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The integral of a continuous derivative over a cycle is zero

Statement

Let Γ=∑k<rmkγk be a complex chain which is a cycle, let V⊆C be open with Γ∗⊆V, and let F be a primitive on V of a continuous f, so that F′=f is continuous (A primitive of a complex function on an open set). Then

∫Γf(z) dz=0.

The hypothesis used is that the boundary function of Γ vanishes, which is weaker than requiring every γk to be closed.

Facts & Assumptions

Given: A cycle Γ=∑k<rmkγk with γk:[ak,bk]→C, an open V⊇Γ∗, and a primitive F on V of a continuous f with F′=f continuous.

[L1]

A complex chain is a finite list of pairs (mk,γk), its trace is the union of the γk∗ with mk≠0, its boundary is ∂Γ(q)=∑{mk:γk(bk)=q}−∑{mk:γk(ak)=q}, and it is a cycle when that function vanishes identically (Complex chains, their traces, and cycles).

[L2]

∫Γf dz=∑k<r, mk≠0mk∫γkf dz (Integration over a complex chain and the index of a chain).

[L3]

If F is a primitive of a continuous f on an open set containing the trace of a rectifiable contour γ:[a,b]→C and F′=f is continuous, then ∫γf(z) dz=F(γ(b))−F(γ(a)) (The line integral of a continuous function admitting a primitive is that primitive's endpoint increment along every rectifiable path).

[L5]

For disjoint finite index sets S,T, ∑u∈S∪Tau=∑s∈Sas+∑t∈Tat (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule); sums over finite index sets are well posed and the empty sum is 0 (A finite sum in a commutative monoid indexed by an arbitrary finite set, The cardinality ∣A∣ of a finite set).

[L6]

A primitive of f on V is a holomorphic F with F′=f on V (A primitive of a complex function on an open set).

Proof

technique · direct
1.1givenL1L5L6

Write K={k<r:mk≠0} and Q={γk(ak):k∈K}∪{γk(bk):k∈K}, a finite subset of Γ∗⊆V by [L1]; so F is defined at every point of Q.

1.2givenL1L3L6

For every k∈K the trace γk∗ lies in the open set V on which F is a primitive of the continuous f with continuous F′, so [L3] gives ∫γkf dz=F(γk(bk))−F(γk(ak)).

2.1step 1.2L2L4

By [L2] and step 1.2, ∫Γf dz=∑k∈KmkF(γk(bk))−∑k∈KmkF(γk(ak)), using [L4] to split the sum.

3.1step 1.1step 2.1L1L4L5

The index set K is the disjoint union over q∈Q of {k∈K:γk(bk)=q}, so [L5] and [L4] give ∑k∈KmkF(γk(bk))=∑q∈QF(q)∑{mk:k∈K, γk(bk)=q}, and likewise with ak in place of bk; a term with mk=0 contributes 0 to the boundary sums of [L1], so subtracting gives ∫Γf dz=∑q∈QF(q) ∂Γ(q).

4.1step 3.1L1L5∎

Every ∂Γ(q) vanishes because Γ is a cycle, so the sum of step 3.1 is 0, whence ∫Γf dz=0; the same conclusion holds for the empty cycle, whose defining sum is empty and therefore 0 by [L5].

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