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Show that $$\int_0^\infty \left|\frac{\cos x}{1+x}\right| \ dx$$ diverges

My Try: $$\int_0^\infty \left|\frac{\cos x}{1+x}\right| \ dx \ge \sum_{n=0}^\infty \int_{\pi n-\pi/2}^{\pi n + \pi/2} \left|\frac{\cos x}{1+x}\right| \ dx \ge 2\sum_{n=0} \int_{\pi n}^{\pi n + \pi /2} \frac{\cos x}{x+1} \ dx$$

At this phase I thought about replacing the $\cos(x)$ with a linear function, and create an easy to calculate area (A triangle) but it seems like over-complicating after all. What should I do instead?

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    $\begingroup$ One approach would be to ignore regions where $|\cos x| < \frac{1}{2}$, bound the other regions below by $\frac{1}{2}$, and sum. $\endgroup$ – rogerl Nov 18 '15 at 23:24
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We have $$\int_{n\pi-\frac\pi2}^{n\pi+\frac\pi2}\left|\frac{\cos x}{x+1}\right|\,dx \ge\int_{n\pi-\frac\pi2}^{n\pi+\frac\pi2} \frac{|\cos x|}{n\pi+\frac\pi2+1}\,dx =\frac2{n\pi+\frac\pi2+1}\ ,$$ and the sum of these diverges.

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  • $\begingroup$ Nice! thank you sir! $\endgroup$ – hibye Nov 19 '15 at 1:19
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$\int_0^\infty \frac{|\cos(x)|}{1+x} dx \gt \sum_{n \ge 0} \int_{2n\pi}^{2n\pi+\pi/4} \frac{\cos(x)}{1+x}dx \gt \sum_{n \ge 0} \int_{2n\pi}^{2n\pi+\pi/4} \frac{\sqrt{2}}{2(1+x)}dx \gt \sum_{n \ge 0} \frac{\pi}{4}\frac{\sqrt{2}}{2(1+2\pi n+\frac{\pi}{4})} = +\infty.$

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