Question

There are several ways to compute the classical integral $$ \int_{\mathbb R}e^{-x^2}dx=\sqrt{\pi}. $$ Probably, best known are

(1) squaring the integral with subsequent change of (now two) variables to the polar form, and

(2) the reducing to the Gamma-function at $1/2$.

I am interested though in a "complex" analysis method (namely, a use of the residue theorem) to do the job. The reason is that several integrals like $$ \int_0^\infty e^{-x^2}\cos ax\ dx \qquad\text{or}\qquad \int_0^\infty\sin x^2\ dx $$ can be computed via the residue theorem and the above integral, so I would like to avoid any reference to real analysis. Is there such a complex evaluation though?!

Answer 1

Yes! For a long time that was thought impossible, but then it was found how to do it using a parallelogram as a contour.

Desbrow, Darrell
On evaluating $\int_{-\infty}^\infty e^{ax(x-2b)}dx$ by contour integration round a parallelogram.
Amer. Math. Monthly 105 (1998), no. 8, 726–731.

According to Desbrow, the parallelogram integral evaluation for the probability integral is due to Mirsky, 1949.

Answer 2

In the Portuguese book Variável Complexa by Maria A. Carreira and Maria S. M. de Nápoles, McGraw-Hill, 1997, this is solved in chapter 6, exercice 23. The function $$\begin{equation*} f(z)=\frac{e^{i\pi z^{2}}}{\sin \left( \pi z\right) } \end{equation*}$$ is integrated around the following paralellogram $$\gamma _{1,3}(t) =te^{i\pi /4}\pm 1/2\qquad -r\leq t\leq r$$ $$\gamma _{2,4}(t) =\mp re^{i\pi /4}+t\qquad -1/2\leq t\leq 1/2. $$ The computation shows that $$\begin{equation*} 2\pi i\text{ }\mathrm{res}(f,0)=2i=\lim_{r\rightarrow +\infty }\frac{4i}{ \sqrt{\pi }}\int_{0}^{r/\sqrt{\pi }}e^{-x^{2}}dx. \end{equation*}$$