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Thread: How to show the divergence of integral

  1. #1
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    Exclamation How to show the divergence of integral

    Hello, how to show the integral diverges:

    $\displaystyle \int_{1}^{\infty}$
    Last edited by lebdim; Nov 22nd 2018 at 10:22 AM.
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    MHF Contributor MarkFL's Avatar
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    Re: How to show the divergence of integral

    Without an integrand and differential, that's going to be a bit tricky.
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    Re: How to show the divergence of integral

    @MarkFL, the integral is: $\displaystyle \int_{1}^{\infty}\dfrac{\sqrt{x}}{\sqrt[3]{x^2 -1}}dx$...
    Last edited by lebdim; Nov 22nd 2018 at 11:26 PM.
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    MHF Contributor MarkFL's Avatar
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    Re: How to show the divergence of integral

    Quote Originally Posted by lebdim View Post
    @MarkFL, the integral is: $\displaystyle \int_{1}^{\infty}\dfrac{\sqrt{x}}{\sqrt[3]{x^2 -1}}dx$...
    I would begin with the fact that for $\displaystyle x\in(1,\infty)$ we have:

    $\displaystyle \frac{1}{x^{\Large\frac{1}{6}}}< \frac{\sqrt{x}}{\sqrt[3]{x^2-1}}$

    Now, let's disregard the singularity at the lower bound, and consider:

    $\displaystyle I=\int_2^{\infty} \frac{\sqrt{x}}{\sqrt[3]{x^2-1}}\,dx$

    What does the $\displaystyle p$-test tell us about the convergence of $\displaystyle I$?
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    Re: How to show the divergence of integral

    Ok, thank you. Because $\displaystyle p = \frac{1}{6} < 1$ and then it's divergent.
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    Re: How to show the divergence of integral

    Because $\displaystyle p$-series are convergent for $\displaystyle p > 1$. Thank you!
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