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This question already has an answer here:

I have come across this formula through one of the members of this forum but i dont know how to prove this formula,can someone help me proving this formula.
$\int\limits_{-\infty}^{\infty}e^{-x^2}\color{red}{dx}=\sqrt \pi$

I know that graph of $e^{-x^2}$ is inverted bell shape,but i am intrigued as to how area between x-axis,this inverted bell from minus infinity to plus infinity is $\sqrt\pi$.

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marked as duplicate by mvw, user186170, Did, Community Jul 29 '15 at 8:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ This is a classic, asked many times on this site. Define it as $I$ and consider $I^2$ which is a double integral. Summing up the plane not along the cartesian coordinate axis, but rather using polar coordinates ("Trick of Poisson") will lead from $dx\, dy$ to $r\, dr\, d\phi$ and allow to solve $I^2$ and thus $I$. $\endgroup$ – mvw Jul 29 '15 at 8:22
  • $\begingroup$ Here is an interesting way to show a similar result using the residue theorem: math.stackexchange.com/questions/1266856/… $\endgroup$ – Ron Gordon Jul 29 '15 at 8:38
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Using function $\Gamma$ $$\int_{-\infty }^\infty e^{-x^2}dx=2\int_0^\infty e^{-x^2}dx=2\int_0^\infty u^{\frac{1}{2}}e^{-u}du=2\Gamma\left(\frac{3}{2}\right)=2\frac{\sqrt \pi}{2}=\sqrt \pi$$

Using polar coordinate

Let $I=\int_{-\infty }^\infty e^{-x^2}dx$.

$$I^2=\iint_{-\infty }^\infty e^{-(x^2+y^2)}dxdy=\int_0^{2\pi}d\theta\int_0^\infty re^{-r^2}dr=2\pi\left[-\frac{e^{-r^2}}{2}\right]_0^\infty =\pi,$$ and thus $$I=\sqrt \pi.$$

Using normal law

$$1=\frac{1}{\sqrt{2\pi}}\int_{-\infty }^\infty e^{\frac{-x^2}{2}}dx\underset{u=x/\sqrt{2}}{=}\frac{\sqrt 2}{\sqrt{2\pi}}\int_{-\infty }^\infty e^{-u^2}du=\frac{1}{\sqrt \pi}\int_{-\infty }^\infty e^{-u^2}du\implies \int_{-\infty }^\infty e^{-u^2}du=\sqrt\pi.$$

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    $\begingroup$ Why post this archi-duplicate answer? $\endgroup$ – Did Jul 29 '15 at 8:35
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    $\begingroup$ And further, the "solution" involving the Gamma function is circular. We prove that Gamma function results by using the gaussian integral whose result you are allegedly deriving. $\endgroup$ – Ron Gordon Jul 29 '15 at 8:36
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$$ I=\int\limits_{-\infty}^{\infty}e^{-x^2}dx\\I=\int\limits_{-\infty}^{\infty}e^{-y^2}dy\\ I*I=I^2=\int\limits_{-\infty}^{\infty}e^{-x^2}dx\int\limits_{-\infty}^{\infty}e^{-y^2}dy=\\ \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}e^{-x^2}e^{-y^2}dxdy\\=\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy$$ now use polar coordinate $$x^2+y^2=r^2 \\x=rcos \theta\\y=r sin \theta \\ \theta =tan^{-1}\frac{y}{x}\\dxdy =r dr d\theta $$ so $$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy=\\\int\limits_{0}^{2\pi}\int\limits_{0}^{\infty} e^{-(r^2)}rdr d\theta =\\ \int\limits_{0}^{2\pi}(\frac{e^{-r^2}}{-2})_(0,\infty)vd\theta =\\ \int\limits_{0}^{2\pi}(0-(-\frac{1}{2})) d\theta =2\pi (\frac{1}{2}) =\pi\\\overset{I^2=\pi}{\rightarrow} I=\sqrt{\pi}$$

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    $\begingroup$ If you're going to derive a result that has already been derived here and probably in about 457 other places within this site, at least have the decency to format the content properly. $\endgroup$ – Ron Gordon Jul 29 '15 at 8:39
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The classic solution uses polar coordinates, but there are others.

For example, define $$h(t)=\int_0^{+\infty}\frac{e^{-tx^2}}{1+x^2}dx$$ You can show that $h$ is differentiable and that it is a solution of a simple differential equation. By solving this equation you can find the value of $I$.

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