TITLE: Convergence of$\int_{0}^{+\infty}\frac{1}{\left(\log(x)\right)^{4\alpha}}\sin^2\left(\frac{1}{x^{\alpha}}\right)\,dx$
QUESTION [5 upvotes]: 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

REPLY [3 votes]: 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.