I was teaching complex analysis today, and wondered about this question.

Let f(z) = \csc \frac{1}{z}.
\oint_{|z|=1}f(z) dz

I formulated a conjecture, which agrees with my intuition.

Let f be meromorphic in \mathbb{C} and it has finitely many poles a_1,...,a_n in a contour C then:
\oint_C f = 2\pi i \sum_{k=1}^n \mbox{res}(f,a_k).
Now suppose that f has infinitely (but countably many) it seems that the correct result would be:
\oint_C f = 2\pi i \sum_{k=1}^{\infty} \mbox{res}(f,a_k).
However, the expression on the right has two problems.
1)The sequence b_k=\mbox{res}(f,a_k) might now converge.
2)Even if \{ b_k \} convergence it needs to be absolutely convergent because otherwise it is not well-defined by the Riemann rearragnement theorem.

Here is the conjecture.

Let f be meromorphic on an open set containint the interior of the contour C with countably infinite many poles at a_k. If the sequence b_k = \mbox{res}(f,a_k) is absolutely convergent with limit L then the round integral over C of f is 2\pi i \cdot L.