Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.