I believe it does, but i would like some help formulating a proof.
Does $\int_{0}^{\infty} \cos (x^2) dx$ diverge absolutely?
5
$\begingroup$
integration
limits
convergence-divergence
improper-integrals
-
0For 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 $\cos(a^2)=0$. – 2012-08-06
1 Answers
5
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 $\int_0^{+\infty}|\cos(x^p)|dx$, $p>0$.
-
0tha$n$$k$s for the help, and to be frank this was a question from a past e$x$am and not homework, but the hints help all the same! – 2012-08-06