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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.

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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.) –  snarski Jan 4 '13 at 22:05

3 Answers 3

up vote 5 down vote accepted

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

Using partial integration we get

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


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

Since $c=\hat{f}(0) = \int_{\mathbb{R}} f(x) \, dx$ one can determine the constant $c$ (correct result: $c=\sqrt{\pi}$). Moreover, 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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Thank you, this is starting to make a lot of sense now!! –  user55225 Jan 4 '13 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 $$ \begin{align} \int_{-\infty}^\infty e^{-n^2(x-m)^2}\,e^{-ix\xi}\,\mathrm{d}x &=\int_{-\infty}^\infty e^{-n^2x^2}\,e^{-i(x+m)\xi}\,\mathrm{d}x\\ &=e^{-im\xi}\int_{-\infty}^\infty e^{-n^2x^2}\,e^{-ix\xi}\,\mathrm{d}x\\ &=\frac{e^{-im\xi}}{n}\int_{-\infty}^\infty e^{-x^2}\,e^{-ix\xi/n}\,\mathrm{d}x\\ &=\frac{e^{-im\xi}}{n}\sqrt{\pi}\,e^{-\xi^2/(4n^2)}\tag{2} \end{align} $$

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How exactly is the third equation justified by contour integration? –  Brian Bi Feb 22 '14 at 22:38
@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 0:02
Oh, that makes sense. Sorry for being so dim lol –  Brian Bi Feb 23 '14 at 0:58
would the downvoter care to comment? –  robjohn Jun 18 '14 at 23:10

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

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