How would one prove that $\int_2^{\infty}\left|\frac{\cos\sqrt{x}}{x^{\alpha}\ln x}\right|dx$ diverges with $\frac{1}{2}\leq \alpha \leq 1$?
It converges for $\alpha > 1$ as $\int_2^{\infty}\left|\frac{\cos\sqrt{x}}{x^{\alpha}\ln x}\right|dx \leq \int_2^{\infty}\left|\frac{1}{x^{\alpha}\ln x}\right|dx$.
With the change of variable $x=t^2$ the given integral becomes
$$ I(\alpha)=\int_{\sqrt{2}}^{+\infty}\frac{\left|\cos t\right|}{t^{2\alpha-1}\log t}\,dt $$ and the non-negative function $f(t)=\left|\cos t\right|$ has mean value $\frac{2}{\pi}$, hence by integration by parts the previous integral is convergent iff $$ J(\alpha)=\frac{2}{\pi}\int_{\sqrt{2}}^{+\infty}\frac{dt}{t^{2\alpha-1}\log t} \stackrel{t\mapsto e^u}{=} \frac{2}{\pi}\int_{\frac{1}{2}\log 2}^{+\infty}e^{(2-2\alpha)u}\frac{du}{u} $$ is convergent, i.e. iff $\color{red}{\alpha > 1}$.
