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Sum and integral

  1. Oct 29, 2005 #1
    let be the function w(x) that only takes discrete values in the sense that is only defined for x=n being n an integer..then my question is if the integral:


    would be equal to the sum of the serie:

    [tex]\sum_{n=-\infty}^{\infty}w(n)f(n) [/tex]

    if the sum and integral would be equal imply that the function

    [tex]w(x)=g(x)\sum_{-\infty}^{\infty}\delta(x-n) [/tex]

    Last edited: Oct 29, 2005
  2. jcsd
  3. Oct 29, 2005 #2


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    Yes, if you mean the Stieljes integral.

    Where the Riemann integral [itex]\int_a^b f(x)dx[/itex] is defined by partitioning the interval from a to b into many small intervals, {xi}, choosing x* in each interval, forming the sum [itex]\Sigma f(x*)(x_{n+1}- x{n})[/itex] and taking the limit as the interval is partioned into more and more intervals, the Stieljes (or Riemann-Stieljes) integral, [itex]\int_a^b f(x)d\alpha(x)[/itex] does the same thing but uses the sum [itex]\Sigma f(x*)(\alpha(x_{n+1})- \alpha(x{n}))[/itex] where [itex]\alpha(x)[/itex] can be any increasing function. If [itex]\alpha(x)[/itex] is differentiable, that gives simply the Riemann integral [itex]\int_a^bf(x)\frac{d\alpha}{dx}dx[/itex] but if [itex]\alpha[/itex] is a step function, say the greatest integer function, then it gives the sum [itex]\Sigma_{n=a}^{b} f(n)[/itex].
    Last edited: Oct 30, 2005
  4. Oct 29, 2005 #3


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    It sounds like your initial sentence is saying:

    Let w be a function whose domain is the integers. I.E. w(x) is defined only when x is an integer.

    In that case, the integral

    [tex]\int_{-\infty}^{+\infty} w(x) f(x) \, dx[/tex]

    is not even defined.


    Let me pose another question, the one I think you meant to ask:

    If we're given a function g that is defined on the integers (and not always zero), can we find a function w such that:

    [tex]\int_{-\infty}^{+\infty} w(x) f(x) \, dx
    \sum_{n = -\infty}^{+\infty} g(n) f(n)

    is true for all functions f? The answer is no.

    However, if we let w be a distribution (or generalized function), then we can find such a w, and it can be given by the sum

    w(x) = \sum_{n=-\infty}^{+\infty} g(n) \delta(x - n)

    If g is also defined at every real number, the above expression is indeed the same as

    w(x) = g(x) \sum_{n=-\infty}^{+\infty} \delta(x - n)
    Last edited: Oct 29, 2005
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