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Thread: Convergence of an integral

  1. #1
    MHF Contributor arbolis's Avatar
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    Convergence of an integral

    Show whether or not, by bounding (from below of from bottom, depends of the appropriate choice) it, if the integral converges or not :
    \int_1^5 \frac{dx}{\sqrt{x^4-1}}. I'm sure it converges, so I wanted to bound it from above but couldn't, due to the indetermination of the lower limit of the integral. Can you help me?
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  2. #2
    Senior Member JaneBennet's Avatar
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    For t>1, note that 0<\frac{1}{\sqrt{t^4-1}}<\frac{1}{\sqrt{t-1}}.

    Hence

    0\ <\ \int_t^5{\frac{dx}{\sqrt{x^4-1}}}\ <\ \int_t^5{\frac{dx}{\sqrt{x-1}}}

    Since \int_1^5{\frac{dx}{\sqrt{x-1}}}=\lim_{t\to1^+}{\int_t^5{\frac{dx}{\sqrt{x-1}}}}=\lim_{t\to1^+}{\left[2\sqrt{x-1}\right]_t^5}=4, we can say that \int_1^5{\frac{dx}{\sqrt{x^4-1}}} converges since it is positive and bounded above.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    Show whether or not, by bounding (from below of from bottom, depends of the appropriate choice) it, if the integral converges or not :
    \int_1^5 \frac{dx}{\sqrt{x^4-1}}. I'm sure it converges, so I wanted to bound it from above but couldn't, due to the indetermination of the lower limit of the integral. Can you help me?
    Just note that

    \sqrt{x^4-1}\geq\sqrt{x-1}~\forall{x}\in{D}

    Where D is the region you are talking about.

    EDIT: Wow, I am slow at typing. Jane Benett did what I did and beyond =D
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