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

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

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.
• 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 $\sin^2(1) \ne 0$. Aug 19 '18 at 20:55