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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

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# I think this is a dominated convergence theorem question

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