Revision as of 10:16, 18 December 2014

Problem

Fourier transform of the Gauss function

Show that $$\frac{1}{\sqrt{2\pi t}} \int_{-\infty}^\infty e^{-\frac{x^2}{2t}} e^{-ikx} dx = e^{-\frac{1}{2}tk^2} , \forall k \in \mathbb{R}$$

Proof

$$ \begin{align} I \ &= \frac{1}{\sqrt{2\pi t}} \int_{-\infty}^\infty e^{-\frac{x^2}{2t}} e^{-ikx} dx \\ &= \frac{1}{\sqrt{2\pi t}} \int_{-\infty}^\infty e^{-\frac{1}{2t}(x^2 + 2ikt)} dx \\ &= \frac{1}{\sqrt{2\pi t}} \int_{-\infty}^\infty e^{-\frac{1}{2t}((x + ikt)^2 + k^2t^2)} dx \\ &= \frac{1}{\sqrt{2\pi t}} e^{-\frac{k^2t}{2}} \int_{-\infty}^\infty e^{-\frac{1}{2t}(x + ikt)^2} dx . \end{align} $$

Now, we substitute \(u = x + ikx\) with \(du = dx\) and find $$ \begin{align} I \ &= \frac{1}{\sqrt{2\pi t}} e^{-\frac{k^2t}{2}} \int_{-\infty}^\infty e^{-\frac{1}{2t}u^2} du\\ &= \frac{1}{\sqrt{2\pi t}} e^{-\frac{k^2t}{2}} \sqrt{2\pi t} \\ &= e^{-\frac{k^2 t}{2}} . \end{align} $$

Where we have used that $$\int_{-\infty}^\infty e^{-x^2 a} dx = \sqrt{\frac{\pi}{a}}$$

Thus, the proof follows. $$\raggedleft{} \square$$

(No, I don't know how to right align a square. Change this if you do!) --Nik (talk) 11:16, 18 December 2014 (CET)