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  • $\displaystyle \int_0^\infty \frac{\arctan\frac{x^3}{1+x^2}}{x^2} \, dx$

So i know that this one converge from 1 to infinity (by Dirichlet rule), but i'm not sure about the 0 to 1 segment. I kind of think that it converge as well, but can't prove it myself. Any suggestions?

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2 Answers 2

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HINT

Note that for $x\to 0$

$$\frac{\arctan\frac{x^3}{1+x^2}}{x^2}=\frac{\arctan\frac{x^3}{1+x^2}}{\frac{x^3}{1+x^2}}\frac{x}{1+x^2}\to 1\cdot 0=0$$

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HINT: The function is positive in that interval. Find the maximum (not at 0). It is a finite value. The integral from $0$ to $1$ must be less than the maximum multiplied by $1-0$

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