Fourier Transform--Gaussian

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

$$F(k) = \int_{-\infty}^\infty e^{-ax^2}e^{ikx}\,dx$$

$$= \int_{-\infty}^\infty e^{-ax^2}[\cos (kx)+i\sin (kx)]\,dx$$

$$= \int_{-\infty}^\infty e^{-ax^2} \cos (kx)\,dx+i\int_{-\infty}^\infty \,e^{-ax^2}\sin (kx)\,dx.$$

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)

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

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.