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Large terminal integration

  1. Sep 5, 2015 #1
    1. The problem statement, all variables and given/known data

    ∫1/((x^1/3)-2)dx between b=666 and a=8

    Aim is to prove divergence or convergence

    2. Relevant equations

    3. The attempt at a solution

    Okay so I know I can prove for 4x1/3 < x when 8 < x but I don't know where to go from here. There is obviously very large terminal values, and i know you could say evalute each terminal values using fundamental theorem of calculus..

    Any easier way to do it?
  2. jcsd
  3. Sep 5, 2015 #2
    The integrand is not continuous over [a, b], so you can't use the fundamental theorem of calculus (directly).

    Try the substitution u^3 = x and write the integrand in terms of u. Make sure you change the limits of integration, too. Then try v = u - 2. Then things will look a little more obvious.
  4. Sep 5, 2015 #3


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    The original problem appears to be to integrate from 2 to 666. There is, of course, a problem at x= 2 but since the problem does NOT require integration to infinity, I do not understand Ray Vickson's "(ii) whether the function grows small quickly enough to yield a finite integration as the upper limit goes to infinity."
  5. Sep 5, 2015 #4

    Ray Vickson

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    I had mid-read the original problem; there is no issue at the lower end ##x = 2##. And, somehow, I thought that the OP was trying to figure out whether the integral ##\int_ a^{\infty} f(x) \, dx## exists, by looking at ##\int_a^b f(x) \, dx## for some large, finite ##b = 666## and trying to get a conclusion or insight from that. However, when I go back and re-read the original, I see that no such statement seems to have been made, so the question is actually a more-or-less pointless exercise except, of course, for actually doing the integral, which is elementary, but tedious. Anyway, I deleted my previous post.
  6. Sep 5, 2015 #5


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    But there is an issue at x= 2 where the integrand is not defined.
  7. Sep 5, 2015 #6


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    I agree that there is no problem at x = 2, but the original integral is over the interval [8, 666], and there is a problem at x = 8.
    No, the original integral is ##\int_8^{666}\frac {dx}{\sqrt[3]{x} - 2}##
  8. Sep 5, 2015 #7
    Okay essentially all I want to prove is convergence or divergence of the integral over the [8,666]. I know no problem exists for x=2 but rather the problem exists at x=8

    Using substitution as u= x1/3 and x= x3 i get ∫3*u2/(u-2)

    I than reapply v = u-2 and and therefore u = v + 2 and integrate respectively. But my challenge is, think it is asking you to use some sort of comparison rather than evaluating the integral to calculate the convergence/divergence? Would the comparison test work?
    Last edited: Sep 6, 2015
  9. Oct 2, 2015 #8
  10. Oct 3, 2015 #9


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    Once you have the integral [itex]3\int \frac{u^2}{u- 2} du[/itex] do the division to get [itex]3\int (v+ 2+ \frac{4}{v-2})dx[/itex]. The first two terms are easily integrable and you can "compare" the last to integrating [itex]\frac{1}{x}[/itex] in the vicinity of x= 0.
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