Is the Poisson Sum Formula Equivalent to the Integral of a Function?

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

[tex]\sum_{n=-\infty}^{\infty}f(n)= \int_{-\infty}^{\infty}dx f(x) \omega (x)[/tex]

with [tex]\omega (x) = \sum_{n=-\infty}^{\infty}e^{2i \pi nx}[/tex]

Then my question is if we would have that:

[tex]\sum_{n=-\infty}^{\infty} \frac{ f(n)}{ \omega (n)} = \int_{-\infty}^{\infty} dx f(x)[/tex] ??
 
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You should stop and ask if:

[tex]\omega (x) = \sum_{n=-\infty}^{\infty}e^{2i \pi nx}[/tex]

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