How to calculate $\int \limits_{-\infty}^{+\infty} e^{-\frac{(x-m)^2}{2\sigma^2}}dx$ My textbook says that the value of the integral is $\sigma \sqrt{2 \pi}$. I suppose it should somehow be related to the Gaussian integral $\int \limits_{-\infty}^{+\infty} e^{-x^2}dx=\sqrt{\pi}$, but I have no clue how to connect two these facts. Thanks in advance.
 A: $$I=\int \limits_{-\infty}^{+\infty} e^{-\frac{(x-m)^2}{2\sigma^2}}dx$$
Substitute with u :
$$u=\frac {(x-m)}{\sqrt 2 \sigma} \implies dx=du\sqrt 2 \sigma$$
$$I=\sqrt 2 \sigma\int \limits_{-\infty}^{+\infty} e^{-u^2}du$$
$$I=\sqrt {2\pi} \sigma$$
A: First, notice that 
$$\int \limits_{-\infty}^{+\infty} e^{-\frac{(x-m)^2}{2\sigma^2}}dx 
= \int \limits_{-\infty}^{+\infty} e^{-\frac{x^2}{2\sigma^2}}dx.$$
Now do a u-sub.
A: Your guess is right , it is related to the Gaussian integral.
First notice that
$\int_{-\infty}^{+\infty} e^{-\frac{(x-m)^2}{2\sigma^2}}dx$ = $\int_{-\infty}^{+\infty} e^{-\frac{x^2}{2\sigma^2}}dx$ 
Now use polar coordinates as you used would do in evaluating $e^{-x^2}$
The general form of the Gaussian integral $\int_{-\infty}^{+\infty} e^{-a(x+b)^2}dx = \sqrt{\frac\pi a}$
In your case  $a= \frac1{2\sigma^2}$
So your integral evaluates to $\sigma\sqrt{2\pi}$
A: Let $u= \frac{x- m}{\sqrt{2}\sigma}$.  Then $du= \frac{1}{\sigma\sqrt{2}} dx$ so $dx= \sigma\sqrt{2} du$.  When x equals $\infty$ or $-\infty$ so does u so $\int_{-\infty}^{\infty}e^{-\frac{(x- m)^2}{2\sigma^2}}dx= \sigma\sqrt{2}\int_{-\infty}^{\infty}e^{-u^2} du$.
  Because $e^{-u^2}$ is symmetric about u= 0, $\int_{-\infty}^{\infty}e^{-u^2} du= 2\int_0^{\infty} e^{-u^2}du$.  Let $I= \int_0^\infty e^{-u^2}du$.  Then $\int_0^{\infty} e^{v^2}dv= I$ also.
$I^2= \left(\int_{-\infty}^{\infty}e^{-u^2} du\right)\left(\int_{-\infty}^{\infty}e^{-v^2} dv\right)= $\int_0^\infty \int_0^\infty e^{-u^2-v^2}dudv$.
That integral is over the first quadrant in the uv coordinate system.  Changing to polar coordinates we have $I^2= \int_0^{\pi/2}\int_0^\infty e^{-r^2} rdrd\theta= \frac{\pi}{2}\int_0^\infty r e^{-r^2} dr$.
Now that we have that "r" in the integrand we can make the substitution $v= r^2$ so that $dv= 2rdr$ and $\frac{1}{2}dv= r dr$.  
$I^2= \frac{\pi}{4}\int_0^\infty e^{-r}dr= \left[-\frac{\pi}{4}e^{-r}\right]_0^\infty$
As r goes to infinity, $e^{-r}$ goes to 0 and at r= 0, we have $-\frac{\pi}{4}$ so $I^2= 0- (-\frac{\pi}{4})= \frac{\pi}{4}$ and $I= \int_0^\infty e^{-u^2}du= \frac{\sqrt{\pi}}{2}$. 
And $\int_{-\infty}^\infty e^{-\frac{(x- m)^2}{\sqrt{2}\sigma}}dx= \sigma\sqrt{2}\int_{-\infty}^\infty e^{-u^2}du= \sigma\sqrt{2}I= \frac{\sigma\sqrt{\pi}}{\sqrt{2}}$
