Fourier transform of sinc function.

Question

Let us consider the sinc function: \begin{equation} {\rm{sinc}}(x)= \begin{cases} \frac{ \sin(\pi x)}{\pi x} \qquad &x \not= 0,\\ 1\qquad & x=0, \end{cases} \end{equation} What is the fourier transform, so-defined: $$\int_{-\pi}^{\pi} f(x) e^{-\imath k x}dx$$ of sinc function? I can't calculate this integral: $$\int_{-\pi}^{\pi} {\rm{sinc}}(x) e^{-\imath k x}dx$$ Any suggepstion please?

Answer 1

Let $f(x) = \text{sinc}(x)$. We can rewrite

$$ f(x) = \text{sinc}(x) = \frac{\sin(\pi x)}{\pi x} = \frac{1}{2\pi}\frac{e^{i \pi x}-e^{-i \pi x}}{i x} = \frac{1}{2 \pi}\int \limits_{- \pi}^{\pi}e^{i \omega x} \,d \omega = \mathcal{F}^{-1}(1_{[-\pi, \pi]}).$$