a Fourier transform (sinc) let $K(u) = \frac{\sin(u)}{\pi u}$ 
show that Fourier transform of $K$ is 
$ \hat{K}(\omega) = \textbf{1}_{|\omega|\leq 1} $
Some help would be appreciated
 A: \begin{eqnarray*}
\tilde{K}(\omega ) &=&\int du\exp [i\omega u]K\left( u\right) =\int_{\Gamma }du\exp [i\omega u]\frac{\sin u%
}{\pi u} \\
&=&\frac{1}{2\pi i}\int_{\Gamma }du\frac{1}{u}\exp [i\omega u]\{\exp
[iu]-\exp [-iu]\} \\
&=&\frac{1}{2\pi i}\int_{\Gamma }du\frac{1}{u}\{\exp [i(\omega +1)u]-\exp
[i(\omega -1)u]\}=\mathcal{I}_{1}+\mathcal{I}_{2}
\end{eqnarray*}
where $\Gamma $ is the real line modified by a semi-circle around the origin
in the upper half-plane (note that the integrand is well-behaved in $u=0$).
For $\omega <-1$ both integrands can be continued in the lower half plane
and have the same residue and hence $\tilde{K}(\omega )=0$. If $\omega >1$
then both integrands can be continued in the upper half plane and have no
residues so again $\tilde{K}(\omega )=0$. In between only one of the $%
\mathcal{I}_{j}$'s is non-zero and actually equals 1, the result you wanted
to obtain.
A: Let $G(\omega)={\bf 1}_{[-1,1]}(\omega)$, then $G\in L^1(\Bbb{R})\cap L^2(\Bbb{R})$, and
$${\cal F}^{-1}(G)(u)=\frac{1}{2\pi}\int_{-1}^1e^{i\omega u}d\omega=\frac{\sin(u)}{\pi u}
=K(u)
$$
(${\cal F}^{-1}$ is the inverse Fourier transform.) But Fourier transform is bijective on $L^2(\Bbb{R})$, thus from ${\cal F}^{-1}(G)=K$ we conclude that $G={\cal F} (K)$.$\qquad\square$
