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?
 A: Because $x \mapsto e^x$ is convex, $\displaystyle \left( 1- \frac{u}{n} \right)^n \leq e^{-u} \leq \left( 1+ \frac{u}{n} \right)^{-n}$, so $\displaystyle \int_0^{\sqrt{n}} \left( 1- \frac{x^2}{n} \right)^n dx \leq \int_0^{\sqrt{n}} e^{-x^2} dx \leq \int_0^{\sqrt{n}} \left( 1+ \frac{x^2}{n} \right)^{-n} dx$.
We use the change of variables $x= \sqrt{n} \sin(\theta)$: $\displaystyle \int_0^{\sqrt{n}} \left( 1- \frac{x^2}{n} \right)^n dx = \int_0^{\pi/2} \cos^{2n+1} (\theta) d\theta= \sqrt{n} W_{2n+1}$, where $W_{2n+1}$ is a Wallis integral.
If $p \geq \sqrt{n}$, $\displaystyle \int_0^{\sqrt{n}} \left(1+ \frac{x^2}{n} \right)^{-n} dx \geq \int_0^p \left(1+ \frac{x^2}{n} \right)^{-n} dx$. We use the change of variables $x=\sqrt{n} \tan (\theta)$ to find: $\displaystyle \int_0^{\arctan(p)} \sqrt{n} \cos^{2n+2} (\theta) d \theta \underset{p \to + \infty}{\longrightarrow} \sqrt{n} W_{2n+2}$.
Finally, $\displaystyle \sqrt{n} W_{2n+1} \leq \int_0^{\sqrt{n}} e^{-x^2}dx \leq \sqrt{n} W_{2n+2}$. But $\displaystyle W_n \underset{n \to + \infty}{\sim} \sqrt{\frac{\pi}{2n}}$ so $\displaystyle \int_0^{+ \infty} e^{-x^2}dx= \frac{\sqrt{\pi}}{2}$.
With this method, an asymptotic development of $W_n$ gives an asymptotic development of the integral.
A: $$ \int_{-\infty} ^\infty e^{-x^2} \; dx  =I$$
$$ \int_{-\infty}^\infty e^{-y^2} \; dx  =I$$
$$\int_{-\infty}^\infty e^{-x^2}\times e^{-y^2} \; dx  =I^2$$
$$\int_{-\infty}^\infty e^{-(x^2+y^2)}\; dx  =I^2$$
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=${\sqrt{\pi}/2}$
A: 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;
A: 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.)
A: Consider the mapping $\eta\! : \mathbb{R}^2\to \mathbb{R}$ given by
$$
\eta((x,y)) = \sqrt{x^2+y^2},\quad (x,y)\in\mathbb{R}^2.
$$
(1) Show that the image-measure $\lambda_2\circ\eta^{-1}$ is the measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ with density
$$
f(z)=2\pi z 1_{(0,\infty)}(z),\quad z\in\mathbb{R},
$$
by using Dynkin's Lemma.
(2) Show that
$$
\int_{\mathbb{R}^2} e^{-x^2-y^2}\,\lambda_2(\mathrm{d}x,\mathrm{d}y)=\pi
$$
by using the formula for integration under measureable transformations.
(3) Use Tonelli's theorem to conclude that
$$
\int_{\mathbb{R}}e^{-x^2}\,\lambda(\mathrm{d}x)=\sqrt{\pi},
$$
and now your result follows.
A: First, we notice that

$$n!~=~\int_0^\infty e^{-\sqrt[n]x}~dx\quad\iff\quad\tfrac1n!~=~\int_0^\infty e^{-x^n}dx\quad\rightarrow\quad\tfrac12!~=~\int_0^\infty e^{-x^2}dx.$$

Then, we further notice that

$$\int_0^1\Big(1-\sqrt[n]x\Big)^m~dx~=~\int_0^1\Big(1-\sqrt[m]x\Big)^n~dx~=~\frac1{C_{m+n}^n}~=~\frac1{C_{m+n}^m}~=~\frac{m!~n!}{(m+n)!},$$

from where we ultimately deduce that

$$\frac\pi4~=~\int_0^1\sqrt{1-x^2}~dx~=~\frac{\Big(\frac12!\Big)^2}{\Big(\frac12 + \frac12\Big)!}~=~\Big(\tfrac12!\Big)^2,$$

which, in its turn, leads us to finally conclude that

$$\int_0^\infty e^{-x^2}dx~=~\tfrac12!~=~\sqrt{\pi\over4}~=~\frac{\sqrt\pi}2.$$

Q.E.D.
A: Let $I=\displaystyle \int_0^{+\infty} e^{-t^2}dt$
Let $f(x)=\displaystyle \int_0^{\tfrac{\pi}{4}} e^{-\tfrac{x^2}{\cos^2t}}dt$ 
(notice : since $f(x)\leq e^{-x^2}$ then $\lim_{x \rightarrow +\infty}f(x)=0$ )
Let $g(x)=\displaystyle \int_0^x e^{-t^2}dt$
First:
For $x>0$
$\displaystyle x\int_0^{\tfrac{\pi}{4}} \dfrac{e^{-x^2\tan^2 t}}{\cos^2t}dt=g(x)$ (1)
(perform change of variable $u=x\tan t$ )
$\displaystyle-2e^{-x^2}g(x)=-2e^{-x^2}\int_0^x e^{-t^2}dt=-2e^{-x^2}x\int_0^{\tfrac{\pi}{4}} \dfrac{e^{-x^2\tan^2 t}}{\cos^2t}dt$ (using (1))
$\displaystyle-2e^{-x^2}g(x)=-2x\int_0^{\tfrac{\pi}{4}} \dfrac{e^{-x^2(1+\tan^2 t)}}{\cos^2t}dt=-2x\int_0^{\tfrac{\pi}{4}} \dfrac{e^{-\tfrac{x^2}{\cos^2t}}}{\cos^2t}dt=f'(x)$ (2)
$(g(x)^2)'=2e^{-x^2}g(x)=-f'(x)$ (using (2))
Therefore:
$g(x)^2+f(x)$ is constant for all $x$
and $g(0)^2+f(0)=\dfrac{\pi}{4}$
so for all $x$
$g(x)^2+f(x)=\dfrac{\pi}{4}$
take the limit of the left member when $x$ goes to infinity:
$I^2=\dfrac{\pi}{4}$
So $I=\dfrac{\sqrt{\pi}}{2}$
