 Solution

Fix real $y$ and use a rectangle with vertical sides at real coordinates $\pm R$. The function $e^{-z^2}$ is entire, so its closed contour integral is zero by Cauchy's integral theorem. The bottom horizontal side contributes ${\int }_{-R}^R{e}^{-x^2}dx$ and the top contributes the negative of ${\int }_{-R}^R{e}^{-(x+iy{)}^2}dx$. This identity applies for either sign of $y$, with the corresponding oriented rectangle.

On a vertical side, $z=\pm R+is$ with $|s|\le |y|$, andThe sum of the two vertical integrals in absolute value is at most $2|y|e^{-R^2+y^2}\to 0$. Both horizontal improper integrals converge absolutely, since $|e^{-(x+iy{)}^2}|=e^{y^2}{e}^{-x^2}$. Letting $R\to \infty$ therefore provesComplete the square in the Fourier transform:The shifted-integral identity applies after scaling by $\sqrt{2}$. The Gaussian integral is $\sqrt{2\pi }$: the square of $\int e^{-u^2}du$ becomes $2\pi {\int }_0^{\infty }{e}^{-r^2}rdr=\pi$ in polar coordinates. Thus the Gaussian Fourier transform yieldsThe eigenvalue reflects the unnormalized transform convention in this question.