$ \int_{-\infty}^{\infty} \frac{\sin^3 x}{x^3} \, dx$ using contour integration. 
$\int_{-\infty}^{\infty}\frac{\sin^3{x}}{x^3}dx$ using contour integration.
Hint: Use this analytic continuation: $h(z) = -\frac{e^{3iz}-3e^{iz}}{4z^3}-\frac{1}{2z^3}$, and then take the imaginary part.

EDIT

*

*I used the complex exponential definition of $\sin$ to show that $\frac{\sin^3{x}}{x^3} = \operatorname{Im}\left(\frac{e^{3ix}-3e^{ix}}{-4x^3}\right)$. Therefore, I am really only itnerested in the first term of $h(x)$ for my result.


*I used a contour that is shown in red - two concentric half-circles with radii $R$ and $r$. The one with radius $R$ I let expand to infinity, while the one with radius $r$ I let shrink to zero.





*I argue that on $\gamma_R$, the integral vanishes because both the exponential terms and the $1/z^3$ terms tend to zero as $R$ increases, and, the integral over $\gamma_r$ vanishes as well since the exponential terms tend to 1 ar $r$ goes to zero. Finally, the total integral is zero by Cauchy.

$$ \int_{-\infty}^{\infty} \frac{\sin^3{x}}{x^3} \, dx
= \operatorname{Im} \int_{-\infty}^{\infty}\left(\frac{e^{3ix}-3e^{ix}}{-4x^3}\right) \, dz $$
I integrate by usage of the hinted analytic continuiation:
\begin{align*}
&\int_{\gamma} \left(-\frac{e^{3iz}-3e^{iz}}{4z^3}-\frac{1}{2z^3}\right) \, dz \\
&\quad= \operatorname{v.p.} \int_{-\infty}^{\infty} \left(\frac{e^{3ix}-3e^{ix}}{-4x^3}-\frac{1}{2x^3}\right) \, dz \\
&\hspace{3em} + \int_{\gamma_R} \left(-\frac{e^{3iz}-e^{iz}}{4z^3}-\frac{1}{2z^3}\right) \, dz
+ \int_{\gamma_r} \left(-\frac{e^{3iz}-3e^{iz}}{4z^3}-\frac{1}{2z^3}\right)dz \\
&\quad= 0
\end{align*}


*I only want the integral of the first term of $h(x)$, I transferred the integrals of $1/x^3$ to the other side and integrated them to get zero again.

$$ \operatorname{v.p.} \int_{-\infty}^{\infty} \left(\frac{e^{3ix}-3e^{ix}}{-4x^3}\right) \, dz
= \operatorname{v.p.} \int_{-\infty}^{\infty}\frac{1}{2x^3}dx
= 0 $$
The correct answer should be $3\pi/4$, which is given in the solutions, and I confirmed with wolfram alpha. However, I have been staring at my method for hours, and I can't find what doesn't check out.
In case something is unclear, or if I made any typos, I also provide a picture of my working.
 A: Consider the function
$$ g(z) = - \frac{e^{3iz} - \bbox[color:red;padding:3px;border:1px red dotted;]{3}e^{iz}}{4z^3} - \frac{1}{2z^3}. $$
Then we have $\operatorname{Im}(g(x)) = \frac{\sin^3 x}{x^3} $ for real $x$. Now using OP's contour and applying the Cauchy integration theorem,
\begin{align*}
0
&= \int_{\gamma} g(z) \, \mathrm{d}z \\
&= \int_{[-R,-r]\cup[r,R]} g(z) \, \mathrm{d}z + \int_{\gamma_r} g(z) \, \mathrm{d}z + \int_{\gamma_R} g(z) \, \mathrm{d}z.
\end{align*}

*

*As $R \to \infty$, as OP argued, we can show that $\int_{\gamma_R} g(z) \, \mathrm{d}z \to 0$.


*Using the power series expansion of the exponential function, it is not hard to check that $g(z)$ has a simple pole at $z = 0$ with
$$\mathop{\mathrm{Res}}_{z=0} g(z) = \frac{3}{4}.$$
Since the change of argument along the path $\gamma_r$ is $-\pi$, by using the well-known lemma (or simply substituting $z = re^{i\theta}$ where $\theta$ varies from $\pi$ to $0$ and letting $r \to 0^+$), it follows that
$$ \lim_{r \to 0^+} \int_{\gamma_r} g(z) \, \mathrm{d}z
= -\pi i \mathop{\mathrm{Res}}_{z=0} g(z)
= -\frac{3\pi i}{4}. $$
Therefore it follows that
$$ 
\int_{-\infty}^{\infty} \frac{\sin^3 x}{x^3} \, \mathrm{d}x
= \lim_{\substack{r \to 0^+ \\ R \to \infty}} \operatorname{Im}\left( \int_{[-R,-r]\cup[r,R]} g(z) \, \mathrm{d}z \right)
= \operatorname{Im}\left(\frac{3\pi i}{4}\right)
= \frac{3\pi}{4}. $$
In light of this solution, we can realize what went wrong in OP's solution:

*

*With the function $h(z)$ given in the hint, we have $\operatorname{Im}(h(x)) \neq \frac{\sin^3}{x^3}$.


*OP made an incorrect claim that the contribution from $\int_{\gamma_r} g(z) \, \mathrm{d}z$ vanishes as $r \to 0^+$, where in fact, it may contribute to the final answer.
A: Here is an alternate approach that is fairly simple, as well.
$$
\begin{align}
\int_{-\infty}^\infty\frac{\sin^3(x)}{x^3}\,\mathrm{d}x
&=\int_{-i-\infty}^{-i+\infty}\frac{\sin^3(z)}{z^3}\,\mathrm{d}z\tag1\\
&=\frac i8\int_{-i-\infty}^{-i+\infty}\frac{e^{3iz}-3e^{iz}+3e^{-iz}-e^{-3iz}}{z^3}\,\mathrm{d}z\tag2\\
&=\frac i8\int_{\gamma_+}\frac{e^{3iz}-3e^{iz}}{z^3}\,\mathrm{d}z+\frac i8\int_{\gamma_-}\frac{3e^{-iz}-e^{-3iz}}{z^3}\,\mathrm{d}z\tag3\\
&=-\frac{2\pi}8\frac{(3i)^2-3i^2}2\tag4\\[3pt]
&=\frac{3\pi}4\tag5
\end{align}
$$
Explanation:
$\text{(1)}$: the integral along $[-\infty,\infty]$ is the same as along $[-i-\infty,-i+\infty]$
$\phantom{\text{(1): }}$since the integrals along $[R,R-i]$ and $[-R-i,-R]$ both vanish as $R\to\infty$
$(2)$: $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$(3)$: $\gamma_+=[-R-i,R-i]\cup Re^{+i[0,\pi]}-i$ and
$\phantom{\text{(3): }}$$\gamma_-=[-R-i,R-i]\cup Re^{-i[0,\pi]}-i$
$\phantom{\text{(3): }}$where the integrals along the circular arcs vanishes as $R\to\infty $
$(4)$: the integral is $2\pi i$ times the sum of the residues inside $\gamma_+$
$\phantom{\text{(4): }}$there are no singularities inside $\gamma_-$
$\phantom{\text{(4): }}$the residue of $\frac{e^{i\lambda z}}{z^3}=-\frac{\lambda^2}2$
$(5)$: simplify
