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

Question

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

Answer 1

No. Consider

$$ P\left(X=\frac6{\pi^2}\cdot\frac{2^n}{n^2}\right)=2^{-n} $$

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}

as the first and third term yield convergent series whereas the second one yields a multiple of the divergent harmonic series.