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I have to determine whether

$$\int_0^\infty \frac{x\arctan x}{\sqrt[3]{1+x^4}}dx$$

converges or not.

I suspect it doesn't because $\arctan x$ is very close to $\pi/2$ as $x$ goes to infinity, and

$$\int_0^\infty \frac{x}{\sqrt[3]{1+x^4}}dx=\infty$$

because, for example,

$$\frac{x}{\sqrt[3]{1+x^4}}>\frac{x}{\sqrt[3]{2x^4}}=(2x)^{-1/3}$$

for sufficiently large $x$.

But I don't know how to bound $\arctan x$ from below for this to work. I also suspect that there is something more refined than the comparison test to this problem. I don't really know the context in which this integral came up, because I don't attend the course, so I don't know what methods I should try.

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4  
Just note that $\arctan$ is greater than $\pi/4$ for sufficiently large $x$, because it is monotonically increasing. Then your argument goes through fine. –  Potato Jan 8 '13 at 20:13
    
@Potato Right! This is embarrassing... If you post this as an answer, I'll accept it. –  Bartek Jan 8 '13 at 20:16

2 Answers 2

up vote 1 down vote accepted

Just note that arctan is greater than π/4 for sufficiently large x, because it is monotonically increasing. Then your argument goes through fine.

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You are on the right track. Notice that

$$ \frac{x}{\sqrt[3]{1+x^4}} \sim \frac{x}{\sqrt[3]{x^4}},$$

as $ x \to \infty. $

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1  
When your nice hint came here, mine became faint.+ –  Babak S. Jan 8 '13 at 20:23
    
@BabakSorouh: Thank you. This is very nice from you. –  Mhenni Benghorbal Jan 8 '13 at 21:57

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