Show ${\int_{0}^{1}|e^{-2\pi\cdot int} f(t)| dt} = {\int_{0}^{1}|f(t)| dt}$

Question

The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).

For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?

Answer 1

Note that

$$|e^{it}|=1 \hspace{4mm}\forall t\in \mathbb{R}.$$

It is because

$$|e^{it}|=|\cos(t)+i\sin(t)|=\sqrt{\cos^2(t)+\sin^2(t)}=1.$$

Thus,

$$|e^{-2\pi i n t}f(t)|=|e^{-2\pi i n t}||f(t)|=|e^{(-2\pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$

Answer 2

For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here $$\left|e^{-2\pi i n t}f(t)\right|=\left|e^{-2\pi i n t}\right|\left|f(t)\right|=\left|f(t)\right|$$ if $nt\in \mathbb{R}$.