compagnia04

2021-12-31

How to prove
${\int }_{0}^{\mathrm{\infty }}{e}^{-{x}^{2}}dx=\frac{\sqrt{\pi }}{2}$

movingsupplyw1

This is an old favorite of mine.
Define
$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{x}^{2}}dx$
Then ${I}^{2}=\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{x}^{2}}dx\right)\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{y}^{2}}dy\right)$
${I}^{2}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-\left({x}^{2}+{y}^{2}\right)}dxdy$
Now change to polar coordinates
${I}^{2}={\int }_{0}^{+2\pi }{\int }_{0}^{+\mathrm{\infty }}{e}^{-{r}^{2}}rdrd\theta$
The $\theta$ integral just gives $2\pi$, while the r integral succumbs to the substitution $u={r}^{2}$
${I}^{2}=2\pi {\int }_{0}^{+\mathrm{\infty }}{e}^{-u}d\frac{u}{2}=\pi$
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.

Carl Swisher

We can start again with the observation
$\left({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-{x}^{2}}dx\right)\left({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-{y}^{2}}dy\right)=\left({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\left({x}^{2}+{y}^{2}\right)}dxdy\right)=V$
Now V is simply the volume of the body

$0
or, equivalently,
$0<{x}^{2}+{y}^{2}<-\mathrm{ln}z$
$0
This implies that the body is a solid of revolution. Using the disk integration formula, we have
$V={\int }_{0}^{1}\pi \left(-\mathrm{ln}z\right)dz={\left[\pi \left(z-z\mathrm{ln}z\right)\right]}_{0}^{1}=\pi$

Vasquez

Define f and g as:
$f\left(x\right):=\left({\int }_{0}^{x}{e}^{-{t}^{2}}dt{\right)}^{2}$ and $g\left(x\right):=\left({\int }_{0}^{1}\frac{{e}^{-{x}^{2}\left({t}^{2}+1\right)}}{{t}^{2}+1}dt\right)$
Now,
${f}^{\prime }\left(x\right)=2{e}^{-{x}^{2}}{\int }_{0}^{x}{e}^{-{t}^{2}}dt$
and
${g}^{\prime }\left(x\right)={\int }_{0}^{1}\frac{d}{dx}\left[\frac{{e}^{-{x}^{2}\left({t}^{2}+1\right)}}{{t}^{2}+1}\right]dt=-2x{e}^{-{x}^{2}}{\int }_{0}^{1}{e}^{-{x}^{2}{t}^{2}}dt$
So putting t=tx get ${\int }_{0}^{1}{e}^{-{x}^{2}{t}^{2}}dt=\frac{1}{x}{\int }_{0}^{x}{e}^{-{t}^{2}}dt$
Then we get:
${g}^{\prime }\left(x\right)=-2{e}^{-{x}^{2}}{\int }_{0}^{x}{e}^{-{t}^{2}}$
Thus f'(x)+g'(x)=0 for all x, then f(x)+g(x) is an constant function. Also
$f\left(0\right)+g\left(0\right)={\int }_{0}^{1}\frac{1}{{t}^{2}+1}dt=\frac{\pi }{4}$
Then $f\left(x\right)+g\left(x\right)=\frac{\pi }{4}$ for all x.
NSK
Now $\underset{x\to +\mathrm{\infty }}{lim}g\left(x\right)=0$
So
$\frac{\pi }{4}=\underset{x\to +\mathrm{\infty }}{lim}f\left(x\right)+g\left(x\right)=\underset{x\to +\mathrm{\infty }}{lim}f\left(x\right)=\left({\int }_{0}^{\mathrm{\infty }}{e}^{-{t}^{2}}dt{\right)}^{2}$
Thus
${\int }_{0}^{\mathrm{\infty }}{e}^{-{t}^{2}}dt=\sqrt{\frac{\pi }{4}}=\frac{\sqrt{\pi }}{2}$

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