Suppose $F(x)$ be a distribution function such that $\displaystyle \lim_{x\to \infty} \frac{F(x)}{1-x^{-a}} =1$ for some $a>0$. Then for all $\epsilon > 0$ there exists a large enough $K$ such that $$1-\epsilon < \left|\frac{\int_K^\infty \sin (x) \,dF(x) }{ \int_K^\infty \frac{a\sin (x) }{x^{a+1}}\,dx}\right| < \epsilon +1$$
Intuitively it is clear to me. As $F(x) \sim 1-x^{-a}$, $dF(x) \sim \frac{a}{x^{a+1}}$ and boundedness of $\sin x$ probably ensures the result. However I cannot prove this rigourously. Any help/suggestions on this.