(Excercise 11.17 in Baby Rudin) Why the integral $\int_{E} \sin n x dx$ is the imaginary part of the Fourier coefficient?

Question

Exercise 11.17 in Baby Rudin(Principles of Mathematical Analysis):

Suppose $E\subset (-\pi,\pi), m(E) > 0,\delta > 0$. Use the Bessel inequality to prove that there are at most finitely many integers $n$ such that $\sin nx \ge \delta$ for all $x \in E$.

A solution I found is as following:

For any integer with this property we have $\int_{E} \sin n x d x \geq \delta \mu(E)$, and the Bessel inequality implies that this inequality can hold for only a finite number of $n$. (The integral is the imaginary part of the Fourier coefficient of the $\mathscr{L}^{2}$-function $\chi_{E}$.)

Need help to under why "the integral is the imaginary part of the Fourier coefficient of the $\mathscr{L}^{2}$-function $\chi_{E}$". Thanks a lot!

Answer 1

Actually, $\int _E\sin (nx)=-Im (\int_Ee^{-inx} I_E(x) dx)$. But the minus sign does not matter for the proof since we can replace $E$ by $-E=\{-x: x \in E\}$.