Fundamental Theorem of Calculus for Complex Riemann Integrals
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Theorem
Let $\closedint a b$ be a closed real interval.
Let $F, f: \closedint a b \to \C$ be complex functions.
Suppose that $F$ is a primitive of $f$.
Then the complex Riemann integral of $f$ satisfies:
- $\ds \int_a^b \map f t \rd t = \map F b - \map F a$
Proof
Let $u, v: \closedint a b \times \set 0 \to \R$ be defined as in the Cauchy-Riemann Equations:
- $\map u {t, y} = \map \Re {\map F z}$
- $\map v {t, y} = \map \Im {\map F z}$
where:
- $\map \Re {\map F z}$ denotes the real part of $\map F z$
- $\map \Im {\map F z}$ denotes the imaginary part of $\map F z$.
Then:
\(\ds \int_a^b \map f t \rd t\) | \(=\) | \(\ds \int_a^b \map {F'} {t + 0 i} \rd t\) | by assumption | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \paren {\map {\dfrac {\partial u} {\partial t} } {t, 0} + i \map {\dfrac {\partial v} {\partial t} } {t, 0} } \rd t\) | Cauchy-Riemann Equations | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \map {\dfrac {\partial u} {\partial t} } {t, 0} \rd t + i \int_a^b \map {\dfrac {\partial v} {\partial t} } {t, 0} \rd t\) | Definition of Complex Riemann Integral | |||||||||||
\(\ds \) | \(=\) | \(\ds \map u {b, 0} - \map u {a, 0} + i \paren {\map v {b, 0} - \map v {a, 0} }\) | Fundamental Theorem of Calculus | |||||||||||
\(\ds \) | \(=\) | \(\ds \map F b - \map F a\) |
$\blacksquare$
Sources
- 2001: Christian Berg: Kompleks funktionsteori: $\S 2.3$