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I am looking to study the convergence of this integral $$ \int_0^1 \frac{x}{\ln x}$$
One method to prove the convergence is to use the special case of Cauchy definition "what is called in French l'integral de Bertrand" as
The integral $\int_0^{1/e} \frac{1}{x^a (\ln x)^b}$ converges iff $(a <1)$ or ($a=1$ and $b >1$)
so in my question $a=-1$ and the integral converges but I am looking for another method to do so. Is there another simple way to prove the convergence.
We have that
$$\int_0^1 \frac{x}{\ln x}dx=\int_1^\frac12 \frac{x}{\ln x}dx+\int_\frac12^1 \frac{x}{\ln x}dx$$
and since $\lim_{x\to 0^+} \frac{x}{\ln x}=0$ all boils down in $\int_\frac12^1 \frac{x}{\ln x}dx$ and by $x=1-y$ we obtain
$$\int_\frac12^1 \frac{x}{\ln x}dx=\int_0^\frac12 \frac{1-y}{\ln (1-y)}dy$$
which diverges by limit comparison test with $\int_0^\frac12 \frac1ydy$ indeed
$$\lim_{x\to 0^+} \frac{\frac{1-y}{\ln (1-y)}}{\frac1y}=\lim_{x\to 0^+} (1-y)\frac{y}{\ln (1-y)}=1$$
In a similar way we can prove that for $b>1$
$$\int_0^1 \frac {x} { (\ln x)^b }dx$$
the integral converges.
Are you sure that the integral converges ? The integrand is equivalent to $\frac1{x - 1}$ around $1$.
