How to calculate this expectation with logarithm? If $\mathbf{x}\sim\mathbf{N}(\mathbf{0},\mathbf{I})$, and assume that $A$ is a symmetric positive definite matrix, how can I calculate the following two expectations where there is a logarithm in it?
$$
\mathrm{E}_\mathbf{x}\left[\log(\mathbf{x}^\top A\mathbf{x})\right]
$$
and
$$
\mathrm{E}_\mathbf{x}\left[(\mathbf{x}^\top A\mathbf{x})\log(\mathbf{x}^\top A\mathbf{x})\right]
$$
Using the decomposition $A=U\Lambda U^\top$ and $\mathbf{y}=U^\top\mathbf{x}$, the first expectation can be reduced to
$$
\mathrm{E}_\mathbf{y}\left[\log(\mathbf{y}^\top \Lambda\mathbf{y})\right]
$$
$$
=\mathrm{E}_\mathbf{y}\left[\log\left(\sum_i{\lambda_i\mathbf{y}_i^2}\right)\right]
$$
where $y$ has the same distribution as $x$ because $U$ is a orthogonal matrix.
However, since the random variables $y_i$ are in the logarithm function, I cannot decompose the expectation further and such a expectation seems to be difficult to calculate.
Could you please help me with this problem or give some suggestions about it?
Thank you very much!
 A: For the case $\lambda_1=\cdots=\lambda_k=1$, $\sum_i \lambda_i y_i^2$ has a chi-squared distribution with $k$ degrees of freedom.  Plugging in the chi-squared density and cranking up Maple, we get
$$E\Bigl(\log\bigl(\sum_{i=1}^k y_i^2\bigl)\Bigl) = \ln 2+\Psi(k/2),$$
where $\Psi(x)$ is the digamma function (the derivative of the gamma function $\Gamma(x)$).
The first six values are $-\gamma-\ln 2$, $-\gamma+\ln 2$, $-\gamma+2-\ln 2$, $-\gamma+1+\ln 2$, $-\gamma+\frac83-\ln 2$, $-\gamma+\frac32+\ln 2$.
If the $\lambda_i$ are not all equal, $\sum_i \lambda_i y_i^2$ is called a weighted chi-squared distribution (and probably other names too).  I don't know if there is an expression for its density that is useful here.
A: Well, with Maple's help I can do the cases $n=1$ and $n=2$: 
$E[\log(\lambda_1 Y_1^2)] = \log(\lambda_1/2) -\gamma$
$E[\log(\lambda_1 Y_1^2 + \lambda_2 Y_2^2)] = \log((\sqrt{\lambda_1} + \sqrt{\lambda_2})^2/2) - \gamma$ 
For $n=3$ I have the beginning of a series expansion: writing $\lambda_1 = \lambda_3(1 + \epsilon_1)$ and $\lambda_2 = \lambda_3 (1 + \epsilon_2)$,
$$
\eqalign{&E[\log(\lambda_1 Y_1^2 + \lambda_2 Y_2^2 + \lambda_3 Y_3^2)] = \cr
&\ln  \left({\lambda_3}/2 \right) +2-\gamma+ \frac{\epsilon_1 + \epsilon_2}{3} 
 - \frac{3 \epsilon_1^2 + 2 \epsilon_1 \epsilon_2 + 3 \epsilon_2^2}{30} 
 + \frac{5 \epsilon_1^3 + 3 \epsilon_1^2 \epsilon_2 + 3 \epsilon_1 \epsilon_2^2 + 5 \epsilon_2^3}{105} \cr
 &- \frac{\epsilon_1^4}{36} - \frac{\epsilon_1^3 \epsilon_2}{63} - \frac{\epsilon_1^2 \epsilon_2^2}{70} - \frac{\epsilon_1 \epsilon_2^3}{63} - \frac{\epsilon_2^4}{36} + \ldots\cr}
$$
Hmm: it looks like for $\epsilon_2 = 0$ this might be
$$E[\log(\lambda_1 Y_1^2 + \lambda_3 Y_2^2 + \lambda_3 Y_3^2)] = \ln(\lambda_3/2) - \gamma
+ \ln(1 + \epsilon_1) + 2 \arctan(\sqrt{\epsilon_1})/\sqrt{\epsilon_1} $$
