This is a self-review exercise in my textbook I've been unable to solve:
Does $\int_{0}^{\infty}\frac{\sin(x^2-1)}{(x-1)^2}dx$ converge absolutely? Converge?
I tried separating the integral to $\int_{0}^{1}\frac{\sin(x^2-1)}{(x-1)^2}dx+\int_{1}^{\infty}\frac{\sin(x^2-1)}{(x-1)^2}dx$. Intuitively it looks like you can use the comparison test to show $\int_{0}^{1}\frac{\sin(x^2-1)}{(x-1)^2}dx$ doesn't converge absolutely/converge? So the answer is a double no. But intuition isn't a proof, unfortunately, so I'm stuck...
In any case, can anyone explain how this question can be solved?
Thank you