Vtyjkyuk

I’m looking for a real-valued function $f$ on $(0..1)$ such that $x · f · log f = 1$. Is there such function? Is it integrable on $(0..1)$?

Why? This could yield an integrable function $f$ such that $f · log f = 1/x$ is not integrable on $(0..1)$. That’s what I’m really interested in. But that’s a separate question maybe.

Your function, in terms of the Lambert W function, is
$$
f(x) = frac1xW(1/x)
$$
The graph looks like this

Although $f$ goes to $+infty$ as $x to 0$ more slowly than $1/x$, it still goes to $+infty$ fast enough that the integral $int_0^1 f(x);dx$ diverges.

We can see this from the antiderivative. Function
$$
F(x) = frac1W(1/x)-logbig(W(1/x)big) .
$$
satisfies $F'(x) = f(x)$.

Now as $x to 0^+$, we have $W(1/x) to +infty$, so $1/W(1/x) to 0$ and $-log(W(1/x)) to -infty$. So $F(x) to -infty$, which means $int_x^1 f(t) ;dy = F(1) - F(x) to +infty$.

Pick $xin (0,1)$, what you want to find is $y$ such that $$y^y=e^frac 1x$$

Note that $e^frac 1xin (e,infty)$ and that $zmapsto z^z$ maps $[1,infty)$ bijectively onto itself.

Since $(e,infty)subset [1,infty)$, the equation (in $y$) $$y^y=e^frac 1x$$ has a unique solution in $[1,infty)$.