# Is $E(|X\log(X)|)$ finite whenever $E(X) = 1$?

Asking for $$X \ge 0$$ under any probability measure $$P$$, does $$\int X \, dP = 1 \implies \int |X\log(X)|\,dP < \infty?$$

• The reason the space between $X$ and $\log$ was too small, so that you saw $X\text{log}(X)$ instead of $X\log(X),$ is that you typed X\text{log}(X) instead of X\log(X). I corrected that. – Michael Hardy Mar 8 at 19:25

$$P\left(X=\frac6{\pi^2}\cdot\frac{2^n}{n^2}\right)=2^{-n}$$
for all positive integers $$n$$. Then $$E(X)=1$$, but
$$\begin{eqnarray} E(|X\log X|) &=& \sum_{n=1}^\infty2^{-n}\cdot\frac6{\pi^2}\cdot\frac{2^n}{n^2}\log\left(\frac6{\pi^2}\cdot\frac{2^n}{n^2}\right) \\ &=& \sum_{n=1}^\infty\frac6{\pi^2}\cdot\frac1{n^2}\left(\log\frac6{\pi^2}+n\log2-2\log n\right) \\ &=& \infty\;, \end{eqnarray}$$