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Given Integral: $$\int_0^\infty\frac{dx}{x^3+1}$$

I have to test its convergence.

I am having problem in integrating it. So far, I have reduced it to the partial fraction:

$$\lim_{p\to\infty}\int^p_0\frac{dx}{3(x+1)}+\int^p_0\frac{(2-x)\,dx}{3(x^2-x+1)}$$

But How do I proceed after this?

José Carlos Santos
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The Dead Legend
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  • Well, you might need to split up the region. For instance, for $x\in [0,1]$ the integrand is $≤ 1$. Can you handle the region $[1,\infty)$? – lulu Sep 03 '17 at 16:51
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    Why do you mention a series in your title ? It is an integral you are working on... – Jean Marie Sep 03 '17 at 16:53

1 Answers1

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All you need is to use the fact that, in $[1,+\infty)$, $\frac1{x^3+1}<\frac1{x^3}$. Therefore$$\int_1^{+\infty}\frac1{x^3+1}\,\mathrm dx<\int_1^{+\infty}\frac1{x^3}\,\mathrm dx=\frac12.$$

José Carlos Santos
  • 427,504