Modulus of Complex Integral

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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 $z$.


From Modulus and Argument of Complex Exponential:

$z = re^{i \theta}$


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.

$\blacksquare$


Sources