$L_1$ norm of Gaussian random variable Ok. This is bit confusing. Let $g$ be a Gaussian random variable (normalized, i.e. with mean 0 and standard deviation 1). Then, in the expression $$\|g\|_{L_1}=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}|x|\exp(\frac{-x^2}{2})dx =\sqrt{\frac{2}{\pi}},$$ shouldn't the term $|x|$ not appear in the integral?

If $X$ is a random variable with density $f$, and $\phi$ is a measurable function, then $E[\phi(X)]=\int_{\Bbb R}\phi(t)f(t)dt$. As the $L^1$ norm of a random variable $X$ is $E[|X|]$, we have, when $X$ is normally distributed, the announced result.
