Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \frac{\sqrt \pi}{2}$

This is an old favorite of mine.
Define $$I=\int_{-\infty}^{+\infty} e^{-x^2} dx$$ Then $$I^2=\bigg(\int_{-\infty}^{+\infty} e^{-x^2} dx\bigg)\bigg(\int_{-\infty}^{+\infty} e^{-y^2} dy\bigg)$$ $$I^2=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)} dxdy$$ Now change to polar coordinates
$$I^2=\int_{0}^{+2 \pi}\int_{0}^{+\infty}e^{-r^2} rdrd\theta$$ The $\theta$ integral just gives $2\pi$, while the $r$ integral succumbs to the substitution $u=r^2$ $$I^2=2\pi\int_{0}^{+\infty}e^{-u}du/2=\pi$$ So $$I=\sqrt{\pi}$$ and your integral is half this by symmetry

I have always wondered if somebody found it this way, or did it first using complex variables and noticed this would work.

A variation on Ross Millikan's answer.

We can start again with the observation $$\left(\int_{-\infty}^{\infty}e^{-x^2}dx\right)\left(\int_{-\infty}^{\infty}e^{-y^2}dy\right)=\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy\right)=V.$$ Now $V$ is simply the volume of the body $$-\infty < x,y < \infty,\qquad 0 < z < e^{-(x^2+y^2)},$$ or, equivalently, $$0 < x^2+y^2 < -\ln z,\qquad 0 < z < 1.$$ This implies that the body is a solid of revolution. Using the disk integration formula, we have $$V=\int_{0}^{1}\pi(-\ln z)dz=[\pi(z-z\ln z)]_{0}^1=\pi.$$

This is Exercise 7.19 of Apostle's Mathematical Nalysis book (second edition))

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Define $f$ and $g$ as:

$$f(x):=\left(\int_0^x e^{-t^2}dt\right)^{2} \quad\text{and}\quad g(x):=\left(\int_{0}^{1}\frac{e^{-x^{2}(t^{2}+1)}}{t^{2}+1}dt\right)$$

Now, $$f'(x)=2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}dt$$ and

$$g'(x)=\int_0^1 \frac{\partial}{\partial x}\left[\frac{e^{-x^2(t^2+1)}}{t^2+1}\right]dt = -2xe^{-x^{2}}\int_{0}^{1}e^{-x^{2}t^{2}}dt$$

So putting $t=tx$, get $\displaystyle\int_{0}^{1}e^{-x^{2}t^{2}}dt= \frac{1}{x}\displaystyle\int_{0}^{x}e^{-t^{2}}dt$

Then we get:

$$g'(x)=-2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}dt$$

Thus $f'(x)+g'(x)=0$ for all $x$, then $f(x)+g(x)$ is an constant function. Also $$f(0)+g(0)=\displaystyle\int_{0}^{1}\frac{1}{t^{2}+1}dt = \displaystyle\frac{\pi}{4}$$

Then $f(x)+g(x)=\displaystyle\frac{\pi}{4}$ for all $x$.

Now $\lim_{x \to{+}\infty}{g(x)}=0$

So $$\displaystyle\frac{\pi}{4} = \lim_{x \to{+}\infty}{f(x)+g(x)}=\lim_{x \to{+}\infty}{f(x)}= \left(\int_{0}^{\infty}e^{-t^{2}}dt\right)^{2}$$

Thus

$$\int_{0}^{\infty}e^{-t^{2}}dt=\sqrt{\frac{\pi}{4}}= \frac{\sqrt{\pi}}{2}$$

The end.