= 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 $\int_{-R}^Re^{-x^2}dx$ and the top contributes the negative of $\int_{-R}^Re^{-(x+iy)^2}dx$. This identity applies for either sign of $y$, with the corresponding oriented rectangle.
On a vertical side, $z=\pm R+is$ with $|s|\leq|y|$, and
$$
|e^{-z^2}|=e^{-R^2+s^2}\leq e^{-R^2+y^2}.
$$
The sum of the two vertical integrals in absolute value is at most $2|y|e^{-R^2+y^2}\to0$. Both horizontal improper integrals converge absolutely, since $|e^{-(x+iy)^2}|=e^{y^2}e^{-x^2}$. Letting $R\to\infty$ therefore proves
$$
\boxed{\int_{\mathbb R}e^{-(x+iy)^2}\,dx=\int_{\mathbb R}e^{-x^2}\,dx}.
$$
Complete the square in the <Fourier transform>:
$$
\widehat f(y)=e^{-y^2/2}\int_{\mathbb R}e^{-(x+iy)^2/2}\,dx=e^{-y^2/2}\int_{\mathbb R}e^{-x^2/2}\,dx.
$$
The shifted-integral identity applies after scaling by $\sqrt2$. The <Gaussian integral> is $\sqrt{2\pi}$: the square of $\int e^{-u^2}du$ becomes $2\pi\int_0^\infty e^{-r^2}r\,dr=\pi$ in polar coordinates. Thus the <Gaussian Fourier transform> yields
$$
\boxed{\widehat f=\sqrt{2\pi}\,f}.
$$
The <eigenvalue> reflects the unnormalized transform convention in this question.
Back to article page
