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Approximation of integral for small boundary

  1. Jun 4, 2012 #1
    This problem arises in a paper on population genetics (Kimura 1962).

    1. The problem statement
    Let [itex]f(p) = \int_0^p ((1 - x)/x)^k dx[/itex].

    For a small value of p, we have approximately
    f(p) = (p ^ (1-k)) / (1-k)

    How is this obtained?

    2. My attempt at a solution
    I tried to expand the f(p) around p = 0. However, f'(p) = ((1 - p)/p)^k is undefined at p=0. Furthermore, it does not seem that this approach can yield the form p^(1-k) / (1-k). I must be missing something.

    I would appreciate any insights. Thanks.
     
  2. jcsd
  3. Jun 4, 2012 #2
    [itex]\frac{d}{dp} \frac{p^{1-k}}{1-k} = \frac{1}{p^{k}}[/itex]

    And the derivative of your original function you said was [itex](\frac{1-p}{p})^{k}[/itex]



    Does that help at all?
     
  4. Jun 4, 2012 #3
    Thanks for your reply Villyer. I've just solved it, with some help from a friend.

    Here's the solution.

    For [itex] x \approx 0 [/itex], [itex] (\frac{1 - x}{x})^k = x^{-k} [/itex]. Therefore, [itex] f(p) = \int_0^p x^{-k} dx = \frac{1}{1-k} p^{1 - k} [/itex], as required.

    Cheers!
     
  5. Jun 4, 2012 #4
    Exactly.
     
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