3.1 The Cauchy–Goursat Theorem
It is important to know the differential 2-forms even for a single variable complex function. Consider and . We can then define their corresponding differentials:
The antisymmetric properties of differential forms still hold in complex space. By taking the wedge product of the two basis complex differential forms, we get
Analogous to the real case, a 0-form is defined as a scalar-valued function in the form , a 1-form in the form , and a 2-form as . The exterior differential operator for this one-dimensional case is defined as , where and . Occasionally, one will informally use and as an abbreviation for and respectively.
Theorem 3.1.1 (Lusin Area Theorem).
For a region and univalent, the area of the image is equal to
Proof.
We aim to find
By the properties above,
as desired.
Remark.
The Jacobian determinant of with respect to , for a holomorphic function is equal to
by (3.1.6).
Theorem 3.1.2 (Green's Theorem, Complex Form).
Let be bounded with a piecewise smooth boundary . For two scalar functions and satisfying , define the 1-form . Then,
Proof.
For real-valued functions , let
Then,
Each of are real-valued functions that can be represented with a domain of . By definition,
From (3.1.2), we can apply Theorem 1.2.8. For the real component of , we obtain
and for the imaginary component,
and the integrands on the right side both match those of (3.1.3).
The theorem above is only a specific case of the Stokes–Cartan Theorem (Theorem 1.2.11). However, it proves the validity of the treatment of the and operators, and the generalization to forms with basis and .
Theorem 3.1.3 (Cauchy–Pompeiu).
Let be bounded with a piecewise boundary . Let . Then ,
Proof.
Since , such that . Consider the complex differential form
with a singularity at . Consider the region . Since , by applying Green’s Theorem (Theorem 3.1.2),
By properties of , the expression is equal to
The first term in the integrand vanishes as it contains . The second term can be simplified using the fact that , leading to
The rightmost term in (3.1.4) can be parameterized with , . Then,
Because , by Proposition 1.2.2, is Lipschitz continuous on , and such that , . On , we get that . Therefore,
which approaches 0 as . Taking this limit, we obtain
We then aim to prove that
Notice that since , by Theorem 1.2.13, such that , . Then,
By a change of variables to a polar coordinate system centered at , we obtain
and by expansion of the wedge product,
Then from rearranging (3.1.5), we obtain:
Corollary 3.1.1.
Let be a continuously differentiable, compactly supported function. Then
for all where .
Proof.
Choose such that . By the Cauchy–Pompeiu Theorem (Theorem 3.1.3), we have
Then the proof is complete given that vanishes on and by letting .
In complex analysis, when integrating over a region that contains a singularity, it is common to exclude a small disk of radius around the singularity, perform the integration over the punctured region, and then take the limit as . As in the proof above, the displayed estimate for the integral over the removed disk is still necessary in confirmation, although it is typically tacitly elided.
From the above result, we can directly obtain the following theorem:
Theorem 3.1.4 (Cauchy's Integral Formula).
Let be an open region with a piecewise boundary , and let be holomorphic on . Then for all ,
Proof.
By (3.1.5), for , . Applying the Cauchy–Pompeiu Theorem (Theorem 3.1.3), the area integral vanishes, and (3.1.7) consequently follows.
Theorem 3.1.5 (Cauchy's Integral Theorem).
Let be an open region with piecewise boundary . For a function holomorphic over ,
Proof.
Let . Applying Theorem 3.1.4 on with , we obtain
Alternatively, we can use Green’s Theorem (Theorem 3.1.2) with :
Theorem 3.1.6.
For a compactly supported function , a solution satisfying to the non-homogeneous Cauchy–Riemann equation
is
Proof.
Split into and . For all , the integral
is continuous. Since is compactly supported over and continuous, by Theorem 1.2.13, it is bounded. Then the limit
Therefore, (3.1.8) is continuous. A trivial substitution can be used to rewrite
Then,
For a fixed , the value of
tends to as . Because has compact support and is , by Proposition 1.2.2, it is Lipschitz continuous for a constant . Let and let . Then,
and specifically, when ,
As shown above, the integrand is uniformly bounded by , which has a convergent integral of , the limit may commute with the integral in (3.1.9). Let . From the real axis,
From the imaginary axis,
Since and has Lipschitz constant , (3.1.10), (3.1.11) are both continuous (by the same argument for the continuity of ). Thus, . It follows from the two equations that
By Corollary 3.1.1,
Remark.
In the first part, we established that a function with compact support satisfies
If , then the first order derivatives of can be written in the same form ((3.1.10), (3.1.11)) since and are also compactly supported. Then they too are continuous functions, and .
Then using the same argument, In general, for , the same process can be used recursively to find that as well.
If the support of is the union of infinitely many or finitely many disjoint compact sets, then the integral in (3.1.8) can be split into a sum of integrals over each compact set, and the same argument applies to each term.
When Cauchy formalized Theorem 3.1.4, Theorem 3.1.5, he included the necessary condition that . It was later shown that all such holomorphic functions had holomorphic derivatives, and this condition was thus later dropped by Goursat:
Lemma 3.1.1.
Let be a continuous function defined for a region . Let be a rectifiable piecewise smooth curve. Then , there exists a polygonal chain inscribing (each vertex lies on ) where
Proof.
Because , there is a compact set enclosing and is the closure of some open set. By Theorem 1.2.15, , such that satisfying , . Partition into curves between points such that the length of is less than . For all , let denote the straight line segment connecting and . The length of is less than as well. Then let
Over the partition formed with , the integral
can be approximated with the Riemann sum
where
Then the sum above can be written as
and it follows that
and
where is the length of and is the length of . Then,
Lemma 3.1.2 (Goursat).
Given a holomorphic function on a simply connected region , for any piecewise closed curve ,
Proof.
By Lemma 3.1.1, , there is a polygonal chain where
The statement we aim to prove is equivalent to proving that
Since is a closed polygonal chain, we can triangulate the interior. For example, consider Figure 1, where
Thus, if the integral over every triangle in vanishes, then (3.1.12) follows. Consider a triangle in with boundary . Then define to be
We can quadrisect the triangle bounded by into four triangles with boundaries as in Figure 2. Then one of , , , or (denote this to be ) satisfy
and recursively, choose
Let denote the perimeter of . Then, the perimeters of respectively are . As , shrinks to a single point . Then, , .
By the definition of holomorphy, , such that ,
and such that , . By Theorem 3.1.5, since the functions and are both entire,
Then
Because the distance between any two points in the interior of a triangle is always less than its perimeter, using the triangle inequality for complex integrals,
Comparing the above equation with (3.1.15),
Since is rectifiable, is finite, and letting , we find that . Then, for every triangle in , the integral vanishes, and (3.1.14), (3.1.13) follow.
Theorem 3.1.7 (Cauchy–Goursat).
Let be an open region bounded with boundary . Let be a holomorphic function continuous on . Then,
Proof.
Since and is not necessarily holomorphic over , we cannot directly apply Lemma 3.1.2.
First assume has the shape of in Figure 3. That is, consists of , for , and two rectifiable curves and such that , .
For some , , construct a new curve to be the boundary of the region bounded by , , , and such that remains simple. By Lemma 3.1.2,
By Theorem 1.2.15, is uniformly continuous over , and therefore , we can choose so that , is satisfied. Letting with and fixing , we get that
and consequently,
Under the same limit, we get
By the continuity of over a compact set,
Then letting , for the same reason as (3.1.18), (3.1.16), (3.1.17) yield
We are left to show the subsequent limits of the results from (3.1.18). For the left integral, let and .
Then,
Similarly,
The difference between the two is then equal to
The first term vanishes by uniform continuity, through the same argument used for , and the remaining four integrals all tend to because they are taken over degenerating intervals. As , and because . Therefore,
and through similar logic,
Therefore,
Any open region with a simple closed boundary can be broken up into smaller regions with the same form as with finitely many auxiliary lines. Then the conclusion follows.
Remark.
The theorem is also valid for any multiply connected region (and its boundary will consist of multiple curves) as a multiply connected region is equal to the union of several simply connected regions with vertical auxiliary lines between.
Additionally, if is simply connected and is holomorphic on , then for any two points , the integral
is well-defined and independent of the path taken from to . In this sense, holomorphic functions behave analogously to potential fields.
Theorem 3.1.8 (Cauchy–Goursat).
Let be an open region bounded with a simple closed boundary , and let be a holomorphic function continuous on . Then for all ,
Proof.
By the Cauchy–Goursat Theorem (Theorem 3.1.7),
From rearrangement,
Since , as ,
By rearrangement,
Remark.
In the proof of Theorem 3.1.3, we used Lipschitz continuity for a smooth function, which was a stronger condition than necessary. The true necessity of smoothness was to be able to apply Green’s Theorem (Theorem 3.1.2).
This profound theorem is extremely important and helpful in complex integration and essential in the evaluation of integrals, as demonstrated below.
Example 3.1.1.
Evaluate the integral , where .
Proof.
Since , where , the integrand has singularities at every -th root of unity. Then the integral is equal to:
where are the coefficients of the partial fraction decomposition. By the Cauchy–Goursat Formula (Theorem 3.1.8), (3.1.20) becomes:
Observe that since . Therefore,
We have also already seen the utility of parameterization via a polar transformation. Many useful identities in classical calculus can also be derived from concepts in its generalization:
Example 3.1.2.
Prove that ,
Proof.
Consider the integral
Letting , we get . Alternatively, we can expand the integrand and get
When , the integrand is holomorphic. The integral is then equal to
since all the higher order terms vanish:
Therefore,
From simple cancellation, we then have
Example 3.1.3 (Cauchy–Goursat Formula on the Exterior).
Let be a simple closed curve, and suppose that is holomorphic and continuous on , where and respectively denote the interior and exterior as in Theorem 1.2.5.
-
If has a removable singularity at , or if exists and is finite, then ,
-
If encloses the origin, then ,
Proof.
-
By the compactness of , it can be completely contained within a sufficiently large disk centered at the origin (). Then by applying Theorem 3.1.8 or Theorem 3.1.7 on the set , we get that
By letting and letting , we get that
By the continuity of on , it attains its maximum . For sufficiently large , attains a positive minimum. Then the integrand is uniformly bounded in and , and hence the order of the limit and the integral may be exchanged. Hence,
as expected.
-
Under the partial fraction decomposition of (3.1.21), we get that
when . We will analyze the first integral as . By the triangle and reverse triangle inequalities,
By substituting the result into (3.1.22), and letting , we get that
as desired.