}^\infty e^{-n^2x^2}\,e^{-i(x+m)\xi}\,\mathrm{d}x\\ &=e^{-im\xi}\int_{-\infty}^\infty e^{-n^2x^2}\,e^{-ix\xi}\,\mathrm{d}x\\ &=\frac{e^{-im\xi}}{n}\int_{-\infty}^\infty e^{-x^2}\,e^{-ix\xi/n}\,\mathrm{d}x\\ &=\frac{e^{-im\xi}}{n}\sqrt{\pi}\,e^{-\xi^2/(4n^2)}\tag{2} \end{align} $$

robjohn
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  • How exactly is the third equation justified by contour integration? – Brian Bi Feb 22 '14 at 22:38
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    @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 $0$. – robjohn Feb 23 '14 at 00:02
  • Oh, that makes sense. Sorry for being so dim lol – Brian Bi Feb 23 '14 at 00:58
  • would the downvoter care to comment? – robjohn Jun 18 '14 at 23:10
  • Hi @robjohn, would it be correct to say that the integration over the red pieces cancel each other out, as we move along the box contour in a counterclockwise direction? Or, are there some issues at infinity, and we must show the estimates -- that the integrals go to zero, as R grows to infinity? The estimates are easy to give, but I was just wondering about whether the cancellation is true. Also, you fixed your box contour with a pretty specific height, it seems like -- but wouldn't *any* height be sufficient, call it $\beta$ = $Im(z)$? Thanks, – User001 Nov 10 '15 at 22:28
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    @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 $R\to\infty$. – robjohn Nov 11 '15 at 06:52
  • Hi @robjohn, yes, I am tempted to say that the integration over the red pieces cancel, since the integration along the second red piece is just in the opposite direction of integration along the first red piece. And, this kind of cancellation technique comes up in all sorts of elementary complex analysis proofs, too. Like the Cauchy integral theorems that show up pretty early. Or perhaps what we have here, integration along a box contour, with length of the box growing to infinity, is different, because of the growth of the contour to infinity? – User001 Nov 11 '15 at 19:18
  • That's what I am a bit confused about. I mean, writing out the integration along the red pieces just gives integration in the (real) y-variable, with differential dy. The only difference is that their limits of integration are in opposite order. What do you think? Thanks, @robjohn, – User001 Nov 11 '15 at 19:19
  • @LebronJames: $\left|e^{-z^2}\right|\le e^{-R^2+\xi^2/4}$ for $z$ on either of the red segments. Thus, the absolute value of the sum of the integrals on the red segments is less than or equal to $|\xi|\,e^{-R^2+\xi^2/4}\to0$ as $R\to\infty$. – robjohn Nov 12 '15 at 00:51
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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.

AndrewG
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    In [this answer](http://math.stackexchange.com/a/253928), I use this result to show that the Heisenberg Uncertainty Principle is sharp. – robjohn Jan 05 '13 at 20:24
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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.

Bob
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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.

Bob
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