Suppose $X\sim F(n_1,n_2)$, then, what is the distribution of $Y=\log \frac {\exp(X)}{X}$?
I don't even have a clue about this distribution.
Any helpful suggestions and answers are greatly appreciated!
Following Christoph Hanck's suggestion, I got the answer.
Note that $Y=X-\log X \in [1,\inf)$.
The CDF of $Y$ is $$ F(y)=\mathbb{P}(X-\log X \leq y) $$
Since $f(x)=x-\log x$ monotically increases from $(1,\infty)$ and monotically decreases from $(0,1)$.
Then, we derive the range indicated by $X-\log X \leq y$.
For $$ x-\log x=y $$ we can equivalently obtain $$ -xe^{-x}=-e^{-y} $$
Since $-\frac 1 e\leq-e^{-y}<0$, from the wiki page w.r.t. the Lambert $W$ function, we deduce that there are two roots (as expected).
Finally, we have two roots $$ x_1=W_{-1}(-e^{-y})\\x_2=W_{0}(-e^{-y}) $$
Then, \begin{align} F(y)&=\mathbb{P}(W_{-1}(-e^{-y}) \geq X \leq W_{0}(-e^{-y}) )\\ &=\int_{W_{-1}(-e^{-y})}^{W_{0}(-e^{-y})} f_X(x)dx \end{align} and \begin{align} f(y)&=\frac{\partial F(y)}{\partial y}\\ &=f_X\left[ W_{0}(-e^{-y})\right] \frac{\partial W_{0}(-e^{-y})}{\partial y}+f_X\left[ W_{-1}(-e^{-y})\right] \frac{\partial W_{-1}(-e^{-y})}{\partial y} \end{align}
Thanks for the suggestion from @Xi'an.
A more tractable solution can be found from wiki page
