Does $\int_0^\infty \frac{\sin(\sin(x))}{x}dx$ converge or diverge?

Hint. Show that $$F(x):=\int_{0}^x\sin(\sin(t))dt$$ is bounded in $[0,+\infty)$. Note that $f(t):=\sin(\sin(t))$ is a continuous periodic function of period $2\pi$ and $f(x+\pi)=-f(x)$.

Then, by integration by parts, $$\int_0^\infty \frac{\sin(\sin(x))}{x}dx=\left[\frac{F(x)}{x}\right]_0^{+\infty}+\int_0^\infty \frac{F(x)}{x^2}dx=\int_0^\infty \frac{F(x)}{x^2}dx$$ where $\lim_{x\to 0^+}\frac{F(x)}{x}=\lim_{x\to 0^+}\frac{\sin(\sin(x))}{1}=0$ and the integral on the right is absolutely integrable.