Cauchy's integral formula states that if the complex function $f(z)$ is analytic on a closed domain $D$ of the complex plane and $a$ is in the interior of $D$, then $$f(a) = \frac{1}{2 \pi i}\oint_{\partial D} \frac{f(z)}{z-a} dz.$$
This formula can be interpreted in two ways:
- If we "use the RHS to learn the LHS", then we can think of the formula as telling us that if we only know the values of $f(z)$ on the boundary curve $\partial D$, then that is enough information to fully determine the values of $f(z)$ on the interior of $D$.
- If we "use the LHS to learn the RHS", then we can think of the formula as telling us that if we want to calculate the contour integral $\oint_{\partial D} \frac{f(z)}{z-a} dz$, then we need only know the single value $f(a)$ on the interior. In this interpretation, it's perhaps more natural to think of the "primary function" being integrated as $g(z) := \frac{f(z)}{z-a}$, which is analytic on $D$ except for at a simple pole at $z = a$, and then to rewrite Cauchy's integral formula as $$\oint_{\partial D} g(z)\ dz = 2 \pi i \ \lim_{z \to a} \left[ g(z) (z-a) \right] = 2 \pi i\ \mathrm{Res}(g, a).$$
As pointed out at https://math.stackexchange.com/a/1654381/268333, it's clear how to generalize the second interpretation to the case of a function $g(z)$ that has any finite number of isolated singularities in $D$: via the residue theorem.
Can we also generalize the first interpretation to the case where $f(z)$ may not be analytic on all of $D$? In other words, if we only know the value of $f(z)$ on the boundary curve $\partial D$, then are there any looser requirements for $f$ (e.g. that it be analytic except at a finite number of isolated singularities in the interior of $D$) that are still enough to allow us to reconstruct it (or at least learn some information about it) on the interior of $D$? What happens if you try to apply Cauchy's integral formula to a function that is not analytic on all of $D$ but has singularities or branch cuts?
Suppose that $f(z)$ is analytic everywhere on $D$ except at a finite number of isolated singularities $z_k$. Then for any $a \in D \setminus \{z_k\}$, we can evaluate the contour integral $\frac{1}{2 \pi i} \oint_{\partial D} \frac{f(z)}{z-a}$ using the residue theorem: \begin{align} \frac{1}{2 \pi i} \oint_{\partial D} \frac{f(z)}{z-a} &= \mathrm{Res}\left( \frac{f(z)}{z-a}, a \right) + \sum_{k} \mathrm{Res}\left( \frac{f(z)}{z-a}, z_k \right) \\ &= f(a) - \sum_{k} \frac{\mathrm{Res}\left(f(z), z_k \right)}{a - z_k}. \end{align} So the original function $f(a)$ gets "corrected" by the function $-\sum_\limits{k} \frac{\mathrm{Res}\left(f, z_k \right)}{a - z_k}$.
If $a$ equals a singularity $z_n$ of $f(z)$, then we instead have \begin{align} \frac{1}{2 \pi i} \oint_{\partial D} \frac{f(z)}{z-z_n} &= \mathrm{Res}\left( \frac{f(z)}{z-z_n}, z_n \right) + \sum_{k \neq n} \mathrm{Res}\left( \frac{f(z)}{z-z_n}, z_k \right) \\ &= f_0(z_n) - \sum_{k \neq n} \frac{\mathrm{Res}\left(f(z), z_k \right)}{z_n - z_k}, \end{align} where $f_0(z_n)$ refers to the zeroth term of the Laurent expansion of $f$ about the point $z_n$. Surprisingly (to me), the integral actually converges at the singularities of the original function $f(z)$. If $z_n$ is a simple pole of $f(z)$, then $\frac{1}{2 \pi i} \oint_{\partial D} \frac{f(z)}{z-a} dz$ is actually analytic at $a = z_n$. But if $z_n$ is a higher-order pole or an essential singularity of $f(z)$, then $\frac{1}{2 \pi i} \oint_{\partial D} \frac{f(z)}{z-a} dz$ remains singular near $a = z_n$, even though it is well-definined and finite (but discontinuous) exactly at $a = z_n$. This simply reflects the fact that $f_0(z)$ is well-defined but discontinuous at a simple pole in a function $f(z)$.
It's interesting to think about what happens if $f(z)$ has a branch cut that's contained entirely within $D$. In that case, I guess that $\oint_{\partial D} \frac{f(z)}{z-a}$ will depend on the monodromy of the branch cut.