$\newcommand\R{\mathbb R}$Let $\int h:=\int_{\mathbb R^d}h(x)\,dx$. 

>**Claim:** For $\int f\,\ln f$ to be finite, it is enough that 
$$f(x)\le\frac C{(1+|x|)^d \ln^a(2+|x|)}\tag{1}$$
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 $a>2$. Switching to polar coordinates, it is then easy to see that 
$$\int f\,\ln f<\infty.\tag{2}$$

On the other hand, consider the probability density function (pdf) $g$ defined by 
$$g(x):=\frac{c_d}{(1+|x|)^{d+1}}$$
for some real $c_d>0$ and all $x\in\R^d$. Then somewhat similarly to (2) we get 
$$\int f\,\ln g\in\R.\tag{3}$$
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  
$$f(x)=\frac{K_d}{(1+|x|)^d \ln^2(2+|x|)}\tag{5}$$
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 $a=2$.