# Evaluating $\int_{-\infty}^{\infty} e^{-{(x-1)}^2} dx$ [duplicate]

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

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

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

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