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I think this is a dominated convergence theorem question

  1. Jan 30, 2012 #1
    Hey All,

    I have the following integral expression:
    [tex]
    y = lim_{h\to0^{+}} \frac{1}{2\pi} \left\{\int^{\infty}_{-\infty} P(\omega)\left[e^{i\omega h} - 1 \right] \right\} \Bigg/ h
    [/tex]

    And I am trying to understand when this expression will be zero.

    I was talking to a mathematician who said that if P(w) decays faster than 1/w then the limit can be brought inside the integral and then the expression will be zero. His explanation for this was that by the "Dominated Theorem of Convergence" as long as the decay rate of the integrand is faster than ln(w) then the resulting decay is fast enough that it dominates the integral and the limit can be brought inside.

    I was looking at the wikipedia page for this http://en.wikipedia.org/wiki/Dominated_convergence_theorem

    which says:

    Suppose that the sequence converges pointwise to a function ƒ and is dominated by some integrable function g in the sense that
    [tex]
    |f_n(x)| \le g(x)
    [/tex]

    for all numbers n in the index set of the sequence and all points x in S. Then ƒ is integrable and
    [tex]
    \lim_{n\to\infty} \int_S f_n\,d\mu = \int_S f\,d\mu.
    [/tex]

    is my case that the P(w) function must be less than some square integrable function that decays faster than 1/w, because anything higher i.e. 1/w^2, 1/w^3 etc... will be square integrable ? (and as far as applied mathematicians / engineers are concerned anything with decay rate faster than 1/w is square integrable)

    Thanks
     
  2. jcsd
  3. Jan 30, 2012 #2

    mathman

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    Gold Member

    Your original statement is incorrect unless you have some condition on P, which you haven't stated.

    Assuming for example that P is bounded and non-zero on a finite interval, then you can take the limit under the integral sign and the integrand becomes iωP(ω). The integral can be anything. The simplest condition on P to get 0 is to assume P is an even function.
     
  4. Jan 31, 2012 #3
    Thanks mathman,

    I think I expressed the question wrong. I don't really care (for the purposes of this post) about when the expression will be zero, I am more interested in understanding what the mathematician was talking about with the theorem of dominated convergence.

    Lets say the function meets your conditions of being bounded and non zero in some interval. Was my description that you need the function being integrated to be 'integrable' (greater than the g(x) function as discussed in wikipedia) (and hence need it to have decay greater than 1/w) in order to bring the limit inside the integral ?

    Thanks
     
  5. Jan 31, 2012 #4

    mathman

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    Science Advisor
    Gold Member

    The dominant convergence in this case requires |P(ω)ω| be dominated by an integrable function.
     
  6. Feb 1, 2012 #5
    Thanks Mathman
     
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