Closed form for integral of integer powers of Sinc function

All sinc function integrals of the type

$$I_n=\int_{0}^{\infty}\frac{\sin^n{(x)}}{x^n}\mathrm{d}x$$

can be expressed using the following general form:

$$\displaystyle \int_{0}^{\infty}\frac{\sin^a{(x)}}{x^b}\mathrm{d}x=\frac{(-1)^{\lfloor(a-b)/2 \rfloor} \cdot \pi^ {1-c}}{ 2^{a-c}(b-1)!}\sum_{k=0}^{\lfloor a/2-c \rfloor } (-1)^k { a \choose k}(a-2k)^{b-1} Log^c(a-2k) $$

where $a$ and $b$ and are positive integers, $c\equiv (a-b) \pmod 2$, and $\lfloor j \rfloor $ denotes the floor function. When $a=b=n$, then $c=0$ and the equation simplifies in

$$\displaystyle I_n=\frac{ \pi}{ 2^{n}(n-1)!}\sum_{k=0}^{\lfloor n/2 \rfloor }(-1)^k { n \choose k}(n-2k)^{n-1} $$

so that you can obtain the rational coefficients for each $n$ by dividing the last expression to $\pi$. This gives you the sequence 1/2, 1/2, 3/8, 1/3, 115/384, 11/40, 5887/23040, 151/630, 259723/1146880, 15619/72576, 381773117/1857945600, 655177/3326400.....

The last equation can also be further simplified in

$$\displaystyle I_n=\frac{n \pi}{ 2^{n}}\sum_{k=0}^{\lfloor n/2 \rfloor } \frac{(-1)^k (n-2k)^{n-1}}{k!(n-k)!} $$