Arc-Length Integral in C

#ComplexAnalysis

Def: Let $f : Ω \subseteq C \to C$ continuous and $γ : [a, b] \subseteq R \to Ω$ a piecewise smooth. We define the **********arc-length integral of $f$ over $γ$ , denoted as

\int_{γ} f (z) | d z | := \int_{a}^{b} f (γ (t)) | γ^{'} (t) | d t

in the special case that $f \equiv 1$ , then

\int_{γ} | d z | = Λ (γ)

Addtionally, if $γ$ is closed it is specially denoted as

\oint_{γ} f (z) | d z |

Prop: Let $f, g : Ω \subseteq C \to C$ , and $γ : [a, b] \to Ω$ , and $δ : [c, d] \to Ω$ be piecewise smooth, then they satisfy the following properties:

| \int_{γ} f (z) | d z | | \leq \int_{γ} | f (z) | | d z |

Let $α, β \in C$ , then

\int_{γ} α f (z) + β g (z) | d z | = α \int_{γ} f (z) | d z | + β \int_{γ} g (z) | d z |

Let $γ (b) = δ (c)$ , we can concatete paths, denoted as $γ * δ$ , and get:

\int_{γ * δ} f (z) | d z | = \int_{γ} f (z) | d z | + \int_{δ} f (z) | d z |

If $δ$ is a reparemetrization of $γ$ , then

\int_{δ} f (z) | d z | = \int_{γ} f (z) | d z |

in particular

\int_{γ^{-}} f (z) | d z | = \int_{γ} f (z) | d z |