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Thread: Convergence of sin(x^2)

  1. #1
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    Convergence of sin(x^2)

    Hello,

    I am reading a calculus book and I am stuck with the following exercise:

    Show if $\displaystyle \lim_{t\rightarrow\infty}\int_1^t \sin(x^2) dx$ is convergent or not.

    Sadly, the solutions at the end of the book only say that it is convergent but don't give any hint on how to show this. I would appreciate any ideas as I have spent quite some time on it already and it nags me a bit. Thanks in advance.
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  2. #2
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    Krizalid's Avatar
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    yes, it converges.

    the key is to use an integration by parts.

    first note that $\displaystyle \int_{1}^{\infty }{\frac{1-\cos x^{2}}{x^{2}}\,dx}<\infty,$ this is easy to verify since the integrand is positive then direct comparison test applies and $\displaystyle \frac{1-\cos x^{2}}{x^{2}}\le \frac{2}{x^{2}}.$

    as for the problem, as i said, integrate by parts and get:

    $\displaystyle \int_{1}^{\infty }{\sin \left( x^{2} \right)\,dx}=\frac{1}{2}\int_{1}^{\infty }{\frac{\left( 1-\cos x^{2} \right)'}{x}\,dx}=\frac{1}{2}\left( \underbrace{\left. \frac{1-\cos x^{2}}{x} \right|_{1}^{\infty }}_{\text{finite}}+\underbrace{\int_{1}^{\infty }{\frac{1-\cos x^{2}}{x^{2}}\,dx}}_{<\,\infty } \right).$

    and we're done!
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  3. #3
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    Thank you!
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