Convergence of $\int_1^{+\infty}\frac{\sin(x)}{x}\arctan(x) \mathop{d}x$

Question

How to prove that integral $$\int_1^{+\infty}\frac{\sin(x)}{x}\arctan(x) \mathop{d}x$$ converges (or does not)?

Answer 1

In fact the integral converges since

$$\frac{\sin x}{x}\arctan x\sim_\infty\frac \pi2\frac{\sin x}{x}$$ and the integral

$$\int_1^\infty \frac{\sin x}{x}dx$$ exists (using an integration by parts).