<s>[INST] Investigate the convergence of the integral Investigate the convergence of the integral 
$$\int_{0}^{1} \left\vert \sin\frac{1}{x} \right\vert\ \frac{1}{x^{\alpha}} dx \ \text{where $\alpha \geq 1$  }$$
I assume that it diverges, so I rated function from below
$|sin\frac{1}{x}| \ \frac{1}{x^{\alpha }} \geq \sin^2(\frac{1}{x}) \ \frac{1}{x^\alpha} = \frac {(1-cos(\frac{2}{x}))}{2x^\alpha } = \frac{1}{2x^\alpha} - \frac {cos(\frac{2}{x})}{x^\alpha}$
The $\int_{0}^{1} \frac{1}{2x^{\alpha}} dx$ diverges
How to show that the integral $\int_{0}^{1}\frac {cos(\frac{2}{x})}{x^\alpha}$ converges? 
 [/INST] Using integration by substitution, we have to study the convergence of the integral
$$I_\alpha=\int_1^\infty x^{\alpha-2}  \left\vert \sin x \right\vert \ dx.$$
For $k \in \mathbb N$, and $x \in [k \pi + \frac{\pi}{6}, k\pi + \frac{5\pi}{6}]$, you have 
$$\left\vert \sin x \right\vert \ge \frac{1}{2}$$ hence
$$g(x) = x^{\alpha-2}\left\vert \sin x \right\vert   \ge \frac{x^{\alpha-2}}{2}.$$
Based on that, you get
$$I_\alpha \ge \frac{\pi}{3} \sum_{k=1}^\infty \frac{1}{\left(k\pi + \frac{5\pi}{6}\right)^{2-\alpha}}.$$
As the right end series is diverging for $2- \alpha \le 1$, i.e. $\alpha \ge 1$, we finally get that the integral is also diverging for $\alpha \ge 1$.
</s>