Given the integral $$\int_{2}^{\infty} \frac{\cos(2x)\cos(6x)}{x\ln x},$$ is there an easier way to show its divergence than starting to break this integral to a sum of smaller integrals using trigonometric identities?
Can I check the limit $\lim_{x\rightarrow \infty}\frac{\cos(2x)\cos(6x)}{x\ln x}*x\ln x$ and given that $\frac{1}{x\ln x}$ is diverging say that $\int_{2}^{\infty} \frac{\cos(2x)\cos(6x)}{x\ln x}$ is diverging?
Both $\cos(4x)$ and $\cos(8x)$ are functions with a bounded primitive, and the function $\frac{1}{x\log x}$ is continuous, decreasing and converging to zero on $[2,+\infty)$, hence the integral $$ \int_{2}^{+\infty}\frac{\cos(2x)\cos(6x)}{x\log x}\,dx $$ is convergent by Dirichlet's test. Numerically,
On the other hand, the integral of the absolute value is (very slowly) diverging, since: $$\int_{e}^{N}\frac{\left|\cos(2x)\cos(6x)\right|}{x\log x}= \frac{3\sqrt{3}}{4\pi}\log\log N +O\left(\frac{1}{N\log N}\right)$$ by the same reason.