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 $C_i$ be parameterized by the smooth path $\gamma_i: \closedint {a_i} {b_i} \to \C$ for all $i \in \set {1, \ldots, n}$.

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


Suppose that $\sigma_i: \closedint {c_i} {d_i} \to \C$ is a reparameterization of $C_i$ for all $i \in \set {1, \ldots, n}$.


Then:

$\ds \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t = \sum_{i \mathop = 1}^n \int_{c_i}^{d_i} \map f {\map {\sigma_i} t} \map {\sigma_i'} t \rd t$

and all complex Riemann integrals in the equation are defined.


Proof

Define $g_i: \closedint {a_i} {b_i} \to \C$ by $\map {g_i} t = \map f {\map {\gamma_i} t} \map {\gamma_i'} t$ for all $i \in \set {1, \ldots, n}$.

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

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

From Continuous Complex Function is Complex Riemann Integrable, we find that $\ds \int_{a_i}^{b_i} \map {g_i} t \rd t$ is defined.

Similarly, it can be shown that $\ds \int_{c_i}^{d_i} \map f {\map {\sigma_i} t} \map {\sigma_i'} t \rd t$ is defined for all $i \in \set {1, \ldots, n}$.

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


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

$\blacksquare$


Sources