Convergence of improper integral of $\ln f(x)$

Question

Is there something know about the convergence of

$\int_0^1 \ln f(x)dx $ for $f(x)$ continous on $\left(0,1\right)$ and both limits exists, i.e. $\lim_{x\to 0} f(x)$ and $\lim_{x\to 1} f(x)$ ?

I tried a lot, but cannot answer this question in general. I also cannot think of a counterexample.

Answer 1

If $f(x)=1$ for every $x$ in $(0,1)$ then both limits of $f$ at $0$ and at $1$ exist and the integral converges. If $f(x)=\mathrm e^{-1/x}$ for every $x$ in $(0,1)$ then both limits of $f$ at $0$ and at $1$ exist and the integral diverges.

Answer 2

As you see in other posts, the answer may vary. I know a test which is depend to some conditions:

Let $\lim_{x\to 0^+}x^p\ln(f(x))=A$. If $p<1$ and you found $A<\infty$ so the integral is converges and if $p\ge1$ while $A\neq 0$ (or $A=\infty)$ then the integral is diverges. We have this when we know that the integrand is unbounded only at the lower limit. The same story is valid when the integrand is unbounded only at the upper limit. In this case we take $A=\lim_{x\to1^-}(1-x)^p\ln(f(x))$. I hope this helps.