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/Gaussian_function

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

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

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