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?

$$ \int_{-\infty} ^\infty e^{-x^2} \; dx  =I$$
$$ \int_{-\infty}^\infty e^{-y^2} \; dx  =I$$
$$\int_{-\infty}^\infty e^{-x^2}\times e^{-y^2} \; dx  =I^2$$
$$\int_{-\infty}^\infty e^{-(x^2+y^2)}\; dx  =I^2$$
Polar Coordinates: $x^2+y^2=R^2,-\pi\leqslant \theta\leqslant +\pi$
$$\int_{-\pi}^{+\pi} \int_{0}^ \infty e^{-(R^2)}\; R\times dR \times d(\theta)  =I^2$$
$$2\times \pi \int_{0}^ \infty e^{-(R^2)}\; R\times dR   =I^2$$
$$-R^2=P\implies -2Rd(R)=d(P)\implies Rd(R)=-d(P)/2$$
$$2\times \pi \int_{-\infty}^ 0 e^{(P)}d(P)/2$$
$$\pi=I^2\Longrightarrow  I=\sqrt{\pi}$$
since $e^{x^2}$ is even $\implies$ ans=${\sqrt{\pi}/2}$
