$$\int_{0}^{\infty}\frac{|\sin x|}{x^2}$$
I have thought a lot but the absolute value just messes every way I try.
Proving that this integral is divergent
2
$\begingroup$
trigonometry
definite-integrals
improper-integrals
integral-equations
divergence
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1Hint: What happens for small $x$? – 2017-01-13
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3The trouble is near $0$, where you could use [this](http://math.stackexchange.com/questions/674153/proof-that-sinx-x-2). – 2017-01-13
1 Answers
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$$\lim_{x\to 0^+}\left(\frac{|\sin x|}{x^2}:\frac{1}{x}\right)=1\ne 0\;\wedge\;\int_0^1\frac{dx}{x}\text{ divergent }$$ $$\Rightarrow \int_0^1\frac{|\sin x|}{x^2}\text{ divergent }\Rightarrow \int_0^{+\infty}\frac{|\sin x|}{x^2}\text{ divergent.}$$