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# Inverse Fourier transform of the Gaussian function

## 1   Derivation

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

The inverse Fourier transform of Eq. \eqref{eq:gaussian} can be written as $$\begin{split} f(x) &=\int_{-\infty}^\infty ae^{-b(u-c)^2}e^{2i\pi ux}\,du\\ &=a\int_{-\infty}^\infty e^{-bu^2+2bcu-bc^2}e^{2i\pi ux}\,du\\ &=ae^{-bc^2}\int_{-\infty}^\infty e^{-bu^2+2bcu}e^{2i\pi ux}\,du\\ &=ae^{-bc^2}\int_{-\infty}^\infty e^{-\pi\left(bu^2/\pi-2bcu/\pi-2iux\right)}\,du\\ &=ae^{-bc^2}\int_{-\infty}^\infty e^{-\pi\left[bu^2/\pi-2(ix+bc/\pi)u\right]}\,du. \end{split} \label{eq:fourier}$$

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

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

## 2   $F(u)=e^{-\pi u^2}$
Plugging $a=1$, $b=\pi$, and $c=0$ into Eq. \eqref{eq:fourier_3} gives us $$\begin{split} f(x) &=\sqrt{\pi/\pi}\,e^{-\pi 0^2+\pi\left(ix\sqrt{\pi/\pi}+0\sqrt{\pi/\pi}\right)^2}\\ &=e^{\pi(ix)^2}\\ &=e^{-\pi x^2}. \end{split} \label{eq:fourier_ex_1}$$
## 3   $F(u)=Ae^{-u^2/(2\sigma^2)}$
Plugging $a=A$, $b=1/(2\sigma^2)$, and $c=0$ into Eq. \eqref{eq:fourier_3} yields $$\begin{split} f(x) &=A\sqrt{2\pi\sigma^2}\,e^{-0^2/(2\sigma^2)+\pi\left(ix\sqrt{2\pi\sigma^2}+0\sqrt{1/(2\pi\sigma^2)}\right)^2}\\ &=\sqrt{2\pi\sigma^2}\,Ae^{\pi\left(ix\sqrt{2\pi\sigma^2}\right)^2}\\ &=\sqrt{2\pi}\,\sigma Ae^{-2\pi^2\sigma^2x^2}. \end{split} \label{eq:fourier_ex_2}$$