# Modulus of Complex Integral

## Theorem

Let $\closedint a b$ be a closed real interval.

Let $f: \closedint a b \to \C$ be a continuous complex function.

Then:

$\ds \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$

where the first integral is a complex Riemann integral, and the second integral is a definite real integral.

## Proof

Define:

$z \in \C$ as the value of the complex Riemann integral:
$z = \ds \int_a^b \map f t \rd t$
$r \in \hointr 0 \to$ as the modulus of $z$
$\theta \in \hointr 0 {2 \pi}$ as the argument of $z$.
$z = re^{i \theta}$

Then:

 $\ds r$ $=$ $\ds z e^{-i \theta}$ Reciprocal of Complex Exponential $\ds$ $=$ $\ds \int_a^b e^{-i \theta} \map f t \rd t$ Linear Combination of Complex Integrals $\ds$ $=$ $\ds \int_a^b \map \Re {e^{-i \theta} \map f t} \rd t + i \int_a^b \map \Im {e^{-i \theta} \map f t} \rd t$ Definition of Complex Riemann Integral

As $r$ is wholly real, we have:

$\ds 0 = \map \Im r = \int_a^b \map \Im {e^{-i \theta} \map f t} \rd t$

Then:

 $\ds r$ $=$ $\ds \int_a^b \map \Re {e^{-i \theta} \map f t} \rd t$ $\ds$ $\le$ $\ds \int_a^b \size {\map \Re {e^{-i \theta} \map f t} } \rd t$ Absolute Value of Definite Integral $\ds$ $\le$ $\ds \int_a^b \size {e^{-i \theta} \map f t} \rd t$ Modulus Larger than Real Part $\ds$ $=$ $\ds \int_a^b \size {e^{-i \theta} } \size {\map f t} \rd t$ $\ds$ $=$ $\ds \int_a^b \size {\map f t} \rd t$ Modulus of Exponential of Imaginary Number is One

As $\ds r = \size {\int_a^b \map f t \rd t}$ by its definition, the result follows.

$\blacksquare$