Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \frac{\sqrt \pi}{2}$ 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.
No entender mucho el idioma pero aqui dejo mi solucion en una imagen didactica.