Getting asymptotic behaviour of an integral?

Question

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?

Answer 1

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. :)