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$$\int_{-\infty}^{\infty} e^{-({x-1)}^2} dx$$

I tried polar coordinates
but couldn't apply the right
variable transformation...

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marked as duplicate by Community Sep 10 '17 at 17:07

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  • $\begingroup$ The same as $\int_{-\infty}^\infty e^{-x^2}\,dx$ surely? $\endgroup$ – Lord Shark the Unknown Sep 10 '17 at 16:56
  • $\begingroup$ Refer to Gaussian integral. $\endgroup$ – Prasun Biswas Sep 10 '17 at 16:57
  • $\begingroup$ Change of variable $X=x-1$ and it will work in polar coordinates. $\endgroup$ – JJacquelin Sep 10 '17 at 16:57
  • $\begingroup$ Start with $u = x-1$ then you have the more familiar Gaussian. $\endgroup$ – Doug M Sep 10 '17 at 16:59