# Prove $\left| \int_a^b f(t) dt \right| \leq \int_a^b \left| f(t) \right| dt$

I've been given a proof that shows the following:

If $f:[a,b]\to \mathbb C$ is a continuous function and $f(t)=u(t)+iv(t)$ then $$\left| \int_a^b f(t) dt \right| \leq \int_a^b \left| f(t) \right| dt$$ The proof begins by letting $\theta$ be the principle argument of the complex number $\int_a^b f(t)dt$ and there is one step in the proof I don't understand; can anyone explain to me why $$\int_a^b e^{-i\theta}f(t) dt=\Re\bigg(\int_a^be^{-i\theta}f(t)dt\bigg)$$

Would this not imply that the imaginary part of the left hand side is equal to $0$? I'm not sure why the LHS and RHS would be equal here.

• The imaginary part of the RHS is $0$, and LHS = RHS.. – user258700 May 5 '16 at 8:33
• I see that the imaginary part of the RHS is $0$, my question is why are the left and right hand side equal? – user330513 May 5 '16 at 8:42
• Sorry, I misread the question. Well in this case the claim is of course not true for arbitrary $f$. Please write down all the assumptions and what exactly is being proved – user258700 May 5 '16 at 8:53
• I've edited the question adding in the rest of the detail. – user330513 May 5 '16 at 9:06
• Since $e^{-i\theta}\int_a^bf(t)dt$ is real and equal to $\int_a^be^{-i\theta}f(t)dt$, then $\int_a^be^{-i\theta}f(t)dt$ is equal to real part of itself. – vnd May 5 '16 at 9:16

If $A = \int_a^b f(t)dt$, notice that, since $\theta$ is the principal argument of $A$,
$$A = |A| e^{i \theta}$$
$$|A| = Ae^{-i\theta} = \left(\int_a^b f(t)dt \right) e^{-i\theta} = \int_a^b e^{-i\theta} f(t)dt$$
Hence $\int_a^b e^{-i\theta} f(t) dt = |A| \in \Bbb R$