Does Integrating a Discrete Function Over All Space Equal Its Infinite Sum?

[eljose]

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:

\int_{-\infty}^{\infty}dxw(x)f(x)

would be equal to the sum of the serie:

\sum_{n=-\infty}^{\infty}w(n)f(n)


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

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



thanks...

 

[HallsofIvy]

Yes, if you mean the Stieljes integral.

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

 

[Hurkyl]

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

\int_{-\infty}^{+\infty} w(x) f(x) \, dx

is not even defined.

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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:

\int_{-\infty}^{+\infty} w(x) f(x) \, dx<br /> =<br /> \sum_{n = -\infty}^{+\infty} g(n) f(n)<br />

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

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

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

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