TITLE: Is there a theorem in Real analysis similar to Cauchy's theorem in Complex analysis?
QUESTION [20 upvotes]: 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?

REPLY [2 votes]: 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.