Setting $t=\rho k$ the integral in question becomes
$$
\int_{1.1\rho}^\infty\frac{\sin t}{\rho^{0.1}t^{1.9}\log (t/\rho)}\;dt.
$$
Fix $\epsilon>0$. Then for $\rho\rightarrow 0$ we have
$$
\int_{1.1\rho}^{\rho^{1/2}} \frac{\sin t}{t^{1.9}\log (t/\rho)}\;dt \ll \int_{1.1\rho}^{\rho^{1/2}} \frac{1}{t^{0.9}}\;dt \ll \rho^{1/20},
$$
$$
\int_{\rho^{1/2}}^\epsilon\frac{\sin t}{t^{1.9}\log (t/\rho)}\;dt \ll\frac{1}{\log\rho^{-1}}\int_{\rho^{1/2}}^\epsilon\frac{dt}{t^{0.9}}\ll\frac{\epsilon^{0.1}}{\log\rho^{-1}},
$$
and
$$
\left|\int_{\epsilon^{-1}}^\infty\frac{\sin t}{t^{1.9}\log (t/\rho)}\;dt\right| \ll
\int_{\epsilon^{-1}}^\infty\frac{1}{t^{1.9}\log \rho^{-1}}\;dt \ll\frac{\epsilon^{0.1}}{\log\rho^{-1}}.
$$
Hence for $\rho<\epsilon^2$ the integral in question equals
$$
\rho^{-0.1}\int_\epsilon^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}\log(t/\rho)}\;dt + \mathcal{O}\left(\frac{\epsilon^{0.1}\rho^{-0.1}}{\log\rho^{-1}}\right).
$$
Now
\begin{split}
\int_\epsilon^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}\log(t/\rho)}\;dt &= \int_\epsilon^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}(\log\rho^{-1} + \mathcal{O}(\log\epsilon^{-1}))}\;dt\\
& = \frac{1}{\log\rho^{-1}} \int_\epsilon^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}}\;dt + \mathcal{O}\left(\frac{\log\epsilon^{-1}}{\log^2\rho^{-1}}\int_\epsilon^{\epsilon^{-1}}\frac{|\sin t|}{t^{1.9}}\;dt \right)\\
&= \left(1+\mathcal{O}\left(\frac{\log\epsilon^{-1}}{\log\rho^{-1}}\right)\right)\frac{1}{\log\rho^{-1}} \int_\epsilon^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}}\;dt\\
&=\left(1+\mathcal{O}\left(\frac{\log\epsilon^{-1}}{\log\rho^{-1}}\right)\right)\frac{1}{\log\rho^{-1}} \int_0^{\infty}\frac{\sin t}{t^{1.9}}\;dt + \mathcal{O}(\epsilon^{0.1}),
\end{split}
provided that $\rho<\epsilon$. We obtain that the integral to be computed equals
\begin{split}
\frac{\rho^{-0.1}}{\log\rho^{-1}}\int_0^\infty\frac{\sin t}{t^{1.9}}\;dt \left(1+ \mathcal{O}\left(\rho^{0.05}+\epsilon^{0.1}+\frac{\log\epsilon^{-1}}{\log\rho^{-1}}\right)\right)\\
= \frac{\rho^{-0.1}}{\log\rho^{-1}}\int_0^\infty\frac{\sin t}{t^{1.9}}\;dt \left(1+ \mathcal{O}\left(\frac{\log\log\rho^{-1}}{\log\rho^{-1}}\right)\right)
\end{split}