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the small-$\rho$ asymptotics of $$I(\rho)=-\int_{1.1}^{\infty}\frac{\sin(k\rho)}{\rho k^{1.9}\ln k}\,dk$$

is governed by the large-$k$ behavior of the integrand, which gives $I(\rho)\propto \rho^{1.9-2}$ up to a logarithmic factor, and a numerical evaluation supports the asymptotic

$$I_{\rm asymp}(\rho)=-\rho^{-0.1}(3.5+0.126\ln\rho)$$

Log-linear plot of $-\rho^{0.1}I(\rho)$ (blue line) and $-\rho^{0.1}I_{\rm asymp}(\rho)$ (orange line)

UPDATE, following Bazin's recent answer:

I have some convergence issues with a numerical evaluation at much smaller $\rho$, but if I can trust the Mathematica output the curve does seem to level off towards an asymptote at zero, as derived by Bazin, possibly as $I(\rho)=-\text{constant}\times(\rho^{0.1}\log\rho)^{-1}$.

the small-$\rho$ asymptotics of $$I(\rho)=-\int_{1.1}^{\infty}\frac{\sin(k\rho)}{\rho k^{1.9}\ln k}\,dk$$

is governed by the large-$k$ behavior of the integrand, which gives $I(\rho)\propto \rho^{1.9-2}$ up to a logarithmic factor, and a numerical evaluation supports the asymptotic

$$I_{\rm asymp}(\rho)=-\rho^{-0.1}(3.5+0.126\ln\rho)$$

Log-linear plot of $-\rho^{0.1}I(\rho)$ (blue line) and $-\rho^{0.1}I_{\rm asymp}(\rho)$ (orange line)

UPDATE, following Bazin's recent answer:

I have some convergence issues with a numerical evaluation at much smaller $\rho$, but if I can trust the Mathematica output the curve does seem to level off towards an asymptote at zero, as derived by Bazin, possibly as $I(\rho)=-\text{constant}\times(\rho^{0.1}\log\rho)^{-1}$.

the small-$\rho$ asymptotics of $$I(\rho)=-\int_{1.1}^{\infty}\frac{\sin(k\rho)}{\rho k^{1.9}\ln k}\,dk$$

is governed by the large-$k$ behavior of the integrand, which gives $I(\rho)\propto \rho^{1.9-2}$ up to a logarithmic factor, and a numerical evaluation supports the asymptotic

$$I_{\rm asymp}(\rho)=-\rho^{-0.1}(3.5+0.126\ln\rho)$$

Log-linear plot of $-\rho^{0.1}I(\rho)$ (blue line) and $-\rho^{0.1}I_{\rm asymp}(\rho)$ (orange line)

the small-$\rho$ asymptotics of $$I(\rho)=-\int_{1.1}^{\infty}\frac{\sin(k\rho)}{\rho k^{1.9}\ln k}\,dk$$

is governed by the large-$k$ behavior of the integrand, which gives $I(\rho)\propto \rho^{1.9-2}$ up to a logarithmic factor, and a numerical evaluation supports the asymptotic

$$I_{\rm asymp}(\rho)=-\rho^{-0.1}(3.5+0.126\ln\rho)$$

Log-linear plot of $-\rho^{0.1}I(\rho)$ (blue line) and $-\rho^{0.1}I_{\rm asymp}(\rho)$ (orange line)

UPDATE, following Bazin's recent answer:

I have some convergence issues with a numerical evaluation at much smaller $\rho$, but if I can trust the Mathematica output the curve does seem to level off towards an asymptote at zero, as derived by Bazin, possibly as $I(\rho)=-\text{constant}\times(\rho^{0.1}\log\rho)^{-1}$.

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

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

the small-$\rho$ asymptotics of $$I(\rho)=-\int_{1.1}^{\infty}\frac{\sin(k\rho)}{\rho k^{1.9}\ln k}\,dk$$

is governed by the large-$k$ behavior of the integrand, which gives $I(\rho)\propto \rho^{1.9-2}$ up to a logarithmic factor, and a numerical evaluation supports the asymptotic

$$I_{\rm asymp}(\rho)=-\rho^{-0.1}(3.5+0.126\ln\rho)$$

Log-linear plot of $-\rho^{0.1}I(\rho)$ (blue line) and $-\rho^{0.1}I_{\rm asymp}(\rho)$ (orange line)