Solution
= Solution
Using the convention
$$
\widehat f(u)=\int_{\mathbb R}f(x)e^{-iux}\,dx,
$$
the <Fourier transform of an interval indicator> is
$$
\widehat f(u)=\int_{-1}^{1}e^{-iux}\,dx
=\frac{e^{iu}-e^{-iu}}{iu}
=\frac{2\sin u}{u}.
$$
The removable value at the origin is obtained by taking the limit. Thus
$$
\boxed{\ \widehat f(u)=2\frac{\sin u}{u},
\qquad \widehat f(0)=2.\ }
$$