On absolute and conditional convergence of trigonometric integral

Question

I am trying to make out if $\int_1^{\infty}{\arctan(\frac{\cos(x)}{x^{2/3}})}dx$ converges absolutely or conditionally. My answer is that there is no convergence at all, because substitution $x^{\frac{2}{3}}=u$ gives $\frac{2}{3}\int_1^{\infty}\frac{\arctan(\frac{\cos(u^{\frac{3}{2}})}{u})dx}{u^{-\frac{1}{2}}}$. But i am not sure.

Answer 1

The integral is not absolutely convergent but it is conditionally convergent. It is enough to study the behaviour of the integrand function $f$ over the intervals $[k\pi,(k+1)\pi]$ for $k\in\mathbb{N}$. Over such an interval, $|f|$ behaves like $x^{-2/3}$, but:

$$ \int_{k\pi}^{(k+1)\pi}\arctan\left(\frac{\cos x}{x^{2/3}}\right)\,dx =\\ \int_{0}^{\pi/2}\arctan\left(\frac{\cos x}{(x+k\pi)^{2/3}}\right)\,dx-\int_{0}^{\pi/2}\arctan\left(\frac{\cos x}{(x+(k+1)\pi)^{2/3}}\right)\,dx$$ behaves like $\frac{1}{k^{5/3}}$, and $\sum_{k\geq 1}\frac{1}{k^{5/3}}$ is convergent.