I encounter contour integrals of the following form, $$ \oint_{|z| = 1} f_q(z) \frac{ dz }{ 2\pi i}\ , \qquad |q| < 1 $$


  1. the meromorphic function $f_q(z)$ contains a lot of simple poles of the form $z_{in} = a_{i} q^{n}, n = 0, 1, 2, ...$ (with $ |a_i| < 1$) inside the unit circle: so there is a non-isolated singularity at the origin.
  2. Inside each annulus $|q|^{n} > |z| > |q|^{n + 1}$, all residues from the poles within cancel each other. Naively, the integral receives contribution only from the non-isolated singularity at the origin.

One such example is $$ \oint \frac{dz}{2\pi i z} \left(\frac{1}{z} - z\right)^2 \frac{(z^2 q;q)^2(z^{-2}q;q)^2}{\prod_{\pm}(z^2 b^\pm q^{1/2};q)(z^{-2} b^\pm q^{1/2};q)} $$ where $(z;q)$ denotes the usual $q$-Pochhammer symbol.

In physics application, we usually simply expand the integrand in $q$-series $f_q(z) = \sum_{k} f^{(k)}(z)q^k$ and then observe that all coefficients of expansion $f^{(k)}(z)$ only have poles at the origin $z = 0$, and one simply extracts the residue there and gets a $q$-series as the final answer.

I wonder if one can properly treat the integral without first performing the $q$-expansion? Are we allowed to talk about "residue" at a non-isolated singularity?


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