Triangle Inequality for Integrals/Complex

Theorem
Suppose that $g : \left[{a \,.\,.\, b}\right] \to \C$ is continuous.

Then:
 * $\displaystyle \left|{\int_a^b g \left({t}\right)\ \mathrm dt}\right| \le \int_a^b \left|{g \left({t}\right)}\right| \ \mathrm dt$

Proof
Let:


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

where $r \ge 0$, $\theta$ are real.

Then:
 * $\displaystyle r = \left| \int_a^b g \left({t}\right)\ \mathrm dt \right|$

and


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

Therefore,

But

Therefore:


 * $\displaystyle r \le \int_a^b |g \left({t}\right)|\ \mathrm dt$

That is:


 * $\displaystyle \left| \int_a^b g \left({t}\right)\ \mathrm dt \right| \le \int_a^b |g \left({t}\right)|\ \mathrm dt$