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?
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Sine the function is not in $L^1$, you can only do it in the distributional sense. The answer, if I remember correctly, is $\chi_{[-1,1]}$, the characteristic function. – Tomas Jan 20 '15 at 07:47
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http://en.wikipedia.org/wiki/Sinc_function#Properties.. Also, I have a FT table here that gives $$f(x) = \bigg(\frac{2\alpha}{\sqrt{2\pi}}\bigg) sinc(\alpha x) \implies \hat f(k) = rect \bigg(\frac{k}{2\alpha}\bigg)$$ – Matthew Cassell Jan 20 '15 at 08:00
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Let $f(x) = \text{sinc}(x)$. We can rewrite
$$ f(x) = \text{sinc}(x) = \frac{\sin(\pi x)}{\pi x} = \frac{1}{2\pi}\frac{e^{i \pi x}-e^{-i \pi x}}{i x} = \frac{1}{2 \pi}\int \limits_{- \pi}^{\pi}e^{i \omega x} \,d \omega = \mathcal{F}^{-1}(1_{[-\pi, \pi]}).$$
Alex Silva
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Then $\mathcal{F} \left(\text{sinc}(x)\right)=\mathcal{F}\left(\mathcal{F}^{-1}(1_{[-\pi, \pi]})\right)=1_{[-\pi, \pi]}$? – Mark Jan 26 '15 at 16:34
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