$\DeclareMathOperator\E{\mathbb E} \DeclareMathOperator\Var{\mathrm{Var}} \newcommand\R{\mathbb R} \DeclareMathOperator\N{\mathcal N} \DeclareMathOperator\tr{\mathrm{tr}}$Suppose $X \sim \N(\mu, \Sigma)$, where $\mathrm{supp}(X) = \R^n$. We can assume $\Sigma$ to be nonsingular. What is $\E \lVert X \rVert$?
If $\Sigma = \sigma^2 I$, then $\lVert X \rVert / \sigma$ follows a noncentral chi distribution, which has known mean (in terms of a generalized Laguerre polynomial). That gives in this case $$ \E \lVert X \rVert = \sigma \sqrt{\frac{\pi}{2}} \; L_{{1/2}}^{{(n/2-1)}}\left({\frac {-\lVert \mu \rVert^{2}}{2 \sigma}}\right)\, ;$$ if we further assume $\mu = 0$ then it becomes a chi distribution and $$ \E \lVert X \rVert = \sigma \sqrt {2} \, \frac{\Gamma((n+1)/2)}{\Gamma(n/2)} .$$
In the general case, we have an easy upper bound via Jensen's inequality. Letting $C^T C = \Sigma$ and $Z \sim \N(0, I)$: \begin{align} \left( \E \lVert X \rVert \right)^2 &< \E \lVert X \rVert^2 \\&= \E \lVert \mu + C Z \rVert^2 \\&= \lVert \mu \rVert^2 + 2 \mu^T C \E Z + \E \tr( Z^T C^T C Z ) \\&= \lVert \mu \rVert^2 + \tr( C^T C \,\E Z Z^T ) \\&= \lVert \mu \rVert^2 + \tr( \Sigma ) ,\end{align} where the inequality is strict since we've assumed $\lVert X \rVert$ is not degenerate.
We also have a lower bound in the same way: since $\lVert \cdot \rVert$ is convex, $\E \lVert X \rVert \ge \lVert \E X \rVert = \lVert \mu \rVert$. (I previously had something more complicated here, based on bounding the variance of the function $z \mapsto \lVert \mu + C z \rVert$, but it gave a worse bound.)
Can we tighten these bounds or, preferably, find an exact expression for $\E \lVert X \rVert$ in the general case? If not, what about the diagonal but non-spherical case, or $\mu = 0$, or other interesting subcases?
There's been some discussion on this site of the distribution of $Y = \lVert X \rVert^2$, particularly here and here. If we have the full distribution of $Y = \lVert X \rVert^2$, we may be able to find $\E \sqrt{Y}$. That distribution is gross, though, and analytical answers seem difficult.
