Finding $E[|X-\mu|]$ dealing with absolute integration. I would like to find $E[|X-\mu|]$ when $X$ is $N(\mu, \sigma^2)$.
I think I need to evaluate the integral:
$$E[|X-\mu|]=\int_{\mathbb{R}} |x-\mu| \frac{1}{\sigma\sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2} dx$$
This is the same as the following since $e^u>0$
$$=\int_{\mathbb{R}} \left|(x-\mu) \frac{1}{\sigma\sqrt{2 \pi}}e^{-\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2}\right| \, dx$$
So, I need to find the solution to:
$$(x-\mu) \frac{1}{\sigma\sqrt{2 \pi}}e^{-\frac{1}{2}\left( \frac{x-\mu}{\sigma} \right)^2}=0$$
To determine where to break the integral. The above is only zero where $x=\mu$. So I break it at $\mu$
$$E[|X-\mu|]=\int_{-\infty}^\mu \left|(x-\mu) \frac{1}{\sigma\sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2}\right| \, dx + \int_\mu^\infty \left|(x-\mu) \frac{1}{\sigma\sqrt{2 \pi}}e^{-\frac{1}{2}\left( \frac{x-\mu}{\sigma} \right)^2}\right| \, dx$$
If $x \in (-\infty, \mu)$ then $x-\mu<0$ so we can multiply the first integral by $-1$ and remove the absolute value bars. 
$$E[|X-\mu|]=-\int_{-\infty}^\mu (x-\mu) \frac{1}{\sigma\sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2} \, dx + \int_\mu^\infty (x-\mu) \frac{1}{\sigma\sqrt{2 \pi}}e^{-\frac{1}{2}\left( \frac{x-\mu} \sigma \right)^2} \, dx$$
now we have:
$$=-\int_{-\infty}^{\mu}  \frac{1}{\sigma\sqrt{2 \pi}}xe^{-\frac{1}{2}\left( \frac{x-\mu}{\sigma} \right)^2}dx +\mu\int_{-\infty}^{\mu}  \frac{1}{\sigma\sqrt{2 \pi}}e^{-\frac{1}{2}\left( \frac{x-\mu}{\sigma} \right)^2}dx +{} $$
$$\int_\mu^\infty \frac{1}{\sigma\sqrt{2 \pi}}xe^{-\frac{1}{2}\left( \frac{x-\mu}{\sigma} \right)^2} \, dx  -\mu\int_\mu^\infty \frac{1}{\sigma\sqrt{2 \pi}}e^{-\frac{1}{2}\left( \frac{x-\mu}{\sigma} \right)^2} \, dx$$
I'm not certain how to deal with these integrals? Any hints?
 A: $X-\mu\sim N(0,\sigma^2)$. thus 
$E(|X-\mu|)=\int_R|x|\frac{1}{\sqrt{2\pi}\sigma}e^{-x^2/2\sigma^2}dx=2\int^\infty_0 x\frac{1}{\sqrt{2\pi}\sigma}e^{-x^2/2\sigma^2}dx$.
The antiderivative is $\frac{-2\sigma^2}{\sqrt{2\pi}}e^{-x^2/2\sigma^2}$. Thus the answer is $2\sigma^2/\sqrt{2\pi}$
A: You omitted some simplifications.  In the first place if $X\sim N(\mu,\sigma^2)$ we can let $Y= X-\mu$ so that $Y\sim N(0,\sigma^2)$ and then we have $\operatorname{E}|X-\mu| = \operatorname{E}|Y|.$
Next, if we let $Z= Y/\sigma$ then we have $Z\sim N(0,1)$ and $\operatorname{E}|Y| = \sigma\operatorname{E}|Z|.$
So the number you seek is $\sigma\operatorname{E}|Z|.$
Next you have
\begin{align}
\operatorname{E}|Z| = {} & \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty |z| e^{-z^2/2} \, dz = \frac 2 {\sqrt{2\pi}} \int_0^\infty z e^{-z^2/2} \,dz \\
& \qquad \text{because we are integrating an even function} \\
& \qquad \text{over a set that is symmetric about 0} \\[15pt]
= {} & \frac 2 {\sqrt{2\pi}} \int_0^\infty e^{-z^2/2} \Big( z\, dz\Big) = \frac 2 {\sqrt{2\pi}} \int_0^\infty e^{-u} \, du = \frac 2 {\sqrt{2\pi}}\cdot 1. \\
& \qquad \text{(Note that as $z$ goes from $0$ to $\infty$, so does $u$, so the} \\
& \qquad \phantom{\text{(}} \text{bounds in this last integral are from $0$ to $\infty$.)}
\end{align}
$\left(\text{And if you like you can write } \dfrac 2 {\sqrt{2\pi}} = \sqrt{\dfrac 2 \pi}. \right)$
