# Does Cauchy's integral formula generalize to non-analytic functions?

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?

• I'm not sure it applies to your specific case, but there's the Cauchy–Pompeiu formula Commented Jan 28 at 18:43
• @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 $D$. Commented Jan 28 at 19:27
• 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. Commented Jan 29 at 21:23
• @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? Commented Jan 30 at 1:57
• 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 Commented Jan 30 at 2:44

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.
• 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 Commented Jan 30 at 16:35