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For the Fourier transform defined as $$\frac {1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) e^{-i\alpha x}\,dx$$ I know there is simple formula for the Fourier transformation and inverse transformation of $$f(x)=e^{-x^2/a}$$ but I can't remember it or derive it on my own. Does anyone have a formula for this?

Olivier Oloa
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Mecury-197
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  • http://math.stackexchange.com/questions/270566/how-to-calculate-the-fourier-transform-of-a-gaussian-function?rq=1 – Yimin May 06 '15 at 19:14
  • @Yimin I'm just discovering your link... hope my answer is a different one from that in the link :) – Olivier Oloa May 06 '15 at 19:36

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Assume $a$ is a real number such that $a>0$. Set $$F(\alpha):=\frac {1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}f(x) e^{-i\alpha x}dx $$ with $\displaystyle f(x):=e^{-x^2/a}$. You may differentiate $F$ giving $$F'(\alpha)=-\frac {i}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}x e^{-x^2/a}e^{-i\alpha x}dx $$ then integrate by parts $$ \begin{align} F'(\alpha)=\left.\frac {i}{2a\sqrt{2\pi}}e^{-i\alpha x}e^{-x^2/a}\right|_{-\infty}^{+\infty}-\frac {\alpha}{2a\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-x^2/a}e^{-i\alpha x}dx =-\frac {\alpha}{2a}F(\alpha) \end{align}$$ observing that $\displaystyle F(0)=\frac{\sqrt{\pi }}{\sqrt{a}}$ you get $$ F(\alpha)=\frac{\sqrt{\pi }}{\sqrt{a}}e^{-\Large \frac{\alpha ^2}{4 a}}. $$

Olivier Oloa
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    I'm not sure all steps here are correct. When I try to evaluate $F(0)$ with wolfram alpha, I get the following result: $F(0) = \sqrt{a/2}$. Are you sure your claim $F(0) = \sqrt{\pi/a}$ is correct? – MaximusIdeal Sep 10 '22 at 06:06
  • It looks like your result applies for $F(\alpha):= \int_{-\infty}^{+\infty}f(x) e^{-i\alpha x}dx$ without the $1/\sqrt{2\pi}$ factor and $f(x) = e^{-ax^2}$. You are using a different Fourier transform convention and a different function $f(x)$ than the one in OP's post. – MaximusIdeal Sep 10 '22 at 20:38