Is there a theorem in Real analysis similar to Cauchy's theorem in Complex analysis?

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?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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

Solution 4:

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 $\Bbb R$