Evaluation of Gaussian integral $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx$ How to prove
 $$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2}$$

Here's a proof that only requires elementary, but clever, calculus manipulations.
Let $I_A = \int_0^A e^{-x^2}dx$.
$$\begin{split} I_A^2 &= \int_0^A \int_0^A e^{-(x^2+y^2)}dxdy\\
&= \int_0^A \int_0^A \sum_{n\geq 0} \frac{(-1)^n}{n!}(x^2+y^2)^ndxdy\\
&=\sum_{n\geq 0} \frac{(-1)^n}{n!}\int_0^A \int_0^A\sum_{k=0}^n{n \choose k}x^{2k}y^{2n-2k}dxdy\\
&=\sum_{n\geq 0} \frac{(-1)^n}{n!}\sum_{k=0}^n{n \choose k}\frac{A^{2k+1}}{2k+1}\frac{A^{2n-2k+1}}{2n-2k+1}\\
&= \sum_{n\geq 0} \frac{(-1)^n}{n!}A^{2n+2}\sum_{k=0}^n{n \choose k}\frac{1}{2k+1}\frac{1}{2n-2k+1}
\end{split}$$
Now, note that
$$\begin{split}
\sum_{k=0}^n{n \choose k}\frac{1}{2k+1}\frac{1}{2n-2k+1} &= \frac 1 {2n+2}\sum_{k=0}^n{n \choose k}\left (\frac{1}{2k+1}+\frac{1}{2n-2k+1}\right)\\
& =\frac 1 {n+1}\sum_{k=0}^n{n \choose k}\frac{1}{2k+1}
\end{split}
$$
Thus,
$$\begin{split} I_A^2 &=
\sum_{n\geq 0} \frac{(-1)^n}{(n+1)!}A^{2n+2}\sum_{k=0}^n{n \choose k}\frac{1}{2k+1}\\
&= \sum_{n\geq 0} \frac{(-1)^n}{(n+1)!}A^{2n+2}\sum_{k=0}^n{n \choose k}\int_0^1x^{2k}dx\\
&= \sum_{n\geq 0} \frac{(-1)^n}{(n+1)!}A^{2n+2}\int_0^1(1+x^2)^ndx\\
&= \int_0^1\frac 1 {1+x^2}\sum_{n\geq 0}\frac{(-1)^n}{(n+1)!}A^{2n+2}(1+x^2)^{n+1}dx\\
&= \int_0^1\frac{1-e^{-A^2(1+x^2)}}{1+x^2}dx\\
&= \frac{\pi}4 + \mathcal O\left(e^{-A^2}\right)
\end{split}$$
Taking the limit as $A\rightarrow+\infty$ yields the result.
