For what values of $a$ does this improper integral converge? $$\text{Let}\;\; I=\int_{0}^{+\infty}{x^{\large\frac{4a}{3}}}\arctan\left(\frac{\sqrt{x}}{1+x^a}\right)\,\mathrm{d}x.$$
I need to find all $a$ such that $I$ converges.

Hint 1: Near $x=0$, $\arctan(x)\sim x$ whereas near $x=+\infty$, $\arctan(x)\sim\pi/2$.
Hint 2: Near $x=0$, consider $a\ge0$ and $a\lt0$. Near $x=+\infty$, consider $a\ge\frac12$ and $a\lt\frac12$.
