Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The fundamental theorem of contour integration says if one has a function and its antiderivative, and integrates the function over a closed loop the result is zero.

Cauchy's theorem (Goursat's Version) says the integral of a function in a holomorphic domain in a closed loop is zero.

Cauchy's theorem is apparently much stronger, the proof is certainly more intricate. Can someone please give trivial and nontrivial examples of integrals that Cauchy's Theorem applies to that FTCI does not?

Having proven Cauchy's Theorem, is the FTCI useful for anything, anymore?

share|improve this question

2 Answers 2

I would state the "fundamental theorem of contour integration" as follows.

Fundamental Theorem of Contour Integration: Let $U\subseteq\mathbb{C}$ be an open set, and let $f\colon U\to \mathbb{C}$ be a continuous function. Then the following conditions are equivalent:

  1. $f$ has an antiderivative on $U$.
  2. For every piecewise differentiable closed curve $\gamma$ in $U$, the integral $\int_\gamma f\,dz = 0$.

Notice that the theorem does not say that all holomorphic functions necessarily satisfy the two equivalent conditions. In fact, there are situations where they won't. For instance, if $U = \mathbb{C}\smallsetminus\{0\}$ and $f(z) = 1/z$, then $f$ does not have an antiderivative on $U$, even though it is holomorphic. The Cauchy-Goursat theorem tells us one situation in which holomorphic functions are guaranteed to satisfy conditions (1) and (2).

Cauchy-Goursat: Let $U\subseteq\mathbb{C}$ be a simply connected open set. Then all holomorphic functions $f\colon U\to \mathbb{C}$ satisfy the equivalent conditions of the previous theorem.

share|improve this answer

The first theorem says the following: If $f:\ \Omega\to{\mathbb C}$ is the derivative of some $F:\ \Omega\to{\mathbb C}$ in a domain $\Omega\subset{\mathbb C}$ then $\int_\gamma f(z)\ dz=0$ for all closed curves in $\Omega$. This is a "complex version" of the well known fact from real multivariable calculus that $\int_\gamma \nabla F\cdot d{\bf x}=0$ for all differentiable $F:\ \Omega\to{\mathbb R}$ and all closed curves $\gamma\subset \Omega$.

Cauchy's theorem is much deeper: It says that a function fulfilling at each point $z_0\in \Omega$ the purely local condition $$\exists \ \lim_{z\to z_0}{f(z)-f(z_0)\over z-z_0}=:f'(z_0)\in{\mathbb C} \ ,$$ has automatically all integrals along closed curves $\gamma\subset \Omega$, however long, equal to zero (if $\Omega$ has no holes).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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