TITLE: Evaluating $\int_0^{\infty} \text{sinc}^m(x) dx$
QUESTION [23 upvotes]: How do I evaluate $$I_m = \displaystyle \int_0^{\infty} \text{sinc}^m(x) dx,$$ where $m \in \mathbb{Z}^+$?
For $m=1$ and $m=2$, we have the well-known result that this equals $\dfrac{\pi}2$. In general, WolframAlpha suggests that is seems to be a rational multiple of $\pi$.
\begin{array}{c|c|c|c|c|c|c|c}
m & 1 & 2 & 3 & 4 & 5 & 6 & 7\\
\hline
I_m & \dfrac{\pi}2 & \dfrac{\pi}2 & \dfrac{3\pi}8 & \dfrac{\pi}3 & \dfrac{115\pi}{384} & \dfrac{11\pi}{40} & \dfrac{5887 \pi}{23040}\\
\end{array}
$(1)$. Can we prove that $I_m$ is a rational multiple of $\pi$ always?
$(2)$. If so, is there a nice formula, i.e., if $I_m = \dfrac{p(m)}{q(m)} \pi$, where $p(m),q(m) \in \mathbb{Z}^+$, are there nice expressions for $p(m)$ and $q(m)$? 
P.S: This integral came up when I was trying my method to answer this question, by writing $\dfrac{\sin(x)}{x+\sin(x)}$ as $$\dfrac{\sin(x)}{x+\sin(x)} = \text{sinc}(x) \cdot \dfrac1{1+\text{sinc}(x)} = \sum_{k=0}^{\infty} (-1)^k \text{sinc}^{k+1}(x)$$

REPLY [2 votes]: Let $f(x)=\sin^m(x)$ where $m>1$. Integration by parts $m-1$ times gives the formula 
$$
\int_0^{\infty}\frac{f(x)}{x^m}dx=\frac{1}{(m-1)!}\int_0^{\infty}\frac{f^{(m-1)}(x)}{x}dx.
$$
Now we have the formulae 
\begin{align}
\sin^{2n}(x)&=\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}(-1)^{n-k}\binom{2n}{k}\cos((2n-2k)x)+\frac{1}{2^{2n}}\binom{2n}{n} \\
\sin^{2n-1}(x)&=\frac{1}{2^{2n-2}}\sum_{k=0}^{n-1}(-1)^{n-k-1}\binom{2n-1}{k}\sin((2n-2k-1)x).
\end{align}
Differentiate the former $2n-1$ times and the latter $2n-2$ times to obtain with the very first equation 
\begin{align}
L(2n)&=\frac{\pi}{2^{2n}(2n-1)!}\sum_{k=0}^{n-1}(-1)^{k}\binom{2n}{k}(2n-2k)^{2n-1} \\
L(2n-1)&=\frac{\pi}{2^{2n-1}(2n-2)!}\sum_{k=0}^{n-1}(-1)^{k}\binom{2n-1}{k}(2n-2k-1)^{2n-2},
\end{align}
since
$$
\int_0^{\infty}\frac{\sin((2n-2k)x)}{x}dx=\int_0^{\infty}\frac{\sin(x)}{x}dx=\frac{\pi}{2}.
$$
This yields in a campact shape 
$$
L(m)=\frac{\pi}{2^{m}(m-1)!}\sum_{k=0}^{\left \lfloor{\frac{m}{2}}\right \rfloor }(-1)^{k}\binom{m}{k}(m-2k)^{m-1}.
$$