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
$$ \mathbf x=\begin{bmatrix} x_{1} \\ x_{2} \\ \vdots\\ x_{n} \end{bmatrix}\quad\text{and}\quad \mathbf P=\begin{bmatrix} \sigma_{1} & 0 & 0 & \dots & 0 \\ 0 & \sigma_{2} & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & \sigma_{n} \end{bmatrix}$$
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
$$\frac{n}{2}\log_2(2 \pi e)+ \frac {1}{2}\log_2|\mathbf P|$$