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

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

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .