# Sum and integral (1 Viewer)

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#### 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...

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#### 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, {xi}, 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 $\Sigma_{n=a}^{b} f(n)$.

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#### Hurkyl

Staff Emeritus
Gold Member
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.

---------------------------------------------------

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:

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

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

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

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