Is there a real function $f$ on $(0..1)$ such that $x·f·\log f = 1$?

Question

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.

Answer 1

Your function, in terms of the Lambert W function, is $$ f(x) = \frac{1}{xW(1/x)} $$ 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 $$ F(x) = \frac{1}{W(1/x)}-\log\big(W(1/x)\big) . $$ 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 $\int_x^1 f(t) \;dy = F(1) - F(x) \to +\infty$.

Answer 2

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

Note that $e^{\frac 1x}\in (e,\infty)$ and that $z\mapsto z^z$ 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 $[1,\infty)$.