$$\int_{0}^{\infty}\frac{|\sin x|}{x^2}$$
I have thought a lot but the absolute value just messes every way I try.
2026-04-12 11:34:34.1775993674
Proving that this integral is divergent
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$$\lim_{x\to 0^+}\left(\frac{|\sin x|}{x^2}:\frac{1}{x}\right)=1\ne 0\;\wedge\;\int_0^1\frac{dx}{x}\text{ divergent }$$ $$\Rightarrow \int_0^1\frac{|\sin x|}{x^2}\text{ divergent }\Rightarrow \int_0^{+\infty}\frac{|\sin x|}{x^2}\text{ divergent.}$$