integration - Evaluating $int_0^{infty} text{sinc}^m(x) dx$

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 $$\dfrac{\sin(x)}{x+\sin(x)} = \text{sinc}(x) \cdot \dfrac1{1+\text{sinc}(x)} = \sum_{k=0}^{\infty} (-1)^k \text{sinc}^{k+1}(x)$$

Answer

Notice $\lim_{x\to 0} \frac{\sin x}{x}$ is bounded at $x = 0$,

$$\begin{align}\int_0^{\infty} \left(\frac{\sin x}{x}\right)^m dx &= \frac12 \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^m dx\tag{*1}\\ &= \lim_{\epsilon\to 0} \frac12 \left(\frac{1}{2i}\right)^m \oint_{C_{\epsilon}} \left(\frac{e^{ix} - e^{-ix}}{x}\right)^m dx\tag{*2} \end{align}$$
We can evaluate the integral $(*1)$ as a limit of a integral over a deformed
contour $C_{\epsilon}$ which has a little half-circle of radius $\epsilon$ at origin:

$$C_{\epsilon} = (-\infty,-\epsilon) \cup \left\{ \epsilon e^{i\theta} : \theta \in [\pi,2\pi] \right\} \cup ( +\epsilon, +\infty)$$

We then split the integrand in $(*2)$ in two pieces, those contains exponential factors $e^{ikx}$ for $k \ge 0$ and those for $k < 0$.

$$(*2) = \lim_{\epsilon\to 0} \frac12 \left(\frac{1}{2i}\right)^m \oint_{C_{\epsilon}} \left( \sum_{k=0}^{\lfloor\frac{m}{2}\rfloor} + \sum_{k=\lfloor\frac{m}{2}\rfloor+1} ^{m} \right) \binom{m}{k} \frac{(-1)^k e^{i(m-2k)x}}{x^m} dx$$

To evaluate the $1^{st}$ piece, we need to complete the contour in upper half-plane. Since the completed contour contains the pole at $0$, we get:

$$\begin{align} \sum_{k=0}^{\lfloor\frac{m}{2}\rfloor} \text{ in }(*2) &= \frac12 \left(\frac{1}{2i}\right)^m (2\pi i)\sum_{k=0}^{\lfloor\frac{m}{2}\rfloor} \binom{m}{k} \frac{(-1)^k i^{m-1}(m-2k)^{m-1}}{(m-1)!}\\ &= \frac{\pi m}{2^m} \sum_{k=0}^{\lfloor\frac{m}{2}\rfloor} \frac{(-1)^k (m-2k)^{m-1}}{k!(m-k)!}\tag{*3}\end{align}$$

To evaluate the $2^{nd}$ piece, we need to complete the contour in lower half-plane instead. Since the completed contour no longer contains any pole, it contributes nothing and hence $I_m$ is just equal to R.H.S of $(*3)$.

Update

About the question whether $I_m$ is decreasing. Aside from the exception $I_1 = I_2$, it is strictly decreasing.

For $m \ge 1$, it is clear $I_{2m} > I_{2m+1}$ because the difference of corresponding integrands is non-negative and not identically zero. For the remaining cases, we have:

$$\begin{align}&I_{2m+1}-I_{2m+2}\\ = & \int_{0}^{\infty} \left(\frac{\sin x}{x}\right)^{2m+1}\left(1 - \frac{\sin x}{x}\right) dx\\ = & \left(\sum_{n=0}^{\infty} \int_{2n\pi}^{(2n+1)\pi}\right) \left(\frac{\sin x}{x}\right)^{2m+1}\left[1 - \frac{\sin x}{x} - \left(\frac{x}{x+\pi}\right)^{2m+1}\left(1 + \frac{\sin x}{x + \pi}\right)\right] dx \end{align}$$

Over the range $\cup_{n=0}^{\infty} (2n\pi,(2n+1)\pi)$, the factor $\left(\frac{\sin x}{x}\right)^{2m+1}$ is positive. The other factor $\Big[\cdots\Big]$ in above integral is bounded below by:

$$\begin{cases} 1 - \frac{\sin x}{x} - \left(\frac{x}{x+\pi}\right)^3\left(1 + \frac{\sin x}{x + \pi}\right), & \text{ for } x \in (0,\pi)\\ 1 - \frac{1}{x} - \frac{x}{x+\pi}\left(1 + \frac{1}{x}\right) = \frac{(\pi - 2)x - \pi}{x(x+\pi)} & \text{ for } x \in \cup_{n=1}^{\infty}(2n\pi,(2n+1)\pi) \end{cases}$$
A simple plot will convince you both bounds are positive in corresponding range. This
implies the integrand in above integral is positive and hence $I_{2m+1} > I_{2m+2}$.