Suppose a meromorphic function $f(z)$ contains infinitely many simple poles $z_i$ inside the unit circle, which converge to the (only) non-isolated singularity at the origin. And I would like to know if there is a way to compute the contour integral $$ I = \oint_{|z| = 1} \frac{dz}{2\pi i z} f(z) \ . $$
A simple case: if $f(z)$ satisfies $f(zq) = f(z)$ with a $|q| < 1$, and let us assume that the simple poles $z_i$ sit within the infinitely many annuli $ |q|^{n + 1} < |z| < |q|^n$, and only finitely many will sit within each annuli. Such $f(z)$ can be rewritten in terms of a sum of the Weierstrass $\zeta$-function, $$ f(z) = C + \frac{1}{2\pi i} \sum_i R_i \zeta(\mathfrak{z} - \mathfrak{z}_i), \quad z_i =e^{2\pi i \mathfrak{z}_i} \ . $$ where the sum is over all poles $z_i$ within $|q| < |z| < 1$ with residues $R_i$. Doing so, the above integral can be performed, and the result can be written using the twisted Eisenstein series $E_1[\theta, \phi]$, $$ I = \sum_i R_i E_1 \left[\begin{matrix} - 1 \\ z_i/z_0 q^{-1/2}\end{matrix}\right] \ . $$ Note that this is not the result of the usual Cauchy theorem, since
all residues $R_i$ sum to zero;
there is no notion of "residue" for a non-isolated singularity $z = 0$.
Now, I wonder in more general case (without the $f(z) = f(zq)$) where we know all the residues $R_i$ of the poles $z_i$, can we say anything about the contour integral? Naively, I would imagine that all the $z_i$ and $R_i$ are enough to determine any property that the non-isolated singularity $z = 0$ may have, and therefore the $I$ as well.