# Does $\int_1^\infty\sin (\frac{\sin x}{x})\mathrm d x$diverge or not?

Does $\int_1^\infty\sin (\frac{\sin x}{x})\mathrm d x$diverge or not? If it converges, does it converge conditionally or absolutely? I guess that it converges conditionally, also,I think it may be related to $\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x}\mathrm d x$ , but I do not know how to start? Any help will be appreciated.

• One hint is that sin(x) is bounded below and above by -1 and 1. What will this tell us about sin(x)/x? – mathreadler Jun 14 '15 at 16:16
• @mathreadler: In itself that doesn't tell us enough -- for example, $\cos(1/x)$ has the same bounds, but $\cos(1/x)/x$ does not have a bounded integral. – Henning Makholm Jun 14 '15 at 16:18
• Hint: For $x$ large, the function behaves as $\sin(x)/x$.. – tired Jun 14 '15 at 16:34
• yes it is not enough in itself. it is just intended to be a hint. – mathreadler Jun 14 '15 at 17:27

For any $n\in\mathbb{Z}^+$ the integral $\int_{0}^{+\infty}\left(\frac{\sin x}{x}\right)^n\,dx$ can be explicitly computed (see here, for instance).
From the approximation $\frac{\sin x}{x}\approx e^{-x^2/6}$, it is expected to be positive and decay like $\sqrt{\frac{3\pi}{2n}}$.
These facts already, together with $\sin z$ being an entire function, give that our integral is a converging one. A more convincing argument, maybe, is that for any $x\geq 1$ the inequality: $$\left(1-\frac{1}{5x^2}\right)\cdot\frac{\sin x}{x}\leq \sin\left(\frac{\sin x}{x}\right)\leq\frac{\sin x}{x}$$ holds. Being equivalent to $\int_{0}^{+\infty}\frac{\sin x}{x}\,dx$, our integral is converging but not absolutely converging.