Fundamental Theorem of Calculus for Complex Riemann Integrals

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Theorem

Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval.

Let $F, f : \left[{a \,.\,.\, b}\right] \to \C$ be complex functions.


Suppose that $F$ is a primitive of $f$.


Then the complex Riemann integral of $f$ satisfies:

$\displaystyle \int_a^b f \left({t}\right) \rd t = F \left({b}\right) - F \left({a}\right)$


Proof

Let $u, v: \left[{a \,.\,.\, b}\right] \times \left\{ {0}\right\} \to \R$ be defined as in the Cauchy-Riemann Equations:

$u \left({t, y}\right) = \operatorname{Re} \left({F \left({z}\right) }\right)$
$v \left({t, y}\right) = \operatorname{Im} \left({F \left({z}\right) }\right)$

where:

$\operatorname{Re} \left({F \left({z}\right)}\right) $ denotes the real part of $F \left({z}\right)$
$\operatorname{Im} \left({F \left({z}\right)}\right) $ denotes the imaginary part of $F \left({z}\right)$.


Then:

\(\displaystyle \int_a^b f \left({t}\right) \rd t\) \(=\) \(\displaystyle \int_a^b F' \left({t + 0i}\right) \rd t\) by assumption
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \left({\dfrac {\partial u} {\partial t} \left({t, 0}\right) + i \dfrac {\partial v} {\partial t} \left({t, 0}\right) }\right) \rd t\) Cauchy-Riemann Equations
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \dfrac {\partial u} {\partial t} \left({t, 0}\right) \rd t + i \int_a^b \dfrac {\partial v}{\partial t} \left({t, 0}\right) \rd t\) Definition of Complex Riemann Integral
\(\displaystyle \) \(=\) \(\displaystyle u \left({b, 0}\right) - u \left({a, 0}\right) + i \left({v \left({b, 0}\right) - v \left({a, 0}\right) }\right)\) Fundamental Theorem of Calculus
\(\displaystyle \) \(=\) \(\displaystyle F \left({b}\right) - F \left({a}\right)\)

$\blacksquare$


Sources