analysis - Improper integral convergence: $int_{0}^{1}{frac{mid{log(x)}mid^a}{sqrt{1-x^2}}}dx$

As its told on the title I want to check the convergence/divergence of the improper integral when $a\in \mathbb{R}$: $$\begin{equation} \int_{0}^{1}{\frac{\mid{\log(x)}\mid^a}{\sqrt{1-x^2}}}dx \end{equation}$$
So, it's improper at $x=0$ and $x=1$, so I split the integral in :
$$\begin{equation} \int_{0}^{\frac{1}{2}}{\frac{\mid{\log(x)}\mid^a}{\sqrt{1-x^2}}}dx + \int_{\frac{1}{2}}^{1}{\frac{\mid{\log(x)}\mid^a}{\sqrt{1-x^2}}}dx \end{equation}$$

I see, that on the first one the integrals it's like $\int\mid{\log(x)}\mid^a dx$ by the comparison limit test, but I don't know how to prove that $\int\mid{\log(x)}\mid^a dx$ converges.

Hopefully you can help me. Much thanks!

Answer

Note that

$$\int_{0}^{\frac{1}{2}}{\frac{\mid{\log(x)}\mid^a}{\sqrt{1-x^2}}}dx$$

converges $\forall a$ by comparison test with $\frac1{\sqrt x}$.

For the second part

$$\int_{\frac{1}{2}}^{1}{\frac{\mid{\log(x)}\mid^a}{\sqrt{1-x^2}}}dx$$

let $1-x^2=y^2$

$$\int_{\frac{\sqrt 3}{2}}^{0}{\frac{\mid{\log(\sqrt{1-y^2}}\mid^a}{y}}\cdot \left(\frac{-y}{\sqrt{1-y^2}}\right)dy = \int_{0}^{\frac{\sqrt 3}{2}}{\frac{\mid{\log(\sqrt{1-y^2}}\mid^a}{\sqrt{1-y^2}}}dy$$

and note that for $y\to 0$

$$\mid\log\left(\sqrt{1-y^2}\right)\mid\sim \frac{y^2}{2}$$

thus the integral converges for $-2a<1$ that is $a>-\frac12$ and diverges for $-2a\ge1$ that is $a\le-\frac12$ by comparison with $y^{2a}$.