How to work out this integral

why $$\int_{-\infty}^{+\infty} \exp \left(\frac{-(x-t)^2}{2}\right)\, dx=\sqrt {2\pi}$$

Here, $\exp$ is exponential.

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Do you know how to integrate using polar coordinates? Also which variable are you integrating with respect to? –  user10444 Feb 9 '14 at 11:52
yes, but I only know when its double integral you can change it to polar coord. –  Dylan Zhu Feb 9 '14 at 11:59
I think it is more $\sqrt{2\pi}$... –  D.L. Feb 9 '14 at 12:01
@D.L. yes, thank you, i correct it –  Dylan Zhu Feb 9 '14 at 12:29

let $y=x-t$ then then integral is $\int_{-\infty}^{\infty} e^{-y^2/2}dy$. Now let $z=y/\sqrt{2}$, then the integral becomes $\sqrt2\int_{-\infty}^{\infty} e^{-z^2}dz=\sqrt {2\pi}$. I asked if you know polar coordinates because to show $\int_{-\infty}^{\infty} e^{-z^2}dz=\sqrt \pi$ you can use polar coordinates here is why. The main idea of that proof is to take the integral $\int_{-\infty}^{\infty} e^{-z^2}dz$ and squaring it and noticing that the square is actually the double integral $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy$ which you can evalute using polar coordinates.
If you know the error function (also called the Gauss error function), the problem is simple since the antiderivative of the integrand is simply $$-\sqrt{\frac{\pi }{2}} \text{erf}\left(\frac{t-x}{\sqrt{2}}\right)$$ and when $y$ goes to infinity $\text{erf}(y)$ has an horizontal asymptote equal to $1$.