# Convergence of an improper integral involving logarithm

I'm in the first year of a Physics degree. I was solving past paper exams so I came across a problem asking about the values of $\alpha \in \mathbb{R}$ for which the following integral is convergent: $$\int_0^{+\infty} \frac{\ln |1-x^2|}{x^{\alpha}}dx.$$ I thought about splitting the integration interval so I have: $$\int_0^1 \frac{\ln |1-x^2|}{x^{\alpha}}dx + \int_1^{+\infty} \frac{\ln |1-x^2|}{x^{\alpha}}dx.$$ For the second integral I thought this way: $$\frac{\ln |1-x^2|}{x^{\alpha}} \sim_{+\infty}\frac{\ln |-x^2|}{x^{\alpha}}=\frac{\ln(x^2)}{x^{\alpha}}=\frac{2\ln x}{x^{\alpha}}=\frac{2}{x^{\alpha}(\ln x)^{-1}}.$$ So it converges iff $\alpha \le 1$, but I'm not sure if it's the right procedure. For the first integral I really have no idea. Can someone please help me?

[Btw I'm not an English native speaker so forgive me my grammar mistakes.]

First notice that $$\frac{\ln{\lvert 1-x^2\rvert}}{x^\alpha} \underset{x\to 1}{\sim} \ln(1+x)+\ln{\lvert 1-x \rvert} \underset{x\to 1}{\sim} \ln{\lvert 1-x\rvert},$$ so our function is integrable in a neighbourhood of $1$ (i.e. for example $\int_{1/2}^{2} \frac{\ln{\lvert 1-x^2 \rvert}}{x^\alpha} \,\mathrm{d}x$ converges).
$$\frac{\ln{\lvert 1-x^2\rvert}}{x^\alpha} \underset{x\to 0}{\sim}\frac{-x^2}{x^\alpha}=-x^{2-\alpha},$$
so the first integral converges iff $2-\alpha>-1$.
The second integral converges iff $\alpha >1$ (you made a mistake here).
• @Michele Indeed, I modified my answer. The integral $\int \ln\lvert 1-x \rvert \,\mathrm{d}x$ is convergent in a neighbourhood of $1$ (you can compute its antiderivative explicitely) so $x=1$ is, in fact, not a problematic point. – tristan Jan 6 '18 at 17:13