$\Gamma$ is $N(0,1)$ and $|\Gamma|$ is Folded Normal.
Show for $|\Gamma|\ge t$ (otherwise $0$) that the expectation $E(|\Gamma|)$ is twice the density of the normal dist.
So far: I have worked out up until the integral in which I get stuck at:
$\int_t^{\infty}z(e^\frac{-(z-\mu)^2}{2\sigma^2}+e^\frac{-(z+\mu)^2}{2\sigma^2}dz)$
This integral doesn't seem correct as it doesn't simplify nicely (ignoring the coefficients for now.)