Fourier transform and characteristic function of an interval

Question

If we define the Fourier transform of a function $f$ as the function $$\mathcal{F}(f)(t) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-ixt}f(x)\,dx,$$ how can I prove that $$\mathcal{F}(h)(t)=\chi_{[-b,b]}(t),$$ where $\chi$ denotes the characteristic function of the interval $[-b,b]$ and $h$ is defined as $$h(x) = \sqrt{2\pi}\big(2\frac{\sin(bx)}{x}\big).$$

Answer 1

Since $$\frac{1}{\sqrt{2\pi}}\int_{\Bbb R}\chi_{[-b,\,b]}(t)e^{ixt}dt=\frac{1}{\sqrt{2\pi}}\int_{-b}^be^{ixt}dt=\frac{e^{ibx}-e^{-ibx}}{\sqrt{2\pi}\,ix}=\frac{1}{\sqrt{2\pi}}\frac{2}{x}\sin bx=\frac{1}{2\pi}h(x),$$inverting the Fourier transform gives$$\chi_{[-b,\,b]}(t)=(2\pi)^{-3/2}\int_{\Bbb R}h(x)e^{-ixt}dx.$$I'd double-check your powers of $2\pi$ if I were you. Note that$$\mathcal{F}(h)(0)=2\int_{\Bbb R}\frac{\sin bx dx}{x}=2\pi=2\pi\chi_{[-b,\,b]}(0)$$for $b>0$.