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Fourier Transform--Gaussian

The Fourier Transform of a Gaussian Function $f(x) \equiv e^{-ax^2}$ is given by

$\displaystyle F(k)$ $\textstyle =$ $\displaystyle \int_{-\infty}^\infty e^{-ax^2}e^{ikx}\,dx$  
  $\textstyle =$ $\displaystyle \int_{-\infty}^\infty e^{-ax^2}[\cos (kx)+i\sin (kx)]\,dx$  
  $\textstyle =$ $\displaystyle \int_{-\infty}^\infty e^{-ax^2} \cos (kx)\,dx+i\int_{-\infty}^\infty \,e^{-ax^2}\sin (kx)\,dx.$  

The second integrand is Even, so integration over a symmetrical range gives 0. The value of the first integral is given by Abramowitz and Stegun (1972, p. 302, equation 7.4.6)

\begin{displaymath}
F(k)=\sqrt{\pi\over a} \,e^{-k^2/4a},
\end{displaymath}

so a Gaussian transforms to a Gaussian.

See also Gaussian Function


References

Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, 1972.




© 1996-9 Eric W. Weisstein
1999-05-26