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

$$\frac{n}{2}\log_2(2 \pi e)+ \frac {1}{2}\log_2|\mathbf P|$$

  • 2
    $\begingroup$ This is more appropriate for math.se. And cross-posting is not appreciated on SE, you have this same question there as well. $\endgroup$
    – Gilles
    Commented Jan 5, 2017 at 10:06
  • $\begingroup$ I still don't get the answer on math.se, and this is one of the time when specific problems related both to mathematics and a particular field is not answered in mathematics. You can still check for no answer for this in mathematics website. Though I agree that it is not right upto an extent $\endgroup$
    – Userhanu
    Commented Jan 5, 2017 at 10:30

1 Answer 1


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.


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