Absolute Value of Complex Integral

Theorem

Suppose that $g : [a,b] \to \C$ is continuous. Then

$\displaystyle \left| \int_a^b g(t)\ \mathrm dt \right| \leq \int_a^b |g(t)|\ \mathrm dt $

Proof

Let

$\displaystyle \int_a^b g(t)\ \mathrm dt = r e^{i \theta} $

where $r\geq 0$, $\theta$ are real. Then

$\displaystyle r = \left| \int_a^b g(t)\ \mathrm dt \right|$

and

$\displaystyle r = e^{-i\theta} \int_a^b g(t)\ \mathrm dt$

Therefore,

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle r\)	\(=\)	\(\displaystyle \Re(r)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \Re\left( e^{-i\theta} \int_a^b g(t)\ \mathrm dt \right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \int_a^b \Re\left( e^{-i\theta} g(t) \right)\ \mathrm dt\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

But

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \Re\left( e^{-i\theta} g(t) \right)\)	\(\leq\)	\(\displaystyle \vert e^{-i\theta}g(t)\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \vert g(t)\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Therefore,

$\displaystyle r \leq \int_a^b |g(t)|\ \mathrm dt $

That is,

$\displaystyle \left| \int_a^b g(t)\ \mathrm dt \right| \leq \int_a^b |g(t)|\ \mathrm dt $

$\blacksquare$