Can someone please find the Fourier transform of $x \mapsto \sin(x)\,\operatorname{rect}\left(\frac{x}{2}\right)$?
I know that $$ \int_{-1}^1 \sin(x)\, e^{-ipx} \,\mathrm dx = 2i\,\frac{p\sin(1)\cos(p)-\cos(1)\sin(p)}{p^2-1}, $$ but when I try to calculate the inverse Fourier transform $$ \int_{-\infty}^\infty\frac{p\sin(1)\cos(p)-\cos(1)\sin(p)}{p^2-1}\, e^{ipx}\,\mathrm dp $$ with Jordan, I arrive at the conclusion that the integral is zero, because the residues of the function are zero. Can someone help me to solve the integral and return to the beginning function?
Informally, using that $\widehat{f \cdot g} = \hat{f} \ast \hat{g}$ (Fourier transform of product is the convolution of the Fourier transforms of the factors) and that $$ \widehat{\sin}(p) = \int_{- \infty}^{\infty} \sin(x) e^{- i p x} \, \text{d}x = \pi i \left( \delta_{- p - 1} - \delta_{1 - p} \right) $$ and $$ \widehat{\text{rect}\left(\frac{\cdot}{2}\right)}(p) = \int_{- \infty}^{\infty} \text{rect}\left(\frac{x}{2}\right) e^{- i p x} \, \text{d}x = \int_{-1}^{1} e^{- i p x} \, \text{d}x = 2 \text{sinc}(p) $$ we obtain $$ \widehat{\sin \cdot \text{rect}\left(\frac{\cdot}{2}\right)}(p) = 2 \pi i \left( \text{sinc}(- p - 1) - \text{sinc}(1 - p) \right). $$