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?

Method 1. Let $I$ denote the integral. Then for $s > 0$, the substitution $t = x\sqrt{s}$ gives
$$ \int_{0}^{\infty} e^{-sx^2} \; dx = \frac{1}{\sqrt{s}} \int_{0}^{\infty} e^{-t^2} \; dt = \frac{I}{\sqrt{s}}. $$
Thus for $u = x^2$,
$$\begin{align*}I^2
&= \int_{0}^{\infty} I e^{-x^2} \; dx
= \int_{0}^{\infty} \frac{I}{2\sqrt{u}}e^{-u} \; du
= \int_{0}^{\infty} \frac{1}{2}e^{-u}\int_{0}^{\infty}e^{-ut^2}\;dtdu\\
&= \int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{2}e^{-u}e^{-ut^2}\;dudt
= \int_{0}^{\infty}\frac{1}{2(t^2+1)}\;dt
= \frac{\pi}{4},
\end{align*}$$
which proves the desired result. Note that this is in fact equivalent to the proof in Wikipedia.
Method 2. We assume that we are aware of the Wallis product
$$ \prod_{n=1}^{\infty} \left(\frac{2n}{2n-1} \cdot \frac{2n}{2n+1}\right) = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2}. $$
Now, it is not hard to find that
$$ f_n(x) = \begin{cases} \left(1 - \tfrac{x^2}{n}\right)^{n} & 0 \leq x \leq \sqrt{n} \\ 0 & \text{otherwise}. \end{cases} $$
increases to $e^{-x^2}$. (Simple application of Bernoulli's inequality will suffice.) Then by monotone convergence theorem,
$$ I := \int_{0}^{\infty} e^{-x^2} \; dx = \lim_{n\to\infty} \int_{0}^{\sqrt{n}} \left(1 - \frac{x^2}{n}\right)^{n} \; dx = \lim_{n\to\infty} \sqrt{n} \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}\theta \; d\theta, $$
where we have used the substitution $x = \sqrt{n}\,\cos\theta$. Now let
$$I_n = \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}\theta \; d\theta. $$
Then for $n \geq 1$,
$$\begin{align*}I_n
&= \left[\sin^{2n}\theta (-\cos\theta)\right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2n \sin^{2n-1}\theta \cos\theta (-\cos\theta) \; d\theta \\
&= \int_{0}^{\frac{\pi}{2}} 2n \sin^{2n-1}\theta (1-\sin^2 \theta) \; d\theta
= 2n(I_{n-1} - I_n)
\end{align*}$$
and we have $I_0 = 1$ and $I_n = \frac{2n}{2n+1} I_{n-1}$. Thus 
$$ I_n = \prod_{k=1}^{n} \frac{2k}{2k+1}. $$
Therefore, by Wallis product,
$$ I^2 = \lim_{n\to\infty} n \prod_{k=1}^{n} \left( \frac{2k}{2k+1} \cdot \frac{2k}{2k+1} \right) = \lim_{n\to\infty} \frac{n}{2n+1} \prod_{k=1}^{n} \left( \frac{2k}{2k-1} \cdot \frac{2k}{2k+1} \right) = \frac{1}{2} \cdot \frac{\pi}{2} $$
and we have the desired result. (Conversely, knowing $I$, it can be used to prove the Wallis product formula.)
