Triangle Inequality for Integrals/Complex

Theorem
Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval.

Let $f : \left[{a \,.\,.\, b}\right] \to \C$ be a continuous complex function.

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

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

Proof
Define $z \in \C$ as the value of the complex integral:
 * $z = \displaystyle \int_a^b f \left({t}\right) \ \mathrm dt$

Define $r \in \left[{0 \,.\,.\, \infty }\right)$ as the modulus of $z$, and $\theta \in \left[{0 \,.\,.\, 2 \pi }\right]$ as the argument of $z$.

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

and:


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

Therefore, using the notation $\operatorname{Re} \left({ r }\right)$ for the real part of $r$:

But:

Therefore:


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

That is:


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