Statement. $F[e^{-\pi x^2}] = e^{-\pi x^2}$
Proof.
$$
\begin{eqnarray}
F[e^{-\pi x^2}](s) &=& \int_{\Bbb R} e^{-\pi x^2} e^{-2\pi i sx} dx \\
&=& \int_{\Bbb R} e^{-\pi x^2 -2\pi i sx} dx \\
&=& e^{\pi (is)^2} \int_{\Bbb R} e^{-\pi x^2 -2\pi i sx – \pi(i s)^2} dx \\
&=& e^{-\pi s^2} \int_{\Bbb R} e^{-\pi(x+i s)^2} dx.
\end{eqnarray}
$$
We can approach this integral via contour integration. Draw the box from $-R \to R \to R + is \to -R + is \to -R$. On the sides, we have $\int_{R}^{R+is} e^{-\pi (x+is)^2} dx$, which gets small when $R$ increases. Thus the top side and the bottom sides cancel out in the limit since the integral over the whole contour vanishes, so $\int_{\Bbb R} e^{-\pi(x+i s)^2} = \int_{\Bbb R} e^{-\pi x^2} = 1$.