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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
up vote 2 down vote accepted

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.

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"you need polar coordinates" ... in fact, there are other ways to do it – GEdgar Feb 9 '14 at 14:49
Even in the paper I linked to there are other ways, I just meant its usually done that way. English isn't my first language so sorry if it sounded like I meant it was the only way. – user10444 Feb 9 '14 at 15:09

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 $1$.

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