Convergence of $\int_{0}^{+\infty}\frac{2\alpha x^{\alpha-1}}{1+x^{2\alpha}}\arctan\left(x^{\alpha}\right)\,dx$ as $\alpha \in \mathbb{R}$

Question

Study the convergence of the following integral as $\alpha \in \mathbb{R}$

$$\int_{0}^{+\infty}\frac{2\alpha x^{\alpha-1}}{1+x^{2\alpha}}\arctan\left(x^{\alpha}\right)\,dx$$

As $\alpha >0$ the only singularity is as $x\to+\infty$, where the integral always converges since $f(x)=\mathcal{O}\left(\frac{1}{x^{\alpha+1}}\right)$

How to interpret it as $\alpha<0$? It seems to have singularity in $0$ when $-\frac{1}{2}<\alpha<0$ and in general, obviously, as $x\to+\infty$

Answer 1

When $\alpha <0$ note that that $arctan (x^{\alpha}) \to \frac {\pi} 2$ as $ x \to 0$. Hence the integrand behaves like $x^{\alpha-1}$ which is not integrable near $0$.