(Recall that a curve is
homotopic to a constant curve if there exists a smooth
homotopy (within ) from the curve to the constant curve. Intuitively, this means that one can shrink the curve into a point without exiting the space.) The first version is a special case of this because on a
simply connected set, every closed curve is
homotopic to a constant curve.
Main example
In both cases, it is important to remember that the curve does not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve:
which traces out the unit circle. Here the following integral:
is nonzero. The Cauchy integral theorem does not apply here since is not defined at . Intuitively, surrounds a "hole" in the domain of , so cannot be shrunk to a point without exiting the space. Thus, the theorem does not apply.
Discussion
As
Édouard Goursat showed, Cauchy's integral theorem can be proven assuming only that the complex derivative exists everywhere in . This is significant because one can then prove
Cauchy's integral formula for these functions, and from that deduce these functions are
infinitely differentiable.
The condition that be
simply connected means that has no "holes" or, in
homotopy terms, that the
fundamental group of is trivial; for instance, every open disk , for , qualifies. The condition is crucial; consider
which traces out the unit circle, and then the path integral
is nonzero; the Cauchy integral theorem does not apply here since is not defined (and is certainly not holomorphic) at .
The Cauchy integral theorem is valid with a weaker hypothesis than given above, e.g. given , a simply connected open subset of , we can weaken the assumptions to being holomorphic on and continuous on
and a
rectifiablesimple loop in .[1]
If one assumes that the partial derivatives of a holomorphic function are continuous, the Cauchy integral theorem can be proven as a direct consequence of
Green's theorem and the fact that the real and imaginary parts of must satisfy the
Cauchy–Riemann equations in the region bounded by , and moreover in the open neighborhood U of this region. Cauchy provided this proof, but it was later proven by Goursat without requiring techniques from vector calculus, or the continuity of partial derivatives.
We can break the integrand , as well as the differential into their real and imaginary components:
In this case we have
By
Green's theorem, we may then replace the integrals around the closed contour with an area integral throughout the domain that is enclosed by as follows:
But as the real and imaginary parts of a function holomorphic in the domain , and must satisfy the
Cauchy–Riemann equations there:
We therefore find that both integrands (and hence their integrals) are zero