a Fourier transform (sinc)

Question

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

Answer 1

\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.

Answer 2

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 $G={\cal F} (K)$.$\qquad\square$