TITLE: Investigate on convergence and absolute convergence
QUESTION [5 upvotes]: How to investigate on convergence and absolute convergence the following integral: $$\int \limits_0^1 \frac{\sin(1/x)}{(\sqrt{x} - x)^n}$$ for all real values of $n$?
I've tried to make a comparison test. In order to do this I needed some upper bound, but I didn't know how to pick it. 
UPD
As @zhw. pointed I made substitution $x = 1/y$ and got $$\int \limits_1^{\pi/2} \frac{\sin y \cdot y^{n - 2}}{(\sqrt{y} - 1)^n}dy + \int \limits_{\pi/2}^{\infty} \frac{\sin y \cdot y^{n - 2}}{(\sqrt{y} - 1)^n}dy$$
For the first integral I want to use Taylor expansion for $\sqrt{y}$ at $y = 1$ and also the numerator can be bounded by some constant: $$\int \limits_1^{\pi/2} \frac{\sin y \cdot y^{n - 2}}{(\sqrt{y} - 1)^n}dy < \int \limits_1^{\pi/2} \frac{c}{(\sqrt{y} - 1)^n}dy \sim \int \limits_1^{\pi/2} \frac{c}{(1+ \frac{y - 1}{2} + O((y - 1)^2) - 1)^n}dy = \int \limits_1^{\pi/2} \frac{c}{(\frac{y - 1}{2} + O((y - 1)^2))^n}dy$$ .
I don't really know how to proceed further. And also I don't like the jump with equivalence.
For the right integral I probably could say that as $y \to \infty \Rightarrow (\sqrt{y} - 1)^n \sim y^{n/2}$, but again I want to be fully correct with this step.
It seems like it is converge for $n < 1$.
I can't say anything about absolute convergence.

REPLY [2 votes]: Hint: Since $\sqrt x - x$ has a simple zero at $1,$ the integral diverges for $n\ge 1.$ For $n<1,$ make the change of variables $x = 1/y.$