Complex evaluation of a classical (real) integral

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.

In the Portuguese book Variável Complexa by Maria A. Carreira and Maria S. M. de Nápoles, McGraw-Hill, 1997, this is evaluated in chapter 6, exercise 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 $$\begin{equation*} 2\pi i\text{ }\mathrm{res}(f,0)=2i=\lim_{r\rightarrow +\infty }\frac{4i}{ \sqrt{\pi }}\int_{0}^{r/\sqrt{\pi }}e^{-x^{2}}dx. \end{equation*}$$

If you take $$g(z)=\sum_{r=0}^{m-1}e^{\pi i a (z+r)^2/m}$$ and integrate $f(z)=\frac{g(z)}{e^{2\pi iz}-1}$ around the same parallelogram, you will get reciprocity law for quadratic Gauss sums: if the product $ma$ is even then $$S(a, m)= \sqrt{\frac ma}\frac{1+i}{\sqrt{2}}\overline{ S(m, a)}, $$ where $$S(a,m)=\sum_{r=0}^{m-1}e^{\pi i a r^2/m}.$$ In particular this observation allows to calculate quadratic Gauss sum (just take $a=2$). This example is taken from Apostol, T. M. Introduction to analytic number theory Springer-Verlag, 1976, section 9.10.