Solution
= Solution
By the even <probability density> of the <standard normal distribution>,
$$
\mathbb E|Z|=\frac2{\sqrt{2\pi}}\int_0^\infty z e^{-z^2/2}\,\mathrm dz=\frac2{\sqrt{2\pi}}\left[-e^{-z^2/2}\right]_0^\infty=\boxed{\sqrt{\frac2\pi}}.
$$
The integral is finite. Equivalently, $|Z|$ has the <half-normal distribution> with unit scale.