<s>[INST] 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?
 [/INST] 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.
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