# Fourier transform of a Gaussian

There is a trick to solve the integral: take the derivative with respect to k.

$\displaystyle F(k)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\exp^{-\lambda x^2} \exp^{i k x}\, dx$

Differentiate with respect to k:

$\displaystyle \frac{dF(k)}{dk}=\frac{i}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} x \exp^{-\lambda x^2} \exp^{i k x}\, dx$

Integrate by parts,

$\displaystyle du=x \exp^{-\lambda x^2}$
$\displaystyle u = \frac{-\exp^{-\lambda x^2}}{2 \lambda}$
$\displaystyle v = \exp^{i k x}$
$\displaystyle dv= i k \exp^{i k x}$

After some manipulation

$\displaystyle \frac{dF(k)}{d k} =\frac{ -k F(k)}{2 \lambda}$

$\displaystyle F(k) = F(0) \exp^{\frac{ -k^2}{4 \lambda}}$

$\displaystyle F(0) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \exp^{-\lambda x^2}\, dx$

$\displaystyle F(k) = \sqrt{\frac{2}{\lambda}} \exp^{\frac{-k^2}{4 \lambda} }$

## References

http://en.wikipedia.org/wiki/Integral_of_a_Gaussian_function