I am trying to calculate this integral by DI method but didn't know how to do it (not looking for using gussian formula) :$ \int_{-\infty }^{+\infty} e^{-x^{2}}\,dx $ . Can anyone help me ?
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@KaviRamaMurthy how ? Guass proves it should be sqrt of pi – Adib Akkari Mar 04 '21 at 09:31
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@AdibAkkari : I expect your integrand is $e^{-x^2}$ – tommik Mar 04 '21 at 09:32
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@tommik sorry my bad. – Adib Akkari Mar 04 '21 at 09:33
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What is the "DI" method ? "not looking for using gussian formula": what ?? – Mar 04 '21 at 09:34
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@fantasie thank you it helped me so much !!! – Adib Akkari Mar 04 '21 at 09:40
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@AdibAkkari: Also see this. – V.G Mar 04 '21 at 09:41
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@LightYagami thank you everyone.... Hope didn't annoyed anyone . I AM A NOOB DON'T LAGH I SEE YOU LAGHING HAHAHAHAHA – Adib Akkari Mar 04 '21 at 09:43
1 Answers
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The proof is very simple and well known. It is enough to calculate first $I^2$ where $I$ is you integral passing in polar coordinates. Can you continue with this hint?
The method using polar coordinates is well explained here
If you are familiar with Gaussian distribution you can by-pass the problem with a substitution
$$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}\underbrace{\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy}_{=1}=\sqrt{\pi}$$
This simply substituting
$$x=\frac{y}{\sqrt{2}}$$
and using the known result because inside the integral you have the Gaussian standard density
tommik
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