There is a much more elaborate formulation of the Cauchy integral formula which involves cycles and an
additional assumption that the function of interest is analytic on an open set which contains the curves
over which the integrals are taken. Thus we abandon the generality which allows the weaker assumption
that the function is continuous on the curves of interest and is only known to be analytic on the
inside.
The reason for this is that a key part of the argument uses Liouville’s theorem. This, and the notion of
winding number are what makes possible the Cauchy integral theorem for a cycle. Recall that the winding
number is defined as
∫
n (γ,z) ≡ -1- --1--dw
2πi γ w− z
where γ is a parametrization of an oriented curve γ^{∗} and z
∕∈
γ^{∗}. This makes perfect sense for any bounded
variation curve γ^{∗}. Note that it is routine to verify that z → n
(γ,z)
is continuous on the complement of
γ^{∗}.
Consider a situation typified by the following picture in which Ω is the open set between the dotted
curves and γ_{j} are closed rectifiable curves in Ω.
PICT
The open set is between the dotted lines and also excludes the points z_{2},z_{3},z_{4}. Note how if you pick
any z
∈∕
Ω, including the z_{k}, then ∑_{k=1}^{4}n
(γ ,z)
k
= 0. This open set is not simply connected as described in
Definition 15.7.3. Its complement relative to the extended complex plane is not connected. Note how this is
manifested by the points left out.
Definition 16.4.1Let
{γk}
_{k=1}^{n}be continuous parametrizations for curves having boundedvariation. Then
{γk}
_{k=1}^{n}is called acycle if whenever, z
∕∈
∪_{k=1}^{n}γ_{k}^{∗},∑_{k=1}^{n}n
(γk,z)
is an integer.Note that, unlike the above picture, there is no reason to believe the γ_{k}^{∗}are closed curves.
Now the following is the general Cauchy integral formula. In the theorem, an assumption is made that
the sum of the winding numbers is an integer. As shown earlier, this means that somehow, you are
probably considering closed curves.
Theorem 16.4.2Let Ω be an open subset of the plane (not necessarily simply connected) and letf : Ω → X be analytic. If γ_{k} :
[a ,b ]
k k
→ Ω,k = 1,
⋅⋅⋅
,m are continuous curves having bounded variationsuch that for all z
∕∈
∪_{k=1}^{m}γ_{k}^{∗},
m∑
n(γk,z) equals an integer
k=1
and for all z
∕∈
Ω,
∑m
n (γk,z) = 0.
k=1
Then for all z ∈ Ω ∖∪_{k=1}^{m}γ_{k}^{∗},
∑m ∑m 1 ∫ f (w)
f (z) n (γk,z) = 2πi w-−-zdw.
k=1 k=1 γk
Proof: Let ϕ be defined on Ω × Ω by
{
f(ww)−−fz(z) if w ⁄= z
ϕ(z,w) ≡ f′(z) if w = z .
Then ϕ is analytic as a function of both z and w and is continuous in Ω × Ω. This follows from the
argument given in Theorem 15.6.1, resulting from the fact that if a function has one derivative on an open
set, then it has them all.
Define
1 ∑m ∫
h (z) ≡ --- ϕ (z,w )dw.
2πik=1 γk
Is h analytic on Ω? To show this is the case, verify
∫
h(z)dz = 0
∂T
for every triangle, T, contained in Ω and apply Corollary 15.5.6. This is an application of the Fubini
theorem of Theorem 15.3.10. By Theorem 15.3.10,
because ϕ is analytic. By Corollary 15.5.6, h is analytic on Ω as claimed.
Now let H denote the set,
{ ∑m }
H ≡ z ∈ ℂ∖ ∪mk=1 γ∗k : n(γk,z) = 0
k=1
{ ∑m }
= z ∈ ℂ∖ ∪mk=1 γ∗k : n(γk,z) ∈ (− 1∕2,1∕2)
k=1
the second equality holding because it is given that the sum of these is integer valued. Thus H
is an open set because z →∑_{k=1}^{m}n
(γk,z)
is continuous. This is obvious from the formula
for n
(γk,z)
. Also, Ω ∪ H = ℂ because by assumption, Ω^{C}⊆ H. Extend h
(z)
to all of ℂ as
follows:
{ ∑m ∫
h(z) ≡ 21πi∫ k=1 γk ϕ (z,w )dw if z ∈ Ω
g(z) ≡ 21πi ∑mk=1 γ fw(−wz)dw if z ∈ H . (16.10)
k
(16.10)
Why is g
(z)
well defined? On Ω ∩ H, z
∈∕
∪_{k=1}^{m}γ_{k}^{∗} and so
∑m ∫ ∑m ∫
g (z) = -1- ϕ (z,w )dw = -1- f (w)−-f-(z)dw
2πik=1 γk 2πik=1 γk w − z
1 ∑m ∫ f (w) 1 m∑ ∫ f (z)
= 2πi w-−-zdw − 2πi w-−-zdw
k=1 γk k=1 γk
-1-∑m ∫ -f (w)
= 2πi γ w − zdw
k=1 k
because z ∈ H. This shows g
(z)
is well defined. Also, g is analytic on Ω because it equals h there. It is
routine to verify that g is analytic on H also because of the second line of 16.10. (See discussion at the end
if this is not clear. )
Therefore, g is an entire function, meaning that it is analytic on all of ℂ.
Now note that ∑_{k=1}^{m}n
(γk,z)
= 0 for all z contained in the unbounded component of ℂ∖∪_{k=1}^{m}γ_{k}^{∗}
which component contains B
(0,r)
^{C} for r large enough. It follows that for
|z|
> r, it must be the case that
z ∈ H and so for such z, the bottom description of g
and so g is bounded and analytic on all of ℂ. By Liouville’s theorem, g is a constant. Hence, from the above
equation, the constant can only equal zero.
For z ∈ Ω ∖∪_{k=1}^{m}γ_{k}^{∗}, since it was just shown that h
(z)
= g
(z)
= 0 on Ω
∑m ∫ ∑m ∫
0 = h (z) =-1- ϕ (z,w )dw = -1- f (w)−-f-(z)dw =
2πik=1 γk 2πik=1 γk w− z
m ∫ m
-1-∑ f (w-)dw − f (z)∑ n(γ ,z).■
2πik=1 γk w − z k=1 k
In case it is not obvious why g is analytic on H, use the formula. It reduces to showing
that
∑m ∫ f (w)
z → γ w− z dw
k=1 k
is analytic. You can show this by taking a limit of a difference quotient and argue that the
limit can be taken inside the integral. Taking a difference quotient and simplifying a little, one
obtains