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Study the convergence of the following integral as $\alpha >0$

$$\int_{0}^{+\infty}\frac{1}{\left(\log(x)\right)^{4\alpha}}\sin^2\left(\frac{1}{x^{\alpha}}\right)\,dx$$

We have to study the function in the neighbourhoods of $0$, $1$ and $+\infty$

As $x\to +\infty$ we have $f(x) = \mathcal{O}\left(\frac{1}{x^{2\alpha}\log^{4\alpha}(x)}\right)$, which converges as $\alpha>=\frac{1}{2}$

How to handle this function in the neighbourhoods of $0$ and $1$? In the first case could Abel-Dirichlet work? I am not sure that $\sin^2\left(\frac{1}{x}\right)$ has a bounded primitive

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Observe that, as $x \to 1$, $$ \frac{1}{\left(\log(x)\right)^{4\alpha}} \sim \frac{1}{\left(1-x\right)^{4\alpha}} $$ giving a convergence of the integral iff $$ \alpha < \frac14 $$ which is in contradiction with $$ \alpha \ge \frac12. $$ The given integral never converges.

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  • $\begingroup$ As $x \to 1$, $$ \frac{1}{\left(\log(x)\right)^{4\alpha}}\sin^2\left(\frac{1}{x^{\alpha}}\right) \sim \frac{\sin^2(1)}{\left(1-x\right)^{4\alpha}} $$ with $\sin^2(1) \ne 0$. $\endgroup$ Aug 19 '18 at 20:55

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