Gaussian integral is of the form (modulo constant) $I = \int_{-\infty}^\infty e^{- x^2} dx$ How to evaluate this? here residue theorem cannot help you. you need to use a little trick.

Consider the double integral $I^2 = \int_{-\infty}^\infty\int_{-\infty}^\infty e^{- x^2 -y^2 } dx dy$ then switch to radial coordinate $I^2 = \int_{r=0}^{\infty}\int_{\theta=0}^{2\pi} e^{-r^2} r drd\theta = 2 \pi \int_{r=0}^{\infty} e^{-r^2} r dr$ subsitatue $r^2=u$, we get $2rdr = du$, and $I^2 = \pi \int_0^\infty e^{-u} du = \pi$ ok, we get $I = \sqrt{\pi}.$

OK, not bad. How about more general case? For $a>0$, consider $I_a = \int_{-\infty}^\infty e^{- a x^2} dx$ we can change variable, let $u = \sqrt{a} x$, then $dx = (1/\sqrt{a}) du$, $I_a = (1/\sqrt{a}) \int_{-\infty}^\infty e^{- u^2} du = (1/\sqrt{a}) I = (1/\sqrt{a}) \sqrt{\pi}.$

How about $a = r e^{i \theta}$? and $0<\theta«1$? and then making $\theta$ larger, and larger?