Getting asymptotic behaviour of an integral? I am interested in the $\rho\sim0$ asymptotics of the following expression
$$
\int_{1.1}^{\infty}\frac{\sin(k\rho)}{k^{1.9}\rho}\frac{1}{\log\frac{1}{k}}\,dk
$$
any ideas of how to tackle this?
 A: As long as $k \rho < \epsilon$ for some $\epsilon > 0$, we can use the fact that $\sin (k \rho) \sim k \rho$ to write 
$$\int_{1.1}^{\frac{\epsilon}{\rho}} \frac{1}{k^{0.9}}\frac{1}{\log \frac{1}{k}} \, dk + \int_{\frac{\epsilon}{\rho}}^{\infty} \frac{\sin(k \rho)}{k^{1.9}\rho}\frac{1}{\log \frac{1}{k}} \, dk$$
After factoring out the $\rho$ in the denominator of the second integral and using the fact that $-1 \leq \sin x \leq 1$, we can bound the integral and get
$$\int_{1.1}^{\frac{\epsilon}{\rho}} \frac{1}{k^{0.9}}\frac{1}{\log \frac{1}{k}} \, dk + O(\rho^{-1})$$
According to wolfram alpha, the antiderivative of $-\frac{1}{k^{0.9} \log(k)}$ is $-\operatorname{Ei}(0.1 \log(k)) + C$, so that first integral works out to be
$$-\operatorname{Ei}\left(0.1 \log \left( \frac{\epsilon}{\rho} \right) \right) + \operatorname{Ei}\left(0.1 \log \left( 1.1\right) \right)$$
The second term is a constant that is approximately $-4.06643$. You can apply some bounds on the Exponential Integral to go further. I tried and ended up with something like $O \left( \frac{ \rho^{0.1} }{ \log \rho } \right)$, but I wasn't confident about some of the steps I took and it's late, so I'll leave the rest as an exercise for the reader. :)
A: Let $u = \rho k$:
$$I = \int_{11/10}^\infty \frac {\sin \rho k} {\rho k^{19/10} \ln (1/k)} dk =
\frac {\rho^{-1/10}} {\ln \rho} \int_{11 \rho/10}^\infty
 \frac {\sin u} {u^{19/10} \left(1 - \frac {\ln u} {\ln \rho} \right)} du.$$
The lower limit of integration can be extended to zero, and the integrand can be expanded into powers of $\ln u/\ln \rho$, which gives
$$\int_0^\infty u^a \ln^i u \sin u \,du =
\frac {d^i} {d a^i} \int_0^\infty u^a \sin u \,du =
\frac {d^i} {d a^i} \left( \Gamma(1 + a) \cos \frac {\pi a} 2 \right), \\
I \sim \rho^{-1/10} \sum_{i = 0}^\infty
 \frac {d^i} {d a^i} \left( \Gamma(1 + a) \cos \frac {\pi a} 2 \right)
  \Big\rvert_{a = -19/10}
 \ln^{-i - 1} \rho.$$
