Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
1  
What have you tried so far? –  Trevor Wilson Dec 22 '12 at 5:15

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 $a\lt\frac12$.

share|improve this answer
    
$\arctan(x)\sim \dfrac{\pi}{2},\;\; x\to {+\infty}$ –  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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.