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Improper integral question

  1. May 13, 2009 #1
    Establish convergence/divergence of the following improper integral:

    integral 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, as hard as I have tried.

    Thanks, any help is much appreciated.
     
  2. jcsd
  3. May 13, 2009 #2

    Mark44

    Staff: Mentor

    Is there some reason you wrote /x - 5/ using the slashes?
    Why did you choose 1 as a limit of integration in two of your integrals? There is nothing unusual happening at x = 1.
    In your integrals, what do you have for the integrands? Please show us the work you did.

    BTW, there is no such word as "intergral."
     
  4. May 13, 2009 #3
    By /x-5/, I mean the absolute value of (x-5).

    I chose 1 because don't you have to break up the original improper integral into appropriate intervals to see what's going on? There are unusual things happening at x=0 and x=5, and 1 was chosen to 'link up' these separate intervals (for want of a better phrase).

    So I have this:
    [0 to 1] 1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) ) + [1 to 5]1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) ) + [5 to infinity]1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) )

    The first two integrals I have shown to be convergent (by comparison), but I am stuck on the last one. Mabye I am going about this the wrong way altogether, but I'm sort of following a similar example from my lecture notes.

    (Sorry about the messiness, I am trying to learn Latex)
     
  5. May 13, 2009 #4
    1+sqrt(x)>sqrt(x), and for large x, x-5>0.5x
     
  6. May 13, 2009 #5
    P.S. You also should break [5,infty) into two intervals.
     
  7. May 13, 2009 #6

    Mark44

    Staff: Mentor

    OK, I understand now why you are using 1 as a limit of integration. For absolute values, you can use this character: |.

    To help you out with your LaTeX, your last integral is
    [tex]\int_5^{\infty}\frac{1}{x^{1/3}|x - 5|^{1/3}(1 + \sqrt{x})^{0.7}}dx[/tex]
    You'll need to break this into two integrals, say from a to 6 and from 6 to b, and look at the limit of each as a approaches 5 from above and as b approaches infinity. You can click on the integral and see the LaTeX code, and can copy and paste it to create your own.

    Since in both integrals, x >= 5, you can drop the absolute value signs so you have (x^2 - 5x)^(1/3). I doubt that the integral can be found by ordinary means, so you'll probably need to use another comparison.

    I'll need to think about this one...
     
  8. May 13, 2009 #7

    Mark44

    Staff: Mentor

    The denominator is roughly comparable with x61/60, with that exponent obtained by adding 2/3 and 7/20. Since this exponent is > 1, the integral of this function, from 6 to infinity, converges. If you can show that your denominator is > x61/60, and thereby that your integrand is < 1/x61/60, you're home free.
     
  9. May 14, 2009 #8
    Ahh okay, that makes alot more sense now, thanks alot Mark.

    When I finish writing it up I will post my final explanation.
     
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