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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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  • $\begingroup$ @KaviRamaMurthy how ? Guass proves it should be sqrt of pi $\endgroup$ – Adib Akkari Mar 4 at 9:31
  • $\begingroup$ @AdibAkkari : I expect your integrand is $e^{-x^2}$ $\endgroup$ – tommik Mar 4 at 9:32
  • $\begingroup$ @tommik sorry my bad. $\endgroup$ – Adib Akkari Mar 4 at 9:33
  • $\begingroup$ What is the "DI" method ? "not looking for using gussian formula": what ?? $\endgroup$ – user65203 Mar 4 at 9:34
  • $\begingroup$ @fantasie thank you it helped me so much !!! $\endgroup$ – Adib Akkari Mar 4 at 9:40
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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

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  • $\begingroup$ Vey simple once you know the trick. Just impossible for a beginner. $\endgroup$ – user65203 Mar 4 at 9:34
  • $\begingroup$ how is that didn't get it .... $\endgroup$ – Adib Akkari Mar 4 at 9:34
  • $\begingroup$ @YvesDaoust : sorry, I will post a sketch now $\endgroup$ – tommik Mar 4 at 9:35
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    $\begingroup$ Do you expect the OP to know about multiple integrals and Jacobians ? $\endgroup$ – user65203 Mar 4 at 9:35

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