Evaluating the contour integral $\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\mathrm dx$ 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!
 A: $$
{\rm J}\left(\alpha\right)
\equiv.
\int_{-\infty}^{\infty}{\sin^{3}\left(\alpha x\right) \over x^{3}}\,{\rm d}x\,,
\qquad\qquad
{\rm J}\left(0\right) = 0\,,\quad {\rm J}\left(1\right) = ?
$$
$$
{\rm J}'\left(\alpha\right)
=
{3 \over 2}\int_{-\infty}^{\infty}
{\sin\left(\alpha x\right)\sin\left(2\alpha x\right) \over x^{2}}\,{\rm d}x
=
{3 \over 4}\int_{-\infty}^{\infty}
{\cos\left(\alpha x\right) - \cos\left(3\alpha x\right) \over x^{2}}\,{\rm d}x
$$
$$
{\rm J}''\left(\alpha\right)
=
{3 \over 4}\int_{-\infty}^{\infty}
{-\sin\left(\alpha x\right) + 3\sin\left(3\alpha x\right) \over x}\,{\rm d}x
=
{3 \over 4}\,2\pi\,{\rm sgn}\left(\alpha\right)
=
{3 \over 2}\,\pi\,{\rm sgn}\left(\alpha\right)
$$
$$
{\rm J}'\left(\alpha\right)
=
{3 \over 2}\,\pi\left\vert\alpha\right\vert\,,
\quad
{\rm J}\left(1\right)
=
\int_{-\infty}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x
=
{3 \over 2}\,\pi\int_{0}^{1}\left\vert\alpha\right\vert\,{\rm d}\alpha
=
{3 \over 4}\,\pi
$$
$$
\int_{0}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x
=
{1 \over 2}\,\int_{-\infty}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x
=
{3 \over 8}\,\pi
$$
A: Your coutour will work perfectly, so I wonder why you're hesitating to proceed with your calculation. Anyway, here is a solution:
Since $\sin^3 z = (\sin 3z - 3\sin z)/4$, we have
$$\int_{0}^{\infty} \frac{\sin^3 x}{x^3} \, dx = \frac{1}{8} \Im \lim_{\epsilon \to 0} \int_{\mathbb{R} \backslash (-\epsilon, \epsilon)} \frac{3 e^{iz} - e^{3iz} -2}{z^3} \, dz,$$
where the term $-2$ is introduced in order to cancel out the pole of order 3, without affecting the value of the integral. Consider the counterclockwise-oriented upper semicircle $C$ of radius $R$, centered at the origin, with semicircular indent of radius $\epsilon$. Let $\Gamma_{R}^{+}$ and $\gamma_{\epsilon}^{-}$ denote semicircular arcs of $C$ of radius $R$ and $\epsilon$, respectively. If we put
$$f(z) = \frac{3 e^{iz} - e^{3iz} - 2}{z^3},$$
then we find that


*

*On $\Gamma_R^+$, we have $|f(z)| \leq 6R^{-3}$ and thus
$$\int_{\Gamma_{R}^{+}} f(z) \, dz \to 0 \quad \text{as } R \to \infty.$$

*Notice that
$$ f(z) = \frac{3}{z} + O(1) \quad \text{near } z = 0. $$
So by the direct computation,
$$\int_{\gamma_{\epsilon}^{-}} f(z) \, dz
= -\int_{0}^{\pi} f(\epsilon e^{i\theta}) i\epsilon e^{i\theta} \, d\theta
= -\int_{0}^{\pi} (3i + O(\epsilon)) \, d\theta
\to -3\pi i \quad \text{as } \epsilon \to 0.$$
(This is exactly the same as $-\pi i$ times the residue of $f$ at $z = 0$. The emergence of residue can be attributed to the fact that $f$ has only simple pole at $z = 0$.)
Since $f(z)$ has no pole on the region enclosed by $C$, we have
$$\lim_{\epsilon \to 0} \int_{\mathbb{R} \backslash (-\epsilon, \epsilon)} \frac{3 e^{iz} - e^{3iz} - 2}{z^3} \, dz = 3\pi i.$$
This proves the desired identity.
A: $\int \frac{\sin^3x}{x^3}\,\mathrm dx$
$=\frac{1}{4}\int\frac{3\sin x-\sin3x}{x^3}\,\mathrm dx$
$=\frac{1}{4}\left[(3\sin x-\sin3x)(\frac{-1}{2x^2})+\frac12\int \frac{3\cos x-3\cos3x}{x^2}\right]\,\mathrm dx$
$=\frac14\left[(3\sin x-\sin3x)(\frac{-2}{x^2})+\frac32\left[(\cos x-\cos3x)(\frac{-1}{x})+\int \frac{-\sin x+3\sin3x}{x}\right]\right]\,\mathrm dx$
$=\frac14\left[(3\sin x-\sin3x)(\frac{-2}{x^2})+\frac32(\cos x-\cos3x)(\frac{-1}{x})+\frac32\int \frac{-\sin x+3\sin3x}{x}\right]\,\mathrm dx$
For limits $x=0$ and $x \to \infty$ the first two terms disappear[each vanishes for both the  limits].
The third term, on considering the integral $\int_0^\infty\frac{\sin x}{x}\, \mathrm dx=\pi/2$ evaluates to:
$\frac{3}{8}[-\pi/2+3\pi/2] $
$=\frac{3\pi}{8}$
