Is there a theorem in Real Analysis similar to Cauchy's Theorem/Cauchy's Integral Formula from Complex Analysis?

If not, then why? What is it about the complex space that makes Cauchy's Theorem true?


I actually think the real way to think about it is that Cauchy's Integral formula is a weird complex form of the Mean Value Property for harmonic functions.

Harmonic functions are characterised by the Mean Value Property, i.e., if $U\subseteq \mathbb{R}^n$ is open, $x\in U$ and $f$ is harmonic on $U$, then for all $r>0$ such that $B(x,r)\subseteq U,$ we have

$$ f(x)=\frac{1}{dS(\partial B(x,r))}\int_{\partial B(x,r)} f(y) \, \textrm{d}S(y), $$ where $dS$ denotes the surface measure on $\partial B(x,r)$. In case $n=2,$ this will be a regular curve integral. The density appearing in the Cauchy formula needs to account for the fact that you're trying to compute a complex curve integral as opposed to a real one.

Note that holomorphic functions are, in particular, harmonic, so the above formula holds for them as a statement about real curve integrals.

  • $\begingroup$ Thank you! I like this way of thinking about Cauchy's Integral formula! $\endgroup$ – user Sep 15 at 14:16
  • 4
    $\begingroup$ I think this means that the real analysis version of Cauchy's Theorem would be: For all functions of the form $f(x)=ax+b$, we have that $(f(c)+f(d))/2=f((c+d)/2)$. Which is true, but not very deep. $\endgroup$ – Teepeemm Sep 16 at 3:23
  • 3
    $\begingroup$ Great answer. This highlights what an absurdly strong property complex differentiability is. $\endgroup$ – leftaroundabout Sep 16 at 8:50
  • 4
    $\begingroup$ @Teepeemm You can at least make it sound stronger by saying that Cauchy's theorem is "If $\tfrac{\partial^2 f}{(\partial x)^2} =0$, then $(f(c)+f(d))/2 = f((c+d)/2)$." $\endgroup$ – David E Speyer Sep 16 at 13:41

There is, but it's boring. Supposing $f$ is integrable on $[a,b]$,

$$ \int_a^b f(x) \,\mathrm{d}x + \int_b^a f(x) \,\mathrm{d}x = 0 \text{.} $$ In $\mathbb{C}$ it is possible for a closed path around a point to not pass through the point. In $\mathbb{R}$, the only way to close a path is to retrace your steps.

You could try the integrand $\frac{f(x)}{x-c}$ for $c \in (a,b)$, but unless $f$ happens to have a zero to cancel the pole at $c$, the resulting integral is improper at $c$ and the pair of integrals on either side of $c$ need not cancel (because their endpoints approaching $c$ approach independently). If they do, you recover the above.

A small variant is the Cauchy formula for repeated integration. \begin{align*} f^{(-n)}(x) &= \int_a^x \int_a^{\sigma_1} \cdots \int_a^{\sigma_{n-1}} f(\sigma_n) \,\mathrm{d}\sigma_n \cdots \,\mathrm{d}\sigma_2 \,\mathrm{d}\sigma_1 \\ &= \frac{1}{(n-1)!} \int_a^x (x-t)^{n-1}f(t) \,\mathrm{d}t \text{,} \end{align*} which has some similarities to the Cauchy differentiation formula, $$ f^{(n)}(a) = \frac{n!}{2\pi\mathrm{i}} \oint \frac{f(z)}{(z-a)^{n+1}} \,\mathrm{d}z \text{.} $$ Understanding these are less different than they appear is easier in the context of the Riemann-Liouville differintegral.


The essential difference between real and complex analysis at this level is that a complex function that's differentiable is analytic - it is the limit of its Taylor series in a disk. There are nonconstant infinitely differentiable real functions of a real variable all of whose derivatives at a point are $0$, hence not a power series in any disk.

See The distinction between infinitely differentiable function and real analytic function


The (possible) reasons are:

(1) If $f$ is analytic(differentiable) in $\Bbb C$ then so is its derivative $f^{'}$ which is not true in $\Bbb R$ , example consider the function $f(x)=x^2\sin (\frac{1}{x})$. (2) If you integrate an analytic function over a closed domain in $\Bbb C$ (Cauchy's Theorem) then its integral is $0$ which is not true in $\Bbb R$ , example consider $x^2 $ over $[-1,1]$

NOTE: For point (2) you could use the fact that every analytic function over a simply connected domain has an antiderivative which is not true in $\Bbb R$

  • $\begingroup$ Thank you for your answer. I am interested in the point you made about the integral of $x^2$ on $[-1,1]$. It is (computanionally) clear why the integral is non-zero, but is there perhaps a more insightful reason why this "closed path" in R is not zero- as this property seems to distinguish $\mathbb{R} $ - calculus from $\mathbb{C}$ - Calculus? $\endgroup$ – user Sep 15 at 14:12
  • 1
    $\begingroup$ @krtgdl; It is because every analytic function has an antiderivative $\endgroup$ – Learnmore Sep 15 at 14:16
  • $\begingroup$ Yes, I see! Thank you! $\endgroup$ – user Sep 15 at 14:21
  • 1
    $\begingroup$ @Learnmore: maybe I’m spacing out, but I don’t understand your final sentence; can you give an example of a real analytic function on a simply connected domain that does not have an antiderivative? $\endgroup$ – symplectomorphic Sep 17 at 2:55

Here is another perspective:

Let $U\subset \mathbb{C}$ be a open and bounded with piecewise smooth boundary $\Gamma$ and $f\in C^1(\bar U,\mathbb{C})$, i.e. $f$ is a complex valued function which is real differentiable in a neighbourhood of $\bar U$. Let's see how far we get proving the Cauchy integral formula for this function.

Let $z_0\in U$ and $\epsilon>0$ so small that the ball $B(z_0,\epsilon)\subset U$. Then Stokes' theorem implies (for more details see end of the post) that $$ \frac{1}{2 \pi i}\int_\Gamma \frac{f(z)}{z-z_0}\ d z = \frac{1}{\pi}\int_{U\backslash B(z_0,\epsilon)} \frac{1}{z-z_0}\frac{\partial}{\partial \bar z}f(z) d \lambda(z) + \frac{1}{2\pi \epsilon}\int_{\partial B(z_0,\epsilon)} f(z) d \sigma(z) \tag{$\star$}. $$ Here $\lambda$ is the $2$-dimensional Lebesgue measure, $\sigma$ is the surface measure on $\partial B$ and $\frac{\partial}{\partial \bar z}$ is the Wirtinger derivative, which encodes the Cauchy-Riemann equations. Since $f$ is continuous, the mean on the RHS converges to $f(z_0)$ as $\epsilon\rightarrow 0$, hence $$ \frac{1}{2 \pi i}\int_\Gamma \frac{f(z)}{z-z_0}\ d z = E(f,z_0) + f(z_0) \tag{$\star\star$}, $$ where $E(f,z_0):=\lim_{\epsilon\rightarrow 0}\frac{1}{\pi}\int_{U\backslash B(z_0,\epsilon)} \frac{1}{z-z_0}\frac{\partial}{\partial \bar z}f(z) d \lambda(z)$.

Note that $f$ is holomorphic if and only if $\frac{\partial}{\partial \bar z} f =0$. In this case $E(f,z_0)=0$ and $(\star\star)$ becomes the usual Cauchy integral theorem.

From this perspective it is clear where the reasoning breakes down in the absence of holomorphicity. The obstruction for the Cauchy integral theorem to hold is exactly $E(f,z_0)$, an error term which sort of measures the failure of $f$ to satisfy the Cauchy-Riemann equations.

Here is a proof of $(\star)$ using Stokes' theorem for complex differential forms (this is just for completeness - my argument should be clear without understanding the following calculations): Let $\Omega = U \backslash \bar B(z_0,\epsilon)$ and $\omega = \frac{f d z}{z-z_0}$. Then $d \omega = \frac{1}{z-z_0}\frac{\partial}{\partial \bar z}fd \bar z \wedge d z$ and Stokes' formula yields:

$$ \int_\Omega \frac{1}{z-z_0}\frac{\partial}{\partial \bar z}fd \bar z \wedge d z = \int_\Omega d \omega = \int_{\partial \Omega} \omega = \int_\Gamma \frac{f d z}{z-z_0} - \int_{\partial B(z_0,\epsilon)} \frac{f d z}{z-z_0}. $$ Standard computations show that $2 i d \lambda = d \bar z \wedge d z$ and that $\epsilon d z/(z-z_0)$ pulls back to $d \sigma$ on $\partial B(z_0,\epsilon)$, hence $(\star)$ follows.

Edit: 1) In a previous version I said that it wasn't clear whether the limit as $\epsilon\rightarrow 0$ exists and one would need something like a Lebesgue differentiation theorem. This was rubbish, as the corresponding result is trivial for continuous functions (as we assume $f$ to be). The virtue of the Lebesgue differentiation theorem lies in its validity for $L^1_{\mathrm{loc}}-$functions.

2) As argued above it is obvious that $E(f,z_0)=0$ if $f$ is holomorphic. On the other hand, any function that satisfies the Cauchy integral theorem, i.e. for which $E(f,z_0)=0$ for all $z_0\in U$ is holomorphic.

  • 1
    $\begingroup$ Very insightful answer! Thank you! $\endgroup$ – user Sep 16 at 9:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.