$$\int_{-\infty}^{\infty} e^{-({x-1)}^2} dx$$
I tried polar coordinates
but couldn't apply the right
variable transformation...
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The same as $\int_{-\infty}^\infty e^{-x^2},dx$ surely? – Angina Seng Sep 10 '17 at 16:56
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Refer to Gaussian integral. – Prasun Biswas Sep 10 '17 at 16:57
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Change of variable $X=x-1$ and it will work in polar coordinates. – JJacquelin Sep 10 '17 at 16:57
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Start with $u = x-1$ then you have the more familiar Gaussian. – user317176 Sep 10 '17 at 16:59