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I am trying to calculate the FT of

$f(x)=\frac{1}{\sqrt{2 \pi a}} e^{-\frac{(x-x_0)^2}{2a}}$ where a is some constant.

Using the definition I get $\hat f(\mu)=\int_{-\infty}^{\infty}\frac{1}{\sqrt(2 \pi a)} e^{-\frac{(x-x_0)^2}{2a}} e^{i \mu x} dx$

I have tried to solve the integral, but with no success.

I tried to use WolframAlpha, but I just get "no results...". I would be thankful if someone could help me with this integral (either solving or proofing that this integral does not exist)

John.W
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    you can complete the square in the exponential, change variables, and then use Cauchy's theorem to shift the contour of integration back to the real axis. It boils down to being able to calculate $\int_{\Bbb{R}}e^{-t^2},dt$ (for example using polar coordinates, or if you know some complex analysis, you can use properties of the Gamma function. There are surely a billion other proofs I don't know). Another option is to differentiate with respect to $\mu$, and solve the resulting ODE. I'm sure all of these approaches are explained in detail in other MSE posts. You have to just find them. – peek-a-boo Nov 17 '21 at 18:11
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    As @peek-a-boo noted, defining $t:=(x-x_0)/a$ reduces to the case $x_0=0,,a=1$, which several math.se questions discuss; see e.g. here. – J.G. Nov 17 '21 at 18:15
  • Does this answer your question? Graphical proof of flat/spread-out Fourier transforms of Gaussian distribution – LL 3.14 Nov 17 '21 at 20:27
  • Did you just try to google "Fourier transform of gaussian function" first? – LL 3.14 Nov 17 '21 at 20:28

1 Answers1

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Using a couple of variable substitutions and the fact that $$ \int_{-\infty}^{\infty} e^{-\frac{z^2}{2}} e^{i \zeta z} \,dz = \sqrt{2\pi} e^{-\frac{\zeta^2}{2}} $$ I get $$\begin{align} \hat f(\mu) &= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi a}} e^{-\frac{(x-x_0)^2}{2a}} e^{i \mu x} dx = \{ y=x-x_0 \} \\ &= \frac{1}{\sqrt{2 \pi a}} \int_{-\infty}^{\infty} e^{-\frac{y^2}{2a}} e^{i \mu (y+x_0)} \,dy \\ &= \frac{1}{\sqrt{2 \pi a}} e^{i\mu x_0} \int_{-\infty}^{\infty} e^{-\frac{y^2}{2a}} e^{i \mu y} \,dy = \{ z=y/\sqrt{a} \} \\ &= \frac{1}{\sqrt{2 \pi a}} e^{i\mu x_0} \int_{-\infty}^{\infty} e^{-\frac{z^2}{2}} e^{i \mu z\sqrt{a}} \sqrt{a} \,dz \\ &= \frac{1}{\sqrt{2\pi}} e^{i\mu x_0} \int_{-\infty}^{\infty} e^{-\frac{z^2}{2}} e^{i \mu z\sqrt{a}} \,dz \\ &= \frac{1}{\sqrt{2\pi}} e^{i\mu x_0} \left. \int_{-\infty}^{\infty} e^{-\frac{z^2}{2}} e^{i \zeta z} \,dz \right|_{\zeta=\mu\sqrt{a}} \\ &= \frac{1}{\sqrt{2\pi}} e^{i\mu x_0} \left. \sqrt{2\pi} e^{-\frac{\zeta^2}{2}} \right|_{\zeta=\mu\sqrt{a}} \\ &= \frac{1}{\sqrt{2\pi}} e^{i\mu x_0} \sqrt{2\pi} e^{-\frac{\mu^2 a}{2}} \\ &= e^{i\mu x_0} e^{-\frac{\mu^2 a}{2}}. \\ \end{align}$$ The result seems correct since $f(x)\to\delta(x-x_0)$ as $a\to 0$ and $$ \int_{-\infty}^{\infty} \delta(x-x_0) e^{i\mu x} \, dx = e^{i\mu x_0} $$ which is the limit of the found expression.

md2perpe
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