Triangle Inequality for Integrals/Complex

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,

But

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 $