# Fundamental Theorem of Calculus for Complex Riemann Integrals

## 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 $\displaystyle$ $=$ $\displaystyle F \left({b}\right) - F \left({a}\right)$

$\blacksquare$