Proof that $\vert\int_a^bf(t)dt\vert\le\int_a^b\vert f(t)\vert dt $ for complex integration

Question

The following proof is given in my notes:
Write the complex integral $\int_a^bf(t)dt = Re^{i\theta}$
where $$ R=\int_a^bf(t)dt. \quad(1) $$ On the other hand, $$ R=\int_a^be^{-i\theta}f(t)dt.\quad(2) $$ Write $e^{-i\theta}f(t)=U(t)+iV(t)$ where $U(t)$ and $V(t) $ are real-valued functions. Then because R is real,

$$ R=\int_z^bU(t)dt\quad(3) $$ But now, $$ U(t)= Re\ e^{-i\theta}f(t) \le e^{-i\theta}f(t)=\vert f(t)\vert\quad(4) $$ Therefore, from the properties of real integrals, $$ \int_a^bU(t)dt\le\int\vert f(t)\vert dt $$ which proves the desired result.

I have trouble figuring out how each of the numbered equations actually hold. After writing the integral as $\int_a^bf(t)dt = Re^{i\theta}$ How can we say that (1) and (2) hold. (3) sort of makes sense as R has to be real.

Answer 1

I think that at the beginning is $|\int_a^b f(t)dt|=|Re^{i\theta}|$ and it would be ok. By the way, here's the proof that I studied. It's basically the same proof, but maybe the notation is more clear.

Let $\alpha \in \Bbb R$ be such that $$e^{i\alpha}\int_a^b f(t)dt \in \Bbb R^+$$ (I'm actually rotating $\int_a^b f(t)dt \in \Bbb C$). So we get $$e^{i\alpha}\int_a^b f(t)dt=\int_a^b e^{i\alpha}f(t)dt=\Re\left(\int_a^b e^{i\alpha}f(t)dt\right).$$ We conclude by $$\left| \int_a^b f(t)dt\right|=\left|\int_a^b e^{i\alpha}f(t)dt\right|=\Re\left(\int_a^b e^{i\alpha}f(t)dt\right)=\int_a^b \Re(e^{i\alpha}f(t))dt\le \int_a^b |e^{i\alpha}f(t)|dt=\int_a^b |f(t)|dt$$ because $\Re(z) \le |z|$ by, sustantially, Pythagorean's Theorem.