# $\int _{1}^{+\infty }\sin\left( \dfrac{\sin x}{x}\right) dx$ is convergent but not absolutely convergent

I have known $$\frac{\sin{x}}{x}\>\text{is convergent but not absolutely convergent(Conditional convergence)}$$,but I don't know the connection of these two integrals. I was wondering how to prove the promblem: $$\int _{1}^{+\infty }\sin\left( \dfrac{\sin x}{x}\right) dx$$is convergent but not absolutely convergent.

I know that it can be shown to be convergent using integral by parts and trigonometric inequalities, but I don't know how to show that it is not absolutely convergent

• Euristics that can be made rigorous: as $x\to +\infty$ $|\sin(\frac{\sin(x)}{x})|\sim|\sin(\frac{1}{x})|\sim\frac{1}{x}$. Mar 31, 2022 at 15:46
• thanks, how to make it rigorous? It seems to show the convergence but how to show the absolutely convergence. Apr 1, 2022 at 3:39

For any $$x$$, $$\left\lvert\frac{\sin(x)}{x}\right\rvert\leq1$$. For any $$\theta$$ with $$\left\lvert\theta\right\rvert\leq1$$, $$\sin\lvert\theta\rvert\geq\frac12\lvert\theta\rvert$$.
So \begin{align} \left\lvert\sin\frac{\sin{x}}{x}\right\rvert &=\sin\left\lvert\frac{\sin{x}}{x}\right\rvert\\ &\geq\frac12\left\lvert\frac{\sin{x}}{x}\right\rvert \end{align}
So if you already know $$\int_1^{\infty}\left\lvert\frac{\sin{x}}{x}\right\rvert\,dx=\infty$$, then $$\int_1^{\infty}\left\lvert\sin\frac{\sin{x}}{x}\right\rvert\,dx$$ is divergent as well.