fourier coefficients and characteristic

Question

I am studyng about convergence of fourier. I have the next dude: if f is a characteristic function on (a,b), then

$\left | \int_{-\pi }^{\pi } f(t)sin(\alpha t) dt\right |=\left | \int_{a}^{b}sin(\alpha t)dt \right |$

how show it?

The characteristic function is the fourier transform not?

Answer 1

The indicator function is defined to be $1$ for $x \in A$, $0$ for $x \notin A$. If $A = (a,b) \subset (- \pi, \pi)$ then

\begin{align} \int_{- \pi}^{\pi} \chi_{[a,b]} \sin(\alpha t) dt &= \int_{- \pi}^{a} \chi_{[a,b]} \sin(\alpha t) dt + \int_{a}^{b} \chi_{[a,b]} \sin(\alpha t) dt + \int_{b}^{\pi} \chi_{[a,b]} \sin(\alpha t) dt \\ &= \int_{- \pi}^{a} 0 \cdot \sin(\alpha t) dt + \int_{a}^{b} 1 \cdot \sin(\alpha t) dt + \int_{b}^{\pi} 0 \cdot \sin(\alpha t) dt \\ &= \int_{a}^{b} \sin(\alpha t) dt \end{align}

If $A = (a,b) = (- \pi, \pi)$ then the result is clear.