I am looking for weak conditions when a probability density function $f$ on $\mathbb{R}^d$ has a finite integral


$$
\int\_{\mathbb{R}^d} \log(f(x)) f(x) dx.
$$


Any references would be appreciated.



Integrability of $\int \log(f(x)) f(x) dx$ for a probability density function $f$
$\newcommand\R{\mathbb R}$Let $\int h:=\int\_{\mathbb R^d}h(x)\,dx$.



> 
> **Claim:** For $\int f\,\ln f$ to be finite, it is enough that
> $$f(x)\le\frac C{(e+|x|)^d \ln^a(e+|x|)}\tag{1}$$
> for some real $a>2$, some real $C>0$, and all $x\in\R^d$.
> 
> 
> The condition $a>2$ here cannot be replaced by $a=2$.
> 
> 
> 


Indeed, suppose that (1) holds with $a>2$. Since $\ln f\le\ln C$ and $\int f=1$, we have
$$\int f\,\ln f<\infty.\tag{2}$$


On the other hand, consider the probability density function (pdf) $g$ defined by
$$g(x):=\frac{c\_d}{(e+|x|)^{d+1}}$$
for some real $c\_d>0$ and all $x\in\R^d$. Switching to polar coordinates, we find that
$$\int f\,\ln g\in\R.\tag{3}$$
Also,
$$\int f\,\ln f-\int f\,\ln g=-\int f\,\ln\frac gf\ge-\int f\,\Big(\frac gf-1\Big)
=\int f-\int g\,1(f\ne0)\ge0.$$
Therefore, in view of (3),
$$\int f\,\ln f>-\infty.$$
This together with (2) yields
$$\int f\,\ln f\in\R,\tag{4}$$
if (1) holds with $a>2$.


On the other hand, if the pdf $f$ is given by  

$$f(x)=\frac{K\_d}{(e+|x|)^d \ln^2(e+|x|)}\tag{5}$$
for a real $K\_d>0$ and all $x\in\R^d$, then it is easy to see that
$\int f\,\ln f=-\infty$. So, the condition $a>2$ for (1) cannot be replaced by $a=2$.


The claim is now completely proved.


