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So I want to prove whether the improper integral in title exists or not. Now existence of $\int_{y=1}^{\infty}\frac{\sin y}{y^2}dy $ is as good as existence of the limit $\lim_{N\to \infty}\int_{y=1}^{N}\frac{\sin y}{y^2}dy$. We have,
$|\int_{y=1}^{N}\frac{\sin y}{y^2}dy|\le1-1/N \; \;\forall N\gt 1$
Therefore,$ \int_{y=1}^{N}\frac{\sin y}{y^2}dy$ is bounded and hence the integral in title converges.

But as per the solution, this integral diverges. Can you please help? Thanks a lot in advance.

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Your solution would be correct if you were integrating a non-negative function. But you have only proved that $n\mapsto\int_1^n\frac{\sin y}{y^2}\,\mathrm dy$ is bounded, not that it converges. However, the integral converges. I would prove that it converges absolutely:$$\int_1^\infty\left|\frac{\sin y}{y^2}\right|\,\mathrm dy\leqslant\int_1^\infty\frac1{y^2}\,\mathrm dy=1.$$

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  • $\begingroup$ Is it correct to say that if an integral say that if $\int_{x=1}^{\infty}|f(t)| dt $ converges then $\int_{x=1}^{\infty} f(t) dt$ also converges? $\endgroup$ – Koro Apr 26 '20 at 13:54
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    $\begingroup$ Yes. In other words, every absolutely convergent integral converges. $\endgroup$ – José Carlos Santos Apr 26 '20 at 13:55

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