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$$\text{Let}\;\; I=\int_{0}^{+\infty}{x^{\large\frac{4a}{3}}}\arctan\left(\frac{\sqrt{x}}{1+x^a}\right)\,\mathrm{d}x.$$

I need to find all $a$ such that $I$ converges.

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1 Answer 1

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 $a\lt\frac12$.

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$\arctan(x)\sim \dfrac{\pi}{2},\;\; x\to {+\infty}$ – M. Strochyk Dec 22 '12 at 8:27
@M.Strochyk: The idea is that it is bounded, but thanks for the correction. – robjohn Dec 22 '12 at 9:35

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