$L_1$ 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?

It should! The $L_1$-norm of a random variable $X$ with density $f(x)$ equals
$$\int_{\Omega} |X(w)| d\mu(\omega) = \int_{-\infty}^{+\infty} |x| f(x) dx,$$
where $(\Omega, \mu)$ is the probability space $X$ defined on.
