For what values of $a$ does $\int_{0}^{1}(-\ln x)^adx$ converge? For what values of $a$ does $\int_{0}^{1}(-\ln x)^adx$ converge? I have seen a duplicate of this question but the answer there, though very good and creative, isn't clear about negative values. When $a=0$ it is trivial. I actually did arrive at something for $a<0$: $\int_{0}^{1}(-\ln x)^a dx=[t=-\ln x, x=e^{-1}]=\int_{0}^{\infty}(e^{-t}t^a)dt$. If $a<0$, then $t^a$ is bounded and $\int_{0}^{\omega}{e^{-t}}$ converges. By Abel, the integral converges. Is my proof admissible? Besides, is there a convenient way to treat the case "a>0"?

By replacing $x$ with $e^{-t}$ we have:
$$ \int_{0}^{1}(-\log x)^{\alpha}\,dx = \int_{0}^{+\infty}t^{\alpha}e^{-t}\,dt = \Gamma(\alpha+1) $$
so the integral is converging for any $\alpha>-1$.
