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Math Help - Testing for converging integrals

  1. #1
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    Testing for converging integrals

    The question:

    Use the inequality form of the comparison test to determine whether or not the following improper integral converges

    \int_{1}^{\infty} \frac{1}{\sqrt{1 + x^4}} dx

    My attempt:
    When x is large, dominant term is x^4

    \frac{1}{\sqrt{1 + x^4}} < \frac{1}{\sqrt{x^4}}

    \int_{1}^{\infty} \frac{1}{x^2} dx

    _{1}^{\infty}[\frac{-1}{x}]

    As x -> infinity, function goes to 0:
    0 - (-1)

    So the integral converges (approaches 1). Is this correct? Thanks.
    Last edited by Glitch; October 29th 2010 at 12:06 AM.
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  2. #2
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    Fixed my previous mistake. Still wondering if this is correct.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    no need to compute the last integral since you know integrals of the form \displaystyle\int_1^\infty\frac{dx}{x^p},\,p>1 converge.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    \displaystyle\int_{1}^{\infty} \frac{1}{\sqrt{1 + x^4}} dx=\frac{2 \Gamma(\frac{5}{4})^2}{\sqrt\pi}
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