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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?

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  • $\begingroup$ integral by parts can help? $\endgroup$ – Yimin May 11 '15 at 17:54
  • $\begingroup$ The integral is converging, and the mentioned limit does not even exist. $\endgroup$ – Jack D'Aurizio May 11 '15 at 18:15
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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,

$$\int_{2}^{+\infty}\frac{\cos(2x)\cos(6x)}{x\log x}\,dx = -0.0762839\ldots $$

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.

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  • $\begingroup$ Why does it diverging? Did you calculated the value of the integral in absolute value and concluded that it's diverging? $\endgroup$ – Yinon Eliraz May 11 '15 at 18:21
  • $\begingroup$ @YinonEliraz: the last estimate follows by integration by parts and proves that the integral of the absolute value is diverging since $\log\log N$ is unbounded as $N\to +\infty$. $\endgroup$ – Jack D'Aurizio May 11 '15 at 18:26

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