Q:

Does $\int_{0}^{\infty} \cos (x^2) dx$ diverge absolutely?

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

A:

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 $\int_0^{+\infty}|\cos(x^p)|dx$, $p>0$. 

