# Fundamental Theorem of Calculus for Complex Riemann Integrals

## 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:

$\displaystyle \int_a^b \map f t \rd t = \map F b - \map F a$

## 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:

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

$\blacksquare$