How to calculate $\int \limits_{-\infty}^{+\infty} e^{-\frac{(x-m)^2}{2\sigma^2}}dx$

Question

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.

Answer 1

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.

Answer 2

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 $\sigma\sqrt{2\pi}$

Answer 3

$$I=\int \limits_{-\infty}^{+\infty} e^{-\frac{(x-m)^2}{2\sigma^2}}dx$$ Substitute with u : $$u=\frac {(x-m)}{\sqrt 2 \sigma} \implies dx=du\sqrt 2 \sigma$$

$$I=\sqrt 2 \sigma\int \limits_{-\infty}^{+\infty} e^{-u^2}du$$ $$I=\sqrt {2\pi} \sigma$$

Answer 4

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 $\int_{-\infty}^\infty e^{-\frac{(x- m)^2}{\sqrt{2}\sigma}}dx= \sigma\sqrt{2}\int_{-\infty}^\infty e^{-u^2}du= \sigma\sqrt{2}I= \frac{\sigma\sqrt{\pi}}{\sqrt{2}}$