Find the values of $\alpha$ for which the integral $\int_0^1(-\ln x)^\alpha$ converges

Question

Find the values of $\alpha$ for which the integral $\int_0^1(-\ln x)^\alpha$ converges.

My thoughts:

For $\alpha>0: 0$ is a the problematic point and for $\alpha<0: 1$ is a the problematic point

$(-\ln x)^\alpha$ is a positive function for all $\alpha \in \mathbb{R}$

So I tried the comparison tests, substitute $x=e^t$ and integrating by parts, but couldn't solve it.

Any help appreciated.

Answer 1

By enforcing the substitution $x=e^{-t}$, $$ \int_{0}^{1}\left(-\log t\right)^{\alpha}\,dt = \int_{0}^{+\infty} x^{\alpha}e^{-x}\,dx =\Gamma(\alpha+1)$$ holds for any $\color{red}{\alpha > -1}$, since we have to deal only with the possible singularity of $x^{\alpha}e^{-x}\sim x^{\alpha}$ at the origin. If $\alpha\geq 0$ there is no singularity at all, and if $\alpha\in(-1,0)$ such singularity is integrable. If $\alpha\leq -1$ it is not.