When I try to do some exercises, the following question raises.
How to bound $\displaystyle\int_{1}^x \dfrac{\sin(\lfloor t\rfloor)}{t}dt$ by a good upper bound than $\ln x$?
$\ln x$ is the trivial as $$\left|\displaystyle\int_{1}^x \dfrac{\sin(\lfloor t\rfloor)}{t}dt\right| \leq \int_{1}^x \dfrac{1}{t}dt = \ln x.$$
Actually the limit (as $x\to\infty$) exists. We may write $\int_1^x=\int_1^{\lfloor x\rfloor}+\int_{\lfloor x\rfloor}^x$ and see that the second integral tends to $0$ as $x\to\infty$, and the first integral (after $\int_1^n=\sum_{k=1}^{n-1}\int_k^{k+1}$) is $$\sum_{k=1}^{\lfloor x\rfloor-1}\log\frac{k+1}{k}\ \sin k,$$ which has a finite limit (as $x\to\infty$) by Dirichlet's test.