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Lebesgue integration

  1. Oct 11, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    I have a HW sheet here on the dominated convergence theorem and this problem is giving me a hard time. It simply asks to show that

    [tex]\sum_{k=1}^{+\infty}\frac{1}{k^k}=\int_0^1\frac{dx}{x^x}[/tex]


    3. The attempt at a solution

    Well, according the the cominated convergence thm, if I could find a sequence of functions fn(x) such that fn(x) -->1/x^x and such that

    [tex]\int_0^1 f_n = \sum_{k=1}^n\frac{1}{k^k}[/tex],

    then I would have won. But I've had no luck with finding this sequence. Any hint?
     
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  3. Oct 11, 2007 #2

    morphism

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    Here's something you might find useful:

    [tex]\int_0^1 \frac{dx}{x^x} = \int_0^1 e^{-x \log{x}} \; dx = \int_0^1 \lim_{n\to \infty} \sum_{k=0}^{n} \frac{(-1)^k}{k!} \, (x\log{x})^k \; dx[/tex]

    And maybe some reduction formulae from here.
     
  4. Oct 11, 2007 #3

    quasar987

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    What do you call a "reduction formula"?
     
  5. Oct 11, 2007 #4

    morphism

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    Basically, try to write I_k in terms of I_(k-1), where

    [tex]I_k = \int_0^1 \frac{(-1)^k}{k!} \, (x\log{x})^k \; dx.[/tex]

    Edit:
    You can try to use the Gamma function to help you out a bit.
     
    Last edited: Oct 11, 2007
  6. Oct 12, 2007 #5

    quasar987

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    What I would need to show to get my answer is that

    [tex]I_k=(I_{k-1}+1)^{I_{k-1}+1}[/tex]

    which seems impossible
     
  7. Oct 12, 2007 #6

    quasar987

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    Actually, (xlogx)^k is easily integrated by part k times! thx for the help morphism, this is solved!
     
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