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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

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Hint: as $x \to 1$, $\frac{\sin(x^2-1)}{(x-1)^2} \sim \frac{2}{x-1}$ using $\lim_{t\to 0}\frac{\sin(t)}{t}$.

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Thanks, I'll try using your hint and come back if I run into trouble! – Dever Dec 13 '11 at 3:18
I used your hint and found that $\int_{0}^{1}\frac{sin(x^2-1)}{(x-1)^2}dx$ doesn't absolutely converge. Because $\int_{0}^{1}\frac{|sin(x^2-1)|}{(x-1)^2}dx=-\int_{0}^{1}\frac{sin(x^2-1)}{(x-1)‌​^2}dx$ this means the integral doesn't converge either. And this means $\int_{0}^{\infty}\frac{sin(x^2-1)}{(x-1)^2}dx$ neither converges absolutely nor converges. Is this correct? – Dever Dec 13 '11 at 3:27

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