Zero Simple Staircase Integral Condition for Primitive/Lemma

Lemma for Zero Simple Staircase Integral Condition for Primitive

Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.

Let $C$ be a closed staircase contour in $D$.

Then there exists a contour $C'$ such that:

$\ds \oint_C \map f z \rd z = \oint_{C'} \map f z \rd z$

This contour $C'$ has the property that:

for all $k \in \set {1, \ldots, n - 1}$, the intersection of the images of $C_k$ and $C_{k + 1}$ is equal to their common end point $\map {\gamma_k} {b_k}$.

Proof

The proof is by induction over $n \in \N$, the number of directed smooth curves that $C$ is a concatenation of.

Basis for the Induction

Let $C$ consists of just one directed smooth curve.

Then we write:

$C' = C$

Thus as $k \in \O$, the proposition is vacuously true.

This is the basis for the induction.

Induction Hypothesis

Suppose that:

if $C$ is a concatenation of $n$ directed smooth curves, we can find a contour $C'$ with the properties above.

This is the induction hypothesis.

Induction Step

This is the induction step:

Let $C$ be a concatenation of $n + 1$ directed smooth curves.

Suppose that there exists $k \in \set {1, \ldots, n}$ such that the intersection of the images of $C_k$ and $C_{k + 1}$ contain more elements than their common end point $\map {\gamma_k} {b_k}$, otherwise we can put $C' = C$.

Then, one of the two line segments must contain the other line segment.

So:

$\Img {C_k} \cap \Img {C_{k + 1} } = \gamma_k \sqbrk {\closedint {a'} {b_k} } \cup \gamma_{k + 1} \sqbrk {\closedint {a_{k + 1} } {b'} }$

where either $a' = a_k$, or $b' = b_{k + 1}$.

Define $C_k'$ as the contour that is parameterized as a line segment that goes from $\map {\gamma_k} {a_k}$ to $\map {\gamma_k} {a'}$, and then to $\map {\gamma_{k + 1} } {b_{k + 1} }$.

Then, the image of $C_k'$ is equal to the part of the line segments that does not overlap.

Define:

$\tilde C_k$ as the contour with the parameterization $\gamma_k \restriction_{\closedint {a'} {b_k} }$

and:

$\tilde C_{k + 1}$ as the contour with the parameterization $\gamma_{k + 1} \restriction_{\closedint {a_{k + 1} } {b'} }$.

Here, $\gamma_k \restriction_{\closedint {a'} {b_k} }$ denotes the restriction of $\gamma_k$ to $\closedint {a'} {b_k}$.

Now $\tilde C_k$ is the reversed contour of $\tilde C_{k + 1}$.

So:

\(\ds \oint_C \map f z \rd z\)

\(=\)

\(\ds \oint_{C_1 \cup \ldots \cup C_{k - 1} } \map f z \rd z + \oint_{C_k} \map f z \rd z + \oint_{C_{k + 1} } \map f z \rd z + \oint_{C_{k + 2} \cup \ldots \cup C_{n + 1} } \map f z \rd z\)

Contour Integral of Concatenation of Contours

\(\ds \)

\(=\)

\(\ds \oint_{C_1 \cup \ldots \cup C_{k - 1} } \map f z \rd z + \oint_{C_k'} \map f z \rd z + \oint_{\tilde C_k} \map f z \rd z + \oint_{\tilde C_{k + 1} } \map f z \rd z + \oint_{C_{k + 2} \cup \ldots \cup C_{n + 1} } \map f z \rd z\)

Contour Integral of Concatenation of Contours

\(\ds \)

\(=\)

\(\ds \oint_{C_1 \cup \ldots \cup C_{k - 1} } \map f z \rd z + \oint_{C_k'} \map f z \rd z + \oint_{\tilde C_k} \map f z \rd z - \oint_{\tilde C_k} \map f z \rd z + \oint_{C_{k + 2} \cup \ldots \cup C_{n + 1} } \map f z \rd z\)

Contour Integral along Reversed Contour

\(\ds \)

\(=\)

\(\ds \oint_{C_1 \cup \ldots \cup C_{k - 1} \cup \tilde C_k \cup C_{k + 2} \ldots \cup C_{n + 1} } \map f z \rd z\)

Contour Integral of Concatenation of Contours

which shows that the contour integral of $f$ along $C$ is equal to the contour integral of $f$ along $C = C_1 \cup \ldots \cup C_{k - 1} \cup \tilde C_k \cup C_{k + 2} \ldots \cup C_{n + 1}$.

This new contour $C$ is a staircase contour which is a concatenation of $n$ directed smooth curves.

So, by the induction hypothesis, there exists a contour $C'$ such that:

$\ds \oint_{C'} \map f z \rd z = \oint_{C} \map f z \rd z$

where $C'$ has the desired property.

$\blacksquare$