Problem. Let $X$ be a random variable (either continuous or discrete) that takes nonnegative values. Prove or provide a counterexample to the following statement: There does not exist an X such that $E(X)$ is finite and $E(X\ln(X + 1)) = +\infty$.

I'm shooting for a counterexample first:

For continuous, let $f(x)$ be the pdf of $X$. If we let $g(X) = X\ln(X + 1)$ then we have that $$E(X) = \int_0^{\infty}xf(x)dx < \infty ~~~~~~~~\text{and}~~~~~~~~ E(g(X)) = \int_0^{\infty} x\ln(x+1)f(x)dx = +\infty. \tag{1}$$

For discrete, let $p(x)$ be the pmf, and we have the similar cases

$$E(X) = \sum_{x=0}^{\infty}xp(x) < \infty ~~~~~~~~\text{and}~~~~~~~~ E(g(X)) = \sum_{x=0}^{\infty}x\ln(x+1)p(x) = +\infty. \tag{2}$$

Then we can either find some $f(x)$ or $p(x)$ which satisfies $(1)$ or $(2)$. I've tried to isolate either $x$ or $x\ln(x+1)$ by say letting $f(x) = \frac{1}{x}, \frac{1}{x\ln(x+1)}$, and so on. Does anyone have any ideas, or is it actually true that no such $X$ will satisfy both expectations?

EDIT: Using comments advice, we have the divergence of the r.h.s of (1) when

$$x\ln(x+1)f(x) = \Theta\left(\frac{1}{x}\right).$$

Now we look at the asymptotic proportionality of $f(x)$ such that

$$f(x) = \Theta\left(\frac{1/x}{x\ln(x+1)}\right) = \Theta\left(\frac{1}{x^2\ln(x+1)}\right).$$

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    $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – Stephan Kolassa Sep 28 at 4:57
  • $\begingroup$ Hint: look at integration by parts. $\endgroup$ – Stephan Kolassa Sep 28 at 4:59
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    $\begingroup$ Since $\log(x+1)$ diverges at $+\infty$ it should not be very difficult to find a $t$ counterexample. $\endgroup$ – Xi'an Sep 28 at 5:23
  • $\begingroup$ @StephanKolassa From an interview a few days ago. Not sure if that falls under self-study? Will add if so, let me know. $\endgroup$ – jimmyAtUCLA Sep 28 at 8:52
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    $\begingroup$ Yes: in concluding the integral is infinite, you mistakenly extended the integration down to an improper lower limit, which is irrelevant to the question. This integral does not diverge as the upper limit grows: see wolframalpha.com/input/… for instance. $\endgroup$ – whuber Sep 29 at 13:23

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