Is differential entropy always less than infinity? For an arbitrary continuous random variable, say $X$, is its differential entropy always less than $\infty$?  (It's ok if it's $-\infty$.)  If not, what's the necessary and sufficient condition for it to be less than $\infty$?
 A: I thought about this question some more and managed to find a counter-example, thanks also to the Piotr's comments above.  The answer to the first question is no - the differential entropy of a continuous random variable (RV) is not always less than $\infty$.  For example, consider a continuous RV X whose pdf is
$$f(x) = \frac{\log(2)}{x \log(x)^2}$$
for $x > 2$.
It's not hard to verify that its differential entropy is infinite.  It grows quite slowly though (approx. logarithmically).
For the 2nd question, I am not aware of a simple necessary and sufficient condition.  However, one partial answer is as follows.  Categorize a continuous RV into one of the following 3 Types based on its support, i.e. 
Type 1: a continuous RV whose support is bounded, i.e. contained in [a,b].
Type 2: a continuous RV whose support is half-bounded, i.e. contained in [a,$\infty$) or ($-\infty$,a]
Type 3: a continuous RV whose support is unbounded.
Then we have the following -
For a Type 1 RV, its entropy is always less than $\infty$, unconditionally.
For a Type 2 RV, its entropy is less than $\infty$, if its mean ($\mu$) is finite.
For a Type 3 RV, its entropy is less than $\infty$, if its variance ($\sigma^2$) is finite.
The differential entropy of a Type 1 RV is less than that of the corresponding uniform distribution, i.e. $log(b-a)$, a Type 2 RV, that of the exponential distribution, i.e. $1+log(|\mu-a|)$, and a Type 3 RV, that of the Gaussian distribution, i.e. $\frac{1}{2} log(2{\pi}e\sigma^2)$.
Note that for a Type 2 or 3 RV, the above condition is only a sufficient condition.  For example, consider a Type 2 RV with $$f(x) = \frac{3}{x^2}$$
for $x > 3$.  Clearly, its mean is infinite, but its entropy is 3.1 nats.  Or consider a Type 3 RV with $$f(x) = \frac{9}{|x|^3}$$
for $|x| > 3$.  Its variance is infinite, but its entropy is 2.6 nats.  So it would be great, if someone can provide a complete or more elegant answer for this part.
