So a friend of mine and I were talking about the Residue Theorem and we were wondering if it is actually possible for a closed, rectifiable curve to contain a non-constant sequence of singularities and if in that case the Residue Theorem is applicable.
Let's add some symbolisms: Let $\Omega$ be an open subset of $\mathbb{C}$ and $f:\Omega\setminus\{z_n: n\in\mathbb{N}\}\to\mathbb{C}$ be a holomorphic function defined on $\Omega$ except of the different points $(z_n)$ which are $f$'s singularities. Let $\gamma$ be a closed, rectifiable, simple curve in $\Omega$, such that $z_n \not\in \hat{\gamma}$ for each $n\in\mathbb{N}$, but all $z_n$ are inside $\gamma$, or if you'd like, $\chi(\gamma,z_n)=1$ .
Since the area inside $\gamma$ is a compact set, $(z_n)$ will have a converging sub-sequence, let's call it $(z_{n_k})$, with $z_{n_k}\to z_0$. There is no real interest if the singularities are removable from an index on, so let's suppose that they're all poles or essential singularities. The questions are the following:
1) Is it possible for the function $f$ to be holomorphic on the limit point $z_0$? Now if one says "but $z_0$ belongs to $\Omega$, that changes nothing, cause the question would become "can there be such a function $f$?"
2) Is it possible for the limit point $z_0$ to be in the inside of $\gamma$ and not on $\hat{\gamma}$?
3) If (1) is possible, would the Residue Theorem take the form of "Integral = Series of residues" ?
We know that $z_0$ can't be a singularity, since $f$ can't be defined on any ring $\Delta(z_0,0,r)=\{z: 0<|z-z_0|<r\}$.
The point $z_0$ is a singularity, since $f$ is not holomorphic on any neighborhood of $z_0$. It is a limit point of singularities, and as such, is not an isolated singularity. The classification of singularities as removable, poles or essential makes sense only for isolated singularities.
The standard example of such a situation is $f(z)=\Bigr(\sin\dfrac1z\Bigr)^{-1}$. The points $z_k=1/(k\,\pi)$, $k\in\Bbb Z$, $k\ne0$ are poles, and $0$ is a limit point of poles. There is no Laurent expansion around $0$, and hence no residue at $0$.