Compute the integral of the Gaussian

  • Thread starter Thread starter docnet
  • Start date Start date
  • Tags Tags
    Gaussian Integral
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
docnet
Messages
796
Reaction score
487
Homework Statement
.
Relevant Equations
.
Screen Shot 2021-12-15 at 2.52.46 AM.png


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?
 
Physics news on Phys.org
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.
 
Last edited:
Reply
  • Love
Likes   Reactions: docnet
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: