Mean absolute value of a normal

Idea. Fold the normal density onto the positive half-line and integrate directly: $z\phi (z)$ has an explicit antiderivative.

Step 1. $$\begin{array}{rl}\mathbb{E}|Z|& =2{\int }_0^{\infty }z\frac{e^{-z^2/2}}{\sqrt{2\pi }}dz\\ & =\frac{2}{\sqrt{2\pi }}[-e^{-z^2/2}{]}_0^{\infty }=\frac{2}{\sqrt{2\pi }}.\end{array}$$

Step 2. $$\frac{2}{\sqrt{2\pi }}=\sqrt{\frac{2}{\pi }}\approx \mathrm{0.798.}$$

Uses. For $X\sim N(0,{\sigma }^2)$, $\mathbb{E}|X|=\sigma \sqrt{2/\pi }\approx 0.8\sigma$: the mean absolute deviation of a normal is about 80% of its standard deviation. The same number is $\mathbb{E}[Z\mid Z>0]$.

The trap. $\frac{1}{\sqrt{2\pi }}\approx 0.399$ is $\mathbb{E}[\max (Z,0)]$, half of $\mathbb{E}|Z|$.