Expansion of a function f(x) with poles

[zetafunction]

If a function f(x) have SIMPLE POLES , could in principle this f(x) be expanded into

$$f(x)= \sum_{r}a_{r} (x-r)^{-1}$$

where 'r' are the poles on the complex plane of the function

Another question, would it be possible to relate using the Euler-Mac Laurin resummation, a series of the form $$\sum_{n=0}^{\infty}(-1)^{n}f(n)$$ to the integral

$$\int_{0}^{\infty}f(x)dx$$

 

[g_edgar]

If a function f(x) have SIMPLE POLES , could in principle this f(x) be expanded into

$$f(x)= \sum_{r}a_{r} (x-r)^{-1}$$

where 'r' are the poles on the complex plane of the function

No. But (assuming a finite sum, or a convergent sum) if you subtract this from f(x), you get something with no poles, perhaps reducing to a simpler problem.

 

[Charles49]

The answer to the first question is YES: see the Mittag-Leffler theorem.

I would substitute $$x=\sin\left(\cos(n\pi) x\right)$$ in the integral. Differentiate several times under the integral and add the terms. Substitute $$n=1$$ in the final step. That's how I would go about and see if something happens. See if it is related to the Fourier series of some other function.