TITLE: Evaluating the contour integral $\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\mathrm dx$
QUESTION [9 upvotes]: I am trying to show
$$\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\mathrm dx = \frac{3\pi}{8}.$$
I believe the contour I should use is a semicircle in the upper half plane with a slight bump at the origin, so I miss the singularity.
Lastly I have the hint to consider 
$$\int_{0}^{\infty}\frac{e^{3iz}-3e^{iz}+2}{z^{3}}\mathrm dz$$
around the contour I mentioned. Thanks for any help or hints!

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