Convergence of an improper integral with a parameter

Question

I am to test the convergence of the improper integral $$ \int_{0+}^{1-} \frac{\ln(x)}{(1-x)^a} dx$$ with the parameter $a \in R$. I have some trouble doing this so I'd appreciate a full explanation so that I can solve similar examples myself. Probably I have to split the integral in two, consider a few cases and use the right convergence criterion but what is the correct way to pick these?

Answer 1

For $x>0$ we have that $\log x\le x-1$. With the substitution $1-x=u$, the integral becomes $$ \int_0^1 \frac{\log x}{(1-x)^a}\mathrm d u=\int_0^1 \frac{\log(1-u)}{u^a}\mathrm d u\le \int_0^1 \frac{-u}{u^a}\mathrm d u=-\int_0^1 \frac{1}{u^{a-1}}\mathrm d u $$ Reminding that $\int_0^1 \frac{1}{x^p}\mathrm d x$ converges if $p<1$, we have that integral converges if $a-1<1$, that is $a<2$.