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}

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 $$ \int_\epsilon^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}\log(t/\rho)}\;dt = \int_\epsilon^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}(\log\rho^{-1} + \mathcal{O}(\log\epsilon^{-1}))}\;dt, $$ since in the range of integration we have $|\log t|\leq\log\epsilon^{-1}$. Putting $L=\log\rho^{-1}$ and $\alpha=\log\epsilon^{-1}$ we have for $\alpha<L/2$, i.e. $\rho<\epsilon^2$ $$ \left|\frac{1}{L}-\frac{1}{L+\alpha}\right| = \frac{|\alpha|}{L(L+\alpha)} \ll \frac{|\alpha|}{L^2}, $$ thus \begin{split} \int_\epsilon^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}(L + \mathcal{O}(\alpha))}\;dt &= \int_\epsilon^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}}\left(\frac{1}{L}+\mathcal{O}\left(\frac{\alpha}{L^2}\right)\right)\;dt\\ & = \frac{1}{L}\int_\epsilon^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}}\;dt + \mathcal{O}\left(\frac{\alpha}{L^2}\int_\epsilon^{\epsilon^{-1}}\frac{|\sin t|}{t^{1.9}}\right). \end{split} In the error term we bound $|\sin t|$ by $\min(t, 1)$ and split the integral into the range $[\epsilon, 1]$ and $[1, \epsilon^{-1}]$. Doing so we find that the integral converges, hence the error term is $\mathcal{O}(\frac{\alpha}{L^2})$. In the main term we extend the integral to the range $(0,\infty)$. Doing so introduces an error $$ \frac{1}{L}\int_0^\epsilon\frac{\sin t}{t^{1.9}}\;dt \leq \frac{1}{L}\int_0^\epsilon\frac{1}{t^{0.9}}\;dt\ll\frac{\epsilon^{0.1}}{L} $$ and another one of size $$ \frac{1}{L}\int_{\epsilon^{-1}}^\infty\frac{\sin t}{t^{1.9}}\;dt \leq \frac{1}{L}\int_{\epsilon^{-1}}^\infty\frac{dt}{t^{1.9}} \ll \frac{\epsilon^{0.9}}{L}. $$ Together we obtain $$ \int_\epsilon^{\epsilon^{-1}}\frac{\sin t}{t^{1.9}\log(t/\rho)}\;dt = \frac{1}{L}\int_0^\infty\frac{\sin t}{t^{1.9}}\;dt + \mathcal{O}\left(\frac{\epsilon^{0.1}}{L}+\frac{\alpha}{L^2}\right). $$ Together with the bounds obtained before we get that the original integral equals $$ \frac{\rho^{-0.1}}{L}\int_0^\infty\frac{\sin t}{t^{1.9}}\;dt + \mathcal{O}\left(\frac{\rho^{-0.1}\epsilon^{0.1}}{L}+\frac{\rho^{-0.1}\alpha}{L^2}\right). $$ Putting $\epsilon=L^{10}$, i.e. $\alpha=10\log\log\rho^{-1}$, we get $$ \frac{\rho^{-0.1}}{L}\int_0^\infty\frac{\sin t}{t^{1.9}}\;dt + \mathcal{O}\left(\frac{\rho^{-0.1}\log\log\rho^{-1}}{(\log\rho^{-1})^2}\right). $$ The integral in the main term is positive, thus the main term is larger than the error term by a factor $\frac{\log\log\rho^{-1}}{\log\rho^{-1}}$, and we got an asymptotic formula.

The main source of error is the approximation of the term $\log\rho^{-1}+t$ by $\log\rho^{-1}$. If you need better asymptotics, you would have to use the series expansion of $\frac{1}{1+x}$. In this way you would get an asymptotic series in $\frac{1}{L}$, but the computations would become rather long.