Convergence of improper integral $\int_{0}^{1}\frac{\log(x)}{1-x^2}dx$

Question

Find whether the integral converges or diverges. $$\int_{0}^{1}\frac{\log(x)}{1-x^2}dx$$

I simplified it to $$\int_{0}^{1}\frac{\log(x)}{(1-x)(1+x)}dx$$

Here I have $2$ "bad" bounds (both $0$ and $1$).

I think, to prove such integral converges, I should find something that is equivalent to or less than $\log(x)$ so I can exchange them, maybe $\sqrt{x}$, so it may cancel out the fraction, and I can compare it to $$\int_{0}^{1}\frac{dx}{x^p}$$

Also I had an idea to substitute $x$ with $u+1$ so I can use $\log(1+u) \sim u$ when $u \rightarrow 0$. I'm allowed(not familiar with anything else) to use direct comparision test, limit comparision test, Cauchy test and Abel-Dirichlet's test.

Any hints are much appreciated, I just want to get the big picture.

Thank you.

Answer 1

As $x\to 1$, $${\log(x)\over 1 - x^2} = {\log(x)\over (1 + x)(1-x)}\sim -{\log'(x) \over 2 } = -{1\over 2}$$ As $x\to 0$, $${\log(x)\over 1 - x^2}\sim \log(x),$$ which integrates at $0$. The improper integral exists and is finite.

Answer 2

$$\begin{align} \int _0^1 \frac{\log x}{\color{#08F}{1 - x^2}} dx &= \int _0^1 \log x \Big( \color{#08F}{\sum_{n=0}^{\infty} x^{2n}}\Big) dx = \sum_{n=0}^\infty \int _0^1 x^{2n}\log x dx \\&= \sum_{n=0}^\infty\Bigg[\frac{x^{2n +1}}{2n +1} \log x \Big|_0^1 - \frac{1}{2n+1}\int _0^1 \frac{x^{2n+1}}{x} dx\Bigg]\\&=-\sum_{n=0}^\infty \frac{1}{(2n+1)^2}\\&= \color{red}{ -\frac{\pi^2}{8}}\tag{*}\end{align}$$

Where you may find why $(*)$ holds, here.