How to prove
 $$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2}$$
This is an old favorite of mine.
Define $$I=\int_{-\infty}^{+\infty} e^{-x^2} dx$$ 
Then $$I^2=\bigg(\int_{-\infty}^{+\infty} e^{-x^2} dx\bigg)\bigg(\int_{-\infty}^{+\infty} e^{-y^2} dy\bigg)$$  
$$I^2=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)} dxdy$$  
Now change to polar coordinates
$$I^2=\int_{0}^{+2 \pi}\int_{0}^{+\infty}e^{-r^2} rdrd\theta$$  
The $\theta$ integral just gives $2\pi$, while the $r$ integral succumbs to the substitution $u=r^2$  
$$I^2=2\pi\int_{0}^{+\infty}e^{-u}du/2=\pi$$ 
So $$I=\sqrt{\pi}$$ and your integral is half this by symmetry

I have always wondered if somebody found it this way, or did it first using complex variables and noticed this would work.
