$ {\rm J}\left(\alpha\right) \equiv. \int_{-\infty}^{\infty}{\sin^{3}\left(\alpha x\right) \over x^{3}}\,{\rm d}x\,, \qquad\qquad {\rm J}\left(0\right) = 0\,,\quad {\rm J}\left(1\right) = ? $
$ {\rm J}'\left(\alpha\right) = {3 \over 2}\int_{-\infty}^{\infty} {\sin\left(\alpha x\right)\sin\left(2\alpha x\right) \over x^{2}}\,{\rm d}x = {3 \over 4}\int_{-\infty}^{\infty} {\cos\left(\alpha x\right) - \cos\left(3\alpha x\right) \over x^{2}}\,{\rm d}x $
$ {\rm J}''\left(\alpha\right) = {3 \over 4}\int_{-\infty}^{\infty} {-\sin\left(\alpha x\right) + 3\sin\left(3\alpha x\right) \over x}\,{\rm d}x = {3 \over 4}\,2\pi\,{\rm sgn}\left(\alpha\right) = {3 \over 2}\,\pi\,{\rm sgn}\left(\alpha\right) $
$ {\rm J}'\left(\alpha\right) = {3 \over 2}\,\pi\left\vert\alpha\right\vert\,, \quad {\rm J}\left(1\right) = \int_{-\infty}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x = {3 \over 2}\,\pi\int_{0}^{1}\left\vert\alpha\right\vert\,{\rm d}\alpha = {3 \over 4}\,\pi $
$ \int_{0}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x = {1 \over 2}\,\int_{-\infty}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x = {3 \over 8}\,\pi $