Suppose one has a continuous random variable $X \in (0, 1)$ and one wants to compute $E[\log(X)\log(X)]$. Suppose further that one also knows that $E[X] < \infty$ and $-\infty < E[\log(X)] < \infty$, where all expectations are taken with respect to a measure $\Pi$. Is it possible to claim that $E[\log(X)\log(X)] < \infty$?
I have tried a few things, namely trying to get an integrable upper bound on $\psi(x) = \log(x)\log(x)$, but could not find much in that direction. Also found this really nice paper that provides a bound on the expectation of an increasing convex function $\phi$ [see Theorem 2.1 therein], which unfortunately does not seem useful because $\psi$ is convex but decreasing.
Edits
I just realised that for $0 < x \leq 1$, the usual bounds on $\log(x)$ can be manipulated to get $$ (x-1)\log(x) \leq \log(x)\log(x) \leq \log(x) - \log(x)/x $$
So if one could show that $-\infty < E[\log(X)/X] < \infty$, I guess we'd be done.
In fact the claim is FALSE in general, as demonstrated quite neatly by @pre-kidney in their answer.
Using the bound above and the substitution $Y = -\log(X)$ one has
$$ Y^2 \leq Ye^Y - Y, $$
but it is possible to have $E[Ye^Y] = \infty$.
No, it is not possible to correctly make such a claim. Indeed, here is an explicit counterexample.
Let $X$ be the random variable in $(0,1)$ satisfying
$$ \mathbb P(X<t)=\frac{1}{(1-\log t)^2},\qquad 0< t\leq 1.\tag{1} $$ I claim that $\mathbb EX<\infty$ and $-\infty<\mathbb E[\log X]<\infty$, and yet $\mathbb E[(\log X)^2]=\infty$. To show this, I will use the auxiliary random variable $Y=-\log X$. Substituting $t=e^{-s}$ in $(1)$ yields $$ \mathbb P(Y>s)=\frac{1}{(s+1)^2},\qquad 0\leq s<\infty. $$ Using the tail sum formula for expectation, $$ \mathbb EY=\int_0^{\infty}\mathbb P(Y>s)\ ds=1, $$ whereas $$ \mathbb E[Y^2]=\int_0^{\infty}2s\cdot \mathbb P(Y>s)\ ds=\infty. $$
Finally, since $X\leq 1$ holds with probability $1$, it follows that $\mathbb EX\leq 1$ and in particular $\mathbb EX<\infty$. Thus we have found an $X$ satisfying all of the hypotheses, but with $\mathbb E[(\log X)^2]=\mathbb E[Y^2]=\infty$, contradicting the claim.