Evaluation of Gaussian integral $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx$ How to prove
 $$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2}$$

This is similar to user17762's answer, but uses Plancherel's Theorem instead of Poisson summation.  Define the Fourier transform by
$$\mathcal{F}[f](y) = \hat{f}(y) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x) e^{-i xy}\;dx$$
Now,
$$\mathcal{F}[e^{-\frac{1}{2}x^2}]= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac{1}{2}x^2} e^{-i xy}\;dx = \frac{e^{-\frac{1}{2}y^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac{1}{2}(x+iy)^2}\;dx$$
Now, consider the contour integral of $e^{-\frac{1}{2}z^2}$ over the rectangular contour with corners at $\pm R$ and $\pm R + iy$.  This integral must be $0$, since $e^{-\frac{1}{2}z^2}$ is analytic.  Taking the limits as $R \to +\infty$, the contributions from the vertical edges go to $0$, so we find that
$$\int_{-\infty}^\infty e^{-\frac{1}{2}x^2}\;dx = \int_{-\infty}^\infty e^{-\frac{1}{2}(x+iy)^2}\;dx$$
Thus, 
$$\mathcal{F}[e^{-\frac{1}{2}x^2}](y) = \frac{e^{-\frac{1}{2}y^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{1}{2}x^2}\;dx$$
By Plancherel's Theorem,
$$\int_{-\infty}^\infty e^{x^2}\;dx = \int_{-\infty}^\infty e^{-y^2} \frac{1}{2\pi}\left(\int_{-\infty}^\infty e^{-\frac{1}{2}x^2}\;dx\right)^2\;dy = \frac{1}{2\pi}\left(\int_{-\infty}^\infty e^{-\frac{1}{2}x^2}\;dx\right)^2\int_{-\infty}^\infty e^{-y^2}\;dy$$
dividing through by $\int_{-\infty}^\infty e^{-x^2}\;dx$, we find
$$1 = \frac{1}{2\pi}\left(\int_{-\infty}^\infty e^{-\frac{1}{2}x^2}\;dx\right)^2$$
so,
$$\sqrt{2\pi} = \int_{-\infty}^\infty e^{-\frac{1}{2}x^2}\;dx$$
Changing variables and observing that $\int_0^\infty e^{-x^2}\;dx = \frac{1}{2}\int_{-\infty}^\infty e^{-x^2}\;dx$, we find that
$$\int_0^\infty e^{-x^2}\;dx = \frac{\sqrt{\pi}}{2}$$
