Poisson sum formula

1. Aug 25, 2006

lokofer

If we have (Poisson sum formula) in the form:

$$\sum_{n=-\infty}^{\infty}f(n)= \int_{-\infty}^{\infty}dx f(x) \omega (x)$$

with $$\omega (x) = \sum_{n=-\infty}^{\infty}e^{2i \pi nx}$$

Then my question is if we would have that:

$$\sum_{n=-\infty}^{\infty} \frac{ f(n)}{ \omega (n)} = \int_{-\infty}^{\infty} dx f(x)$$ ??

2. Aug 25, 2006

shmoe

You should stop and ask if:

$$\omega (x) = \sum_{n=-\infty}^{\infty}e^{2i \pi nx}$$

makes any sense at all. You have an infinite sum and the terms aren't tending to zero.