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I believe it does, but i would like some help formulating a proof.

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by "diverge absolutely" do you mean $\int_0^\infty \vert \cos(x^2) \vert\ dx=\infty$? – J. Loreaux Aug 6 '12 at 12:17
yes - i was just not sure how to phrase it correctly or how to add the absolute symbol... – JohnnyE. Aug 6 '12 at 12:24
For example, estimate each "bump" by a triangle inside it, to see $\int_0^{a}$ is greater than $a/2$ when $a$ is a spot with $\cos(a^2)=0$. – GEdgar Aug 6 '12 at 12:25
up vote 5 down vote accepted
  • It's equivalent to the convergence of $\int_\pi^{\infty}\frac{|\cos t|}{\sqrt t}dt$, after having used the substitution $x^2=t$.

  • We have $$ \int_{\pi}^{N\pi}\frac{|\cos t|}{\sqrt t}dt=\sum_{n=1}^{N-1}\int_{n\pi}^{(n+1)\pi}\frac{|\cos t|}{\sqrt t}dt$$ Use $\pi$ periodicity of $|\cos|$ and a substitution $s=t-n\pi$ to get bound which doesn't depend on $n$.

  • Find a good below bound will help to show the divergence.

This argument can be applied for the divergence of $\int_0^{+\infty}|\cos(x^p)|dx$, $p>0$.

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This was a homework question – Norbert Aug 6 '12 at 12:33
@Norbert: you are right, I should have look at the tags. What do you suggest? – Davide Giraudo Aug 6 '12 at 12:34
I suggest to organize this answer as list of hints (as I usually do) – Norbert Aug 6 '12 at 12:35
Not a bad idea! I will do that. – Davide Giraudo Aug 6 '12 at 12:36
thanks for the help, and to be frank this was a question from a past exam and not homework, but the hints help all the same! – JohnnyE. Aug 6 '12 at 12:51

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