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:

  1. 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 $D$.
  2. 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?

  • 4
    $\begingroup$ I'm not sure it applies to your specific case, but there's the Cauchy–Pompeiu formula $\endgroup$
    – dvdgrgrtt
    Commented Jan 28 at 18:43
  • $\begingroup$ @dvdgrgrtt Yeah, I came across that one, but I think that it's generalizing Cauchy's integral formula in a different direction than I'm asking about - from analytic functions to smooth functions, rather than by allowing multiple singularities within the domain $D$. $\endgroup$
    – tparker
    Commented Jan 28 at 19:27
  • 1
    $\begingroup$ If I understand you right, the answer is no in the most literal sense. If $f(z)=\frac{1}{2\pi i}\int_{\partial D}\frac{f(w)}{w-z}\,\mathrm{d}w$ for all $z\in\mathrm{int}(D)$ then $f$ is necessarily analytic, assuming a nice simply connected open domain. We cannot have the integral formula without the analyticity. But maybe you want to know if there is a slightly tweaked integral formula that works. I personally doubt it due to differentiation under the integral sign phenomena that would probably allow analyticity to be deduced, exactly as it is with Cauchy's formula. $\endgroup$
    – FShrike
    Commented Jan 29 at 21:23
  • $\begingroup$ @FShrike Does the integral necessarily converge everywhere on $D$? If so, is there some simple characterization of the function that you do get from applying the formula? I'm trying to understand what happens if you just choose some random function $f(z)$ with poles, etc., choose a closed curve that surrounds multiple poles, and then apply the Cauchy integral formula at every point in the interior. What function do you get? $\endgroup$
    – tparker
    Commented Jan 30 at 1:57
  • $\begingroup$ I don’t see what you mean. In the statement I gave, it is implicit that the integral is convergent (else the equality is senseless) and the function that we get from Cauchy’s formula is just $f$ - by the hypothesis of my statement! If we use some very arbitrary badly behaved function $g$ and then used the Cauchy formula to define a new function $f$, it is guaranteed that $f\neq g$, and then we wouldn’t have this principle of interior values being determined by boundary values. So what I’m saying is, a good answer to your question will necessarily have to have a different formula than Cauchy’s $\endgroup$
    – FShrike
    Commented Jan 30 at 2:44

1 Answer 1


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.

  • 1
    $\begingroup$ for example for $f(z)=\sqrt {z^2-z}$ and any disc (or more generally Joerdan domain) that contains $0,1$ the integral on the boundary is zero for all $a$ inside the disc $\endgroup$
    – Conrad
    Commented Jan 30 at 16:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .