How to solve this entropy integral? I am having the entropy integral below where $\mathbf x$ is a $N$ dimensional Gaussian vector having variance as $\mathbf P$ and mean zero,

$$ -\int\frac {\exp\left(\frac{1}{2}\mathbf x^T\mathbf P^{-1}\mathbf x\right)}{{(2\pi)}^{\frac{1}{2}}|\mathbf P|^{\frac{1}{2}}}\log_2\left(\frac {\exp\left(\frac{1}{2}\mathbf x^T \mathbf P^{-1}\mathbf x\right)}{{(2\pi)}^{\frac{1}{2}}| \mathbf P|^{\frac{1}{2}}}\right) d\mathbf x$$ 

How do I solve this integral ? My approach is,  let 

$$
\mathbf x=\begin{bmatrix}
    x_{1} \\
    x_{2}  \\
    \vdots\\
    x_{n}
\end{bmatrix}\quad\text{and}\quad \mathbf P=\begin{bmatrix}
    \sigma_{1} &amp; 0 &amp; 0 &amp; \dots  &amp; 0 \\
    0 &amp; \sigma_{2} &amp; 0 &amp; \dots  &amp; 0 \\
    \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\
    0 &amp; 0 &amp; 0 &amp; \dots  &amp; \sigma_{n}
\end{bmatrix}$$

I am getting an integral like
$$-\displaystyle\int \frac {\exp\left(\frac{1}{2}\left(\frac{x_1^2}{\sigma_1^2}+\frac{x_2^2}{\sigma_2^2}+\ldots+\frac{x_n^2}{\sigma_n^2}\right)\right)}{{(2\pi)}^{\frac{1}{2}}|\mathbf P|^{\frac{1}{2}}}\log_2\left(\frac {\exp\left(\frac{1}{2}\left(\frac{x_1^2}{\sigma_1^2}+\frac{x_2^2}{\sigma_2^2}+\ldots +\frac{x_n^2}{\sigma_n^2}\right)\right)}{{(2\pi)}^{\frac{1}{2}}|\mathbf P|^{\frac{1}{2}}}\right) d\mathbf x$$ 
but I am confused how to do this $d\mathbf x$ stuff when $\mathbf x$ is a vector. Am I doing it right?

The answer given is


  $$\frac{n}{2}\log_2(2 \pi e)+ \frac {1}{2}\log_2|\mathbf P|$$
The integration over a vector $\mathbf x$ as given by you means something like

$$\int f(\mathbf x) d\mathbf x=\int\dots\int f(x_1,\dots,x_n)dx_1\dots dx_n$$

So, you integrate over each component separately. Now, to solve your integral, you should first split the $\log$ into a difference and evaluate both integrals separately (exploiting the fact that $\exp(ab)=\exp(a)\exp(b)$. Then you will get a lot of similarly structured integrals that depend only on $\sigma_i$. If you solve them, you'll get the solution.