# Fourier Transform of Gaussian function

Approximating the incoming stimulus arising from the sensory neurons as a Gaussian in both space and time. Such a stimulus has the normalized form \begin{eqnarray} &&\phi_n({\bf{r}}, t)=\frac{{\text e}^{-\frac{1}{2}\left(\frac{t-t_{os}}{t_s}\right)^2}}{t_s\sqrt(2\pi)}\frac{{\text e}^{\left(\frac{\left|{\bf r}-{{\bf r}_{os}}\right|}{r_s}\right)^2}}{\pi r_s^2}. \end{eqnarray} where $t_s$ is the characteristics duaration of the stimulus, $t_{os}$ is the transmission delay from cochlea to thalamus, $r_s$ is the spatial width of the stimulus at cortex, and $\bf r_{os}$ is the offset of the centre of the stimulus from the point of measurement. The Fourier transform of this function is \begin{eqnarray} \phi_n({\bf{k}}, \omega)&&={\text{e}}^{-\frac{1}{2}\omega^2 t_s^2}{\text{e}}^{i \omega t_{os}}{\text {e}}^{-(\frac{kr_s}{2})^2}{\text {e}}^{{{\text{i}}\bf{k}.\bf{r_{os}}}} \end{eqnarray} Can anyone please help me to show every step to find the Fourier transform of this function? Thanks in advance!

• There is a problem, did you mean $e^{\displaystyle\color{green}{-}\frac{|r-r_{os}|^2}{r_s^2}}$ ? – reuns Aug 9 '17 at 9:43
• In that case, all you need is to know the Fourier transform of $e^{-t^2}$, then use the linearity, shift and scale rules for the FT which follows from a linear change of variable $t = at'+b$ – reuns Aug 9 '17 at 9:44

IMO one of the most intuitive proof is to set $$F(s) = \int_{-\infty}^\infty e^{-t^2} e^{-2st}dt, \qquad s \in \mathbb{C}$$ Assuming $s \in \mathbb{R}$ then $$F(s) = \int_{-\infty}^\infty e^{s^2} e^{-(t+s)^2}dt =e^{s^2}\int_{-\infty}^\infty e^{-t^2}dt= e^{s^2} F(0)$$ And to convince yourself that $F(s)-e^{s^2}F(0)=0$ stays true for every $s \in \mathbb{C}$, because both sides are complex analytic functions of $s$ (a non-zero power series $\sum_{n=0}^\infty a_n s^n$ cannot vanish for every $s\in \mathbb{R}$).
Thus $\displaystyle\int_{-\infty}^\infty e^{-t^2}e^{-i \omega t}dt = F(i\omega/2)=e^{-(\omega/2)^2} F(0)$ and $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-a^2(t+b)^2}e^{-c^2(r+d)^2}e^{-i k r}e^{-i \omega t}dtdr = \frac{1}{|a|}e^{-(\omega/2)^2/a^2} e^{-i \omega b}\frac{1}{|c|}e^{-(k/2)^2/c^2} e^{-i kd}F(0)^2$$ Finally by inverse Fourier transform we show $F(0) = \sqrt{\pi}$.