Compute the integral of the Gaussian

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    Gaussian Integral
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SUMMARY

The discussion focuses on computing the integral of the Gaussian function using Fourier transforms. The integral is expressed as $$\int_{-\infty}^\infty fe^{-ikx}dx=\sqrt{\pi}\sigma e^{(\frac{\sigma k}{2})^2}e^{-ikx_0}$$, demonstrating the relationship between the Fourier transform and the Gaussian function. Participants explore the process of completing the square and substituting variables to simplify the integral, ultimately leading to the result $$\tilde{f}(k) = \sigma \sqrt{\pi} e^{-ikx_0+\frac{\sigma^2k^2}{4}}$$. The discussion emphasizes the importance of justifying limits when dealing with complex variables.

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why does it say transforms? is there more than one Fourier transform?? we learned in class that the inverse Fourier transform of the Fourier transform of ##f## is ##f##, so there should be just one right? I'm uncertain of how to calulate this integral though.. Mr Wolfram showed me an indefinite integral involving an error function, but there has to be a different way to integrate it because we didn't learn about the error function. Mr Wolfram says

$$\int_{-\infty}^\infty fe^{-ikx}dx=\sqrt{\pi}\sigma e^{(\frac{\sigma k}{2})^2}e^{-ikx_0}$$

how does one go from the beginning to the finish?
 
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Write\begin{align*}
\tilde{f}(k) &= \int_{-\infty}^{\infty} \mathrm{exp}\left( -ikx - \frac{(x-x_0)^2}{\sigma^2} \right) dx \\
&= e^{-ikx_0} \int_{-\infty}^{\infty} \mathrm{exp}\left( -ik(x-x_0) - \frac{(x-x_0)^2}{\sigma^2} \right) dx\end{align*}from here a 'complete-the-square' substitution like ##u = \dfrac{x-x_0}{\sigma} + \dfrac{i\sigma k}{2}## looks helpful, but take some care to justify the limits of the integral over ##u## given that ##u## is complex.
 
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ergospherical said:
Write\begin{align*}
\tilde{f}(k) &= \int_{-\infty}^{\infty} \mathrm{exp}\left( -ikx - \frac{(x-x_0)^2}{\sigma^2} \right) dx \\
&= e^{-ikx_0} \int_{-\infty}^{\infty} \mathrm{exp}\left( -ik(x-x_0) - \frac{(x-x_0)^2}{\sigma^2} \right) dx\end{align*}from here a 'complete-the-square' substitution like ##u = \dfrac{x-x_0}{\sigma} + \dfrac{i\sigma k}{2}## looks helpful, but take some care to justify the limits of the integral over ##u## given that ##u## is complex.
\begin{align} \tilde{f}(k) =& \int_{-\infty}^{\infty} \mathrm{exp}\left( -ikx - \frac{(x-x_0)^2}{\sigma^2} \right) dx \\
=e^{-ikx_0} &\int_{-\infty}^{\infty} \mathrm{exp}\left( -ik(x-x_0) - \frac{(x-x_0)^2}{\sigma^2} \right)dx\\
=e^{-ikx_0} &\int_{-\infty}^{\infty} \mathrm{exp}\left( -\frac{\sigma^2i^2k^2}{4}+\frac{\sigma^2i^2k^2}{4}-ik(x-x_0) - \frac{(x-x_0)^2}{\sigma^2} \right)dx\\
=e^{-ikx_0-\frac{\sigma^2i^2k^2}{4}} &\int_{-\infty}^{\infty} \mathrm{exp}\left( -(\frac{\sigma i k}{2}+\frac{x-x_0}{\sigma})^2 \right)dx\end{align}
let ##u = \dfrac{x-x_0}{\sigma} + \dfrac{i\sigma k}{2}##. since ##u## is complex, it leads to integrating for ##u\in\{(-\infty,\infty)\times \{0\}\}##.
\begin{align}&\sigma e^{-ikx_0-\frac{\sigma^2i^2k^2}{4}}\int_ u e^{-u^2}du\\
=&\sigma e^{-ikx_0-\frac{\sigma^2i^2k^2}{4}} \Big[\frac{\sqrt{\pi}}{2}\text{erf}(u)\Big]_u\\
=&\sigma \sqrt{\pi} e^{-ikx_0+\frac{\sigma^2k^2}{4}}\end{align}
 
Last edited:
is it ok?
:)
 

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