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Math Help - improper intergal question

  1. #1
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    improper intergal question

    Establish convergence/divergence of the following improper integral:

    int from 0 to infinity of 1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) )

    My attempt at a solution was to break it up into 3 intergrals: 0 to 1, 1 to 5, and 5 to infinity...I showed that the first two of these integrals converges by comparing them to 1/x^(1/3) and 1//x-5/^(1/3) respectively. I can't establish convergence/divergence of last intergral from 5 to infinity.

    Thanks, any help is much appreciated.
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  2. #2
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    Quote Originally Posted by davido View Post
    Establish convergence/divergence of the following improper integral:

    \int_0^\infty\!\!\! \frac1{x^{1/3}|x-5|^{1/3}(1 + \sqrt{x})^{0.7}}\,dx

    My attempt at a solution was to break it up into 3 integrals: 0 to 1, 1 to 5, and 5 to infinity...I showed that the first two of these integrals converges by comparing them to 1/x^(1/3) and 1//x-5/^(1/3) respectively. I can't establish convergence/divergence of last integral from 5 to infinity.
    You probably know that \int^\infty \!\!\!x^{\alpha}dx converges if \alpha<-1. So you want to estimate the power of x in the denominator of that fraction, as x\to\infty. The first factor is x^{-1/3}, the second one is also approximately x^{-1/3} (because |x5| will have the same order of magnitude as x, when x is large). Similarly, the third factor is approximately x^{-(1/2)0.7} = x^{-7/20}. So the integrand is approximately x^{-(1/3)-(1/3)-(7/20)} = x^{-61/60}.

    Since 61/60 < 1, you should conclude that the integral converges on the interval [5,\infty), and you prove it by comparing it with \int^\infty \!\!\!x^{-61/60}dx.
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