# What is wrong in this proof of existence of the improper integral $\int_{y=1}^{\infty}\frac{\sin y}{y^2}dy$

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.

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:$$\int_1^\infty\left|\frac{\sin y}{y^2}\right|\,\mathrm dy\leqslant\int_1^\infty\frac1{y^2}\,\mathrm dy=1.$$
• 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? – Koro Apr 26 '20 at 13:54