How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ without changing into polar coordinates? How to prove $ \int_0^\infty e^{-x^2} \; dx  = \frac{\sqrt{\pi}}2$ other than changing into polar coordinates? It is possible to prove it using infinite series?

We use the change of variables $y=x^2$.
$$ \int_0^\infty e^{-x^2} \; dx = \frac{1}{2}\int_{0}^{\infty} y^{-1/2} {\rm e}^{-y} \,dy = \frac{1}{2} \Gamma(\frac{1}{2}) \,,$$
where $\Gamma(x)$ is defined by the integral,
$$\Gamma( x ) = \int_{0}^{\infty} y^{x-1} {\rm e}^{-y} \,dy \,. $$
Series Method
If you are interested in series method, then you can use Watson's lemma;
