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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.]

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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).

We have

$$ \frac{\ln{\lvert 1-x^2\rvert}}{x^\alpha} \underset{x\to 0}{\sim}\frac{-x^2}{x^\alpha}=-x^{2-\alpha}, $$

so the first integral converges iff $2-\alpha>-1$.

The second integral converges iff $\alpha >1$ (you made a mistake here).

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  • $\begingroup$ Thanks. But isn't x=1 a problematic point? In its neighbourhood the logarithm goes to infinity. $\endgroup$ – user519137 Jan 6 '18 at 16:53
  • $\begingroup$ @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. $\endgroup$ – tristan Jan 6 '18 at 17:13

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