Contour Integral is Well-Defined

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Theorem

Let $C$ be a contour defined by a finite sequence $C_1, \ldots, C_n$ of directed smooth curves in the complex plane $\C$.

Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.

Let $f: \Img C \to \C$ be a continuous complex function, where $\Img C$ denotes the image of $C$.


Suppose that $\sigma_k: \closedint {c_k} {d_k} \to \C$ is a reparameterization of $C_k$ for all $k \in \set {1, \ldots, n}$.


Then:

$\ds \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \map f {\map {\gamma_k} t} \map {\gamma_k'} t \rd t = \sum_{k \mathop = 1}^n \int_{c_k}^{d_k} \map f {\map {\sigma_k} t} \map {\sigma_k'} t \rd t$

and all complex Riemann integrals in the equation are defined.


Proof

Define $g_k: \closedint {a_k} {b_k} \to \C$ by $\map {g_k} t = \map f {\map {\gamma_k} t} \map {\gamma_k'} t$ for all $k \in \set {1, \ldots, n}$.

By definition of smooth path, it follows that $\gamma_k$ and $\gamma_k'$ are continuous for all $k \in \set {1, \ldots, n}$.

From Continuity of Composite Mapping/Corollary and Sum Rule for Continuous Complex Functions, it follows that $g_k$ is continuous.

From Continuous Complex Function is Complex Riemann Integrable, we find that $\ds \int_{a_k}^{b_k} \map {g_k} t \rd t$ is defined.

Similarly, it can be shown that $\ds \int_{c_k}^{d_k} \map f {\map {\sigma_k} t} \map {\sigma_k'} t \rd t$ is defined for all $k \in \set {1, \ldots, n}$.

Hence, all complex Riemann integrals in the theorem are defined.


The equality now follows from Contour Integral is Independent of Parameterization.

$\blacksquare$


Sources