Expectation of a function of two different transformations of a random variable

Question

I have a random variable $X$ where the support of $X$ is non-negative and I want to compute the expectation of $aX\log(bX)$ where $a,b>0$

If I can chose how $X$ is distributed, is there a non-degenerate distribution that will enable me to compute the closed form of the expectation?

Answer 1

If $X$ is uniform on $(0,\theta)$, one can introduce $U$ uniform on $(0,1)$ to compute $$ E(aX\log(bX))=a\theta\log(b\theta)E(U)+a\theta E(U\log U), $$ with $$ E(U)=\int_0^1u\,\mathrm du=\frac12,\qquad E(U\log U)=\int_0^1u\log u\,\mathrm du=-\frac14. $$ Likewise, if $X$ is beta $(c,1)$, then $X$ has density $cx^{c-1}$ on $(0,1)$ hence $$ E(aX\log(bX))=a\log(b)E(X)+aE(X\log X), $$ with $$ E(X)=\int_0^1cx^c\,\mathrm dx=\frac{c}{c+1},\qquad E(X\log X)=\int_0^1cx^c\log x\,\mathrm dx=-\frac{c}{(c+1)^2}. $$

Answer 2

I don't know if this is what you want, but without a specific distribution we can't go further than this:

Denote the law of $X$ by $\mathscr{L}_X$ and, if $X$ has a density, then denote this by $f_X$. Then we know that

$$ \mathbb{E}(aX \log(bX)) = a \int_{\mathbb{R}} x \log(bx)\mathscr{L}_X(dx) = a \int_{\mathbb{R}} x \log(bx) f_X (x) (dx). $$