calculus - Proving $int_{0}^{infty} mathrm{e}^{-x^2} dx = frac{sqrt pi}{2}$

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

Answer

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.