Using contour integration, or other means, is there a way to find a general form for $\int_{0}^{\infty}\frac{\sin^{n}(x)}{x^{n}} \, dx$? While studying some CA, I ran across methods of evaluating $$\int_0^\infty \frac{\sin x}{x} \, dx, \;\ \int_0^\infty \frac{\sin^2 x}{x^2} \, dx, \;\ \text{and} \ \int_0^\infty \frac{\sin^3 x}{x^{3}} \, dx.$$
Is there a way to find a closed form for $$\int_0^\infty \frac{\sin^n x}{x^n} \, dx, \ n \in \mathbb{N}_{>0}  ?$$
Rather it be contour integration or some clever method using real analysis. 


$$
   \int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x = \frac{\pi}{2^{n+1} \cdot (n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} (2k-n)^{n-1} \operatorname{sign}(2k-n)
$$

where $\operatorname{sign}(x) = \cases{ 1 & $x > 0$ \\ 0 & $x = 0$\\ -1 & $x < 0$}$.
As to the (probabilistic) proof, notice that $\frac{\sin(t)}{t}$ is the characteristic function of a uniform random variable on $(-1,1)$. The sum of $n$ independent identically distributed such uniform random variables is known as Irwin-Hall random variable $Y_n$, and the integral in question is a multiple of its PDF evaluated at $x=0$:
$$
  \phi_{Y_n}(x) = \frac{1}{2 \pi} \int_{-\infty}^\infty \frac{\sin^n(t)}{t^n} \mathrm{e}^{-i t x} \mathrm{d} t = \frac{1}{\pi} \int_{0}^\infty \frac{\sin^n(t)}{t^n} \cos(t x) \mathrm{d} t
$$
The closed form for the PDF is given on the wikipedia with the reference.

As to more explicit derivation. We first integrate by parts, $n-1$ times, then use binomial theorem for $\sin^n(x)$:
$$ \begin{eqnarray}
   \int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x &=& \int_0^\infty \frac{\mathrm{d}^{n-1}}{\mathrm{d} x^{n-1}}\left( \sin^n(x)  \right) \frac{1}{(n-1)!}\frac{\mathrm{d} x}{x} \\
   &=& \frac{1}{(n-1)!} \int_0^\infty \frac{1}{2^n i^n} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \frac{\mathrm{d}^{n-1}}{\mathrm{d} x^{n-1}}\left( \mathrm{e}^{i (2k-n)x} \right) \frac{\mathrm{d} x}{x} \\
  &=& \frac{1}{(n-1)!} \int_0^\infty \frac{1}{2^n i^n} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k}  \left(i (2k-n)\right)^{n-1}  \mathrm{e}^{i (2k-n)x} \frac{\mathrm{d} x}{x}  \\
  &=& \frac{1}{(n-1)!} \int_0^\infty \frac{1}{2^n} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k}  \left((2k-n)\right)^{n-1} \sin((2k-n)x) \frac{\mathrm{d} x}{x} 
\end{eqnarray}
$$
In the last line, $\mathrm{e}^{i (2k-n) x}$ was expanded use Euler's formula, and since the sum is real, only real summands are retained. Then, integrating term-wise nails it:
$$ 
\begin{eqnarray}
   \int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x &=&  \frac{1}{2^n} \frac{1}{(n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k}  \left((2k-n)\right)^{n-1} \int_0^\infty  \sin((2k-n)x) \frac{\mathrm{d} x}{x}  \\
 &=& \frac{1}{2^n} \frac{1}{(n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k}  \left((2k-n)\right)^{n-1} \frac{\pi}{2} \operatorname{sign}(2k-n)
\end{eqnarray}
$$
