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Fourier transform of the Gaussian function

1   Derivation

Let $f(x)$ denote a generalized Gaussian function \begin{equation} f(x)=ae^{-b(x-c)^2} \label{eq:gaussian} \end{equation} where $b>0$.

The Fourier transform of Eq. \eqref{eq:gaussian} can be written as \begin{equation} \begin{split} F(u) &=\int_{-\infty}^\infty ae^{-b(x-c)^2}e^{-2i\pi ux}\,dx\\ &=a\int_{-\infty}^\infty e^{-bx^2+2bcx-bc^2}e^{-2i\pi ux}\,dx\\ &=ae^{-bc^2}\int_{-\infty}^\infty e^{-bx^2+2bcx}e^{-2i\pi ux}\,dx\\ &=ae^{-bc^2}\int_{-\infty}^\infty e^{-\pi\left(bx^2/\pi-2bcx/\pi+2iux\right)}\,dx\\ &=ae^{-bc^2}\int_{-\infty}^\infty e^{-\pi\left[bx^2/\pi+2(iu-bc/\pi)x\right]}\,dx. \end{split} \label{eq:fourier} \end{equation}

Multiplying the right-hand side of Eq. \eqref{eq:fourier} by $e^{\pi\left(iu\sqrt{\pi/b}-c\sqrt{b/\pi}\right)^2}e^{-\pi\left(iu\sqrt{\pi/b}-c\sqrt{b/\pi}\right)^2}=1$ yields \begin{equation} \begin{split} F(u) &=ae^{-bc^2}e^{\pi\left(iu\sqrt{\pi/b}-c\sqrt{b/\pi}\right)^2}\int_{-\infty}^\infty e^{-\pi\left[bx^2/\pi+2(iu-bc/\pi)x\right]}e^{-\pi\left(iu\sqrt{\pi/b}-c\sqrt{b/\pi}\right)^2}\,dx\\ &=ae^{-bc^2+\pi\left(iu\sqrt{\pi/b}-c\sqrt{b/\pi}\right)^2}\int_{-\infty}^\infty e^{-\pi\left[bx^2/\pi+2(iu-bc/\pi)x+\left(iu\sqrt{\pi/b}-c\sqrt{b/\pi}\right)^2\right]}\,dx\\ &=ae^{-bc^2+\pi\left(iu\sqrt{\pi/b}-c\sqrt{b/\pi}\right)^2}\int_{-\infty}^\infty e^{-\pi\left(\sqrt{b/\pi}\,x+iu\sqrt{\pi/b}-c\sqrt{b/\pi}\right)^2}\,dx. \end{split} \label{eq:fourier_2} \end{equation}

Letting $y=\sqrt{b/\pi}\,x+iu\sqrt{\pi/b}-c\sqrt{b/\pi}$ and $dy=\sqrt{b/\pi}\,dx$ results in \begin{equation} \begin{split} F(u) &=a\sqrt{\pi/b}\,e^{-bc^2+\pi\left(iu\sqrt{\pi/b}-c\sqrt{b/\pi}\right)^2}\int_{-\infty}^\infty e^{-\pi y^2}\,dy\\ &=a\sqrt{\pi/b}\,e^{-bc^2+\pi\left(iu\sqrt{\pi/b}-c\sqrt{b/\pi}\right)^2}. \end{split} \label{eq:fourier_3} \end{equation} because $\int_{-\infty}^\infty e^{-\pi y^2}\,dy=1$ (see the integration of the Gaussian function).

See also the inverse Fourier transform of the Gaussian function.

2   $f(x)=e^{-\pi x^2}$

Plugging $a=1$, $b=\pi$, and $c=0$ into Eq. \eqref{eq:fourier_3} gives us \begin{equation} \begin{split} F(u) &=\sqrt{\pi/\pi}\,e^{-\pi 0^2+\pi\left(iu\sqrt{\pi/\pi}-0\sqrt{\pi/\pi}\right)^2}\\ &=e^{\pi(iu)^2}\\ &=e^{-\pi u^2}. \end{split} \label{eq:fourier_ex_1} \end{equation}

3   $f(x)=\sqrt{2\pi}\,\sigma Ae^{-2\pi^2\sigma^2x^2}$

Plugging $a=\sqrt{2\pi}\,\sigma A$, $b=2\pi^2\sigma^2$, and $c=0$ into Eq. \eqref{eq:fourier_3} yields \begin{equation} \begin{split} F(u) &=\sqrt{2\pi}\,\sigma A\sqrt{\pi/\left(2\pi^2\sigma^2\right)}\,e^{-2\pi^2\sigma^20^2+\pi\left(iu\sqrt{\pi/\left(2\pi^2\sigma^2\right)}-0\sqrt{2\pi^2\sigma^2/\pi}\right)^2}\\ &=\sqrt{2\pi}\,\sigma\sqrt{\pi/\left(2\pi^2\sigma^2\right)}\,Ae^{\pi\left(iu\sqrt{\pi/\left(2\pi^2\sigma^2\right)}\right)^2}\\ &=Ae^{-u^2/(2\sigma^2)}. \end{split} \label{eq:fourier_ex_2} \end{equation}