How is the entropy of the multivariate normal distribution with mean 0 calculated? Here is what I have so far:
$$\begin{align}
h(x) &= - \int \frac{1}{(2\pi)^{\frac{D}{2}}\det\Sigma^{\frac{1}{2}}} \exp(-\frac{1}{2} x^T\Sigma^{-1}x) \ln \frac{1}{(2\pi)^{\frac{D}{2}}\det\Sigma^{\frac{1}{2}}} \exp(-\frac{1}{2} x^T\Sigma^{-1}x) dx \\
&= \frac{1}{2} \ln ((2\pi)^D\det\Sigma) + \frac{1}{2}\int \frac{x^T \Sigma^{-1} x}{(2\pi)^{\frac{D}{2}}\det \Sigma} \exp(-\frac{1}{2} x^T\Sigma^{-1}x)
\end{align}$$
Having solved the 1D case, I assume the trick is to show that the remaining integral is similar to $\Sigma$, but I'm not good with multidimensional calculus.
 A: Since $\Sigma$ is the variance of your function you know that
$$ \Sigma = E[(x-\mu)(x-\mu)^\top]=E[xx^\top]=\int xx^\top N(x;0,\Sigma)\mathrm{d}x $$
where $N(x;0,\Sigma)=\frac{1}{\sqrt{\mathrm{det}(2\pi\Sigma)}}\exp(-\frac{1}{2}x^\top\Sigma^{-1} x)$.
The integral you are stuck on can be written $E[x^\top\Sigma^{-1} x]$.
Now recall that $\mathrm{trace}(AB)=\mathrm{trace}(BA)$ and therefore since $x^\top \Sigma^{-1} x$ is a number we have 
$$
x^\top \Sigma^{-1} x = \mathrm{trace}(x^\top\Sigma^{-1} x)=\mathrm{trace}(\Sigma^{-1}xx^\top).
$$ 
Plugging into your expression we get
$$ E[x^\top\Sigma^{-1} x] = E[\mathrm{trace}(\Sigma^{-1} xx^\top)] = \mathrm{trace}(\Sigma^{-1}E[xx^\top])
$$
Finally, since we know the value of $E[xx^\top]$ we see that this equals
$$ 
\mathrm{trace}(\Sigma^{-1}E[xx^\top])=\mathrm{trace}(\Sigma^{-1}\Sigma)=\mathrm{trace}(I)=D
$$
Plugging into your result you can simplify it all to 
$$
\frac{1}{2}\log(\mathrm{det}(2\pi e\Sigma)), 
$$
since $\mathrm{det}(\alpha \Sigma) = \alpha^D \mathrm{det}(\Sigma)$ for any number $\alpha$.
This is also the answer when the mean is not zero.
It's easy to get lost if you write out the actual expressions as they look rather intimidating.  
