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I need help finding if the improper integral below converges.

$$\int _{ 2 }^{ \infty }{ \frac { dx }{ \sqrt [ 3 ]{ 1-{ x }^{ 4 } } } } $$.

we learnt at class: comparison test ratio test

Thanks in advance.

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  • $\begingroup$ the given integral cannot be real. if we do nnot mind then iitt is asymtoically equivalent to the integral of x ^(-4/3) which exists. $\endgroup$ – Adelafif Jul 13 '16 at 12:33
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As $x\to\infty$, $${1\over \root{3}\of{1-x^4}} = -{1\over \root{3}\of{x^4-1}}\sim -{1\over x^{4/3}}$$ so the integral converges.

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For $x>2, x^{4}-1>\frac{x^{4}}{2}$ holds. Then we have \begin{align} \int_{2}^{\infty}\frac{dx}{\sqrt[3]{x^{4}-1}}<2^{4/3}\int_{2}^{\infty}\frac{dx}{x^{4/3}}=2^{4/3}\times 2\times 2^{1/3}=6 \end{align} and the integral converges.

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