Compute the integral of the Gaussian

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    Gaussian Integral
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The discussion revolves around the computation of the integral of the Gaussian function and the confusion regarding the Fourier transform. Participants question the existence of multiple Fourier transforms, clarifying that the inverse Fourier transform should return the original function. The integral involving the error function is highlighted, with a focus on finding alternative integration methods not covered in class. A detailed step-by-step approach is provided, including a substitution to simplify the integral, ultimately leading to the expression involving the error function and the Gaussian integral. The conversation emphasizes the importance of justifying the limits of integration 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}
 
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is it ok?
:)
 
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...