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Math Help - improper integral

  1. #1
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    improper integral

    show the improper integral is well defined.

    \int_0^{\infty} \frac{x^{2/3}}{e^{2x}-1}\,dx
    Last edited by Chris L T521; November 15th 2009 at 02:17 PM. Reason: TeXified the question.
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  2. #2
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    Quote Originally Posted by stumped765 View Post
    show the improper integral is well defined.

    \int_0^{\infty} \frac{x^{2/3}}{e^{2x}-1}\,dx
    Split the integral into two, \textstyle\int_0^\infty = \int_0^1+\int_1^\infty, then use the comparison test.

    Near x=0, e^{2x}\approx 1+2x. So \frac{x^{2/3}}{e^{2x}-1}\approx \tfrac12x^{-1/3}, and \int_0^1x^{-1/3}dx converges.

    For large x, e^{2x}-1>x^2 (because exponentials grow faster than any power of x). So \frac{x^{2/3}}{e^{2x}-1} < x^{-4/3}, and \int_1^\infty x^{-4/3}dx converges.
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