the small-$\rho$ asymptotics is governed by the large-$k$ behavior of the integrand, which gives a scaling as $\propto 1/[ \rho^{(2-1.9)}\log\rho]$, and numerics suggest that 

$$\lim_{\rho\rightarrow 0}\rho^{0.1}\log\rho
\int_{1.1}^{\infty}\frac{\sin(k\rho)}{k^{1.9}\rho}\frac{1}{\log\frac{1}{k}}\,dk\approx 25$$