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Determine whether the limit exists and evaluate the integral if it does

  1. Dec 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Determine whether the limit

    [itex] \lim_{n \rightarrow \infty} \int_{-1}^1 \frac{1}{\sqrt[3]{1-x^{2n}}} dx [/itex]

    exists and evaluate the integral if it does


    2. Relevant equations

    Dominated convergence theorem (I think) and a power series representation.


    3. The attempt at a solution

    I've been attempting to find a dominating function for [itex] \frac{1}{\sqrt[3]{1-x^{2n}}} [/itex] by looking at the power series representation for [itex] \frac{1}{1-x^{2n}} [/itex], but I'm not seeing it. I'd appreciate any help.
     
    Last edited: Dec 23, 2011
  2. jcsd
  3. Dec 23, 2011 #2

    SammyS

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    What does k have to do with this?

    Why take the limit as k → ∞ ?
     
  4. Dec 23, 2011 #3
    Much more interesting if we consider:

    [tex]\lim_{n\to\infty}\int_{-1}^1 \frac{dx}{\sqrt[3]{1-x^{2n}}}[/tex]

    and in that case isn't:

    [tex]x^{2n}\leq x^2[/tex]

    in that interval?
     
    Last edited: Dec 23, 2011
  5. Dec 23, 2011 #4
    Sorry, that should be an n, not a k.

    It's true that x^{2n} \leq x^2 on [-1,1], so [itex] \frac{1}{\sqrt[3]{1-x^{2n}}} \leq \frac{1}{1-x^2} [/itex] on [-1,1], but [itex] \frac{1}{1-x^2} [/itex] isn't in [itex] L^1(-1,1) [/itex]

    So I'm not sure what to do with it.
     
  6. Dec 23, 2011 #5

    I like Serena

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    What about ##1 \over \sqrt{1-x^2}##?
     
  7. Dec 23, 2011 #6
    Thanks, I think that works perfectly. (I can't believe I didn't see that before - sorry - and that problem was really starting to piss me off.)
     
  8. Dec 23, 2011 #7

    I like Serena

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    Good! :)

    Does that mean that you also evaluated the integral in its limit?
     
  9. Dec 23, 2011 #8
    Yeah, I knew that several hours ago, just not the dominating function. THank you for your help.
     
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