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

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Yes, $I_m$ is always a rational multiple of $\pi$, $$I_m = \frac{\pi m}{2^m} \sum_{k=0}^{\lfloor\frac{m}{2}\rfloor} \frac{(-1)^k (m-2k)^{m-1}}{k!(m-k)!}$$ You can derive the formula using contour integral. – achille hui May 1 '13 at 19:19
@achillehui Thanks. Could you expand on the answer? – user17762 May 1 '13 at 19:21
This amazing paper may also be of interest. – joriki May 1 '13 at 19:27
@joriki Interesting $\left(1/2 + \displaystyle \sum_n \text{sinc}^N(n) = \displaystyle \int_0^{\infty} \text{sinc}^N(x)dx \text{ valid for } N=1,2,\ldots,6 \text{ but not for }N \geq 7 \right)$ paper indeed. Thanks. – user17762 May 1 '13 at 19:37
Another solution is provided by my answer here. – Nick Strehlke May 2 '13 at 7:43

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 $I_{2m+1} > I_{2m+2}$.

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+1. Thanks for taking time on expanding your comment to give a complete answer. – user17762 May 1 '13 at 20:10
Is there an easy way to show that $I_m$ is a decreasing sequence? – user17762 May 1 '13 at 22:56
(+1) Nice answer. – Mhenni Benghorbal May 2 '13 at 13:03
why we dont take the contour in upper half-plan to compute $2^{nd}$ ? – mhd.math Oct 8 '13 at 20:28
@hmedan.mnsh the $2^nd$ piece contains exponential with -ve imaginary coefficient. If you complete the contour in upper half-plane, the contribution from the semi-circle at infinity wont' vanish. – achille hui Oct 8 '13 at 20:39

I hope an alternative method (to that of achille hui) may be of some interest. It does not use contour integration and is based on the trick employed here. Instead of the initial integral we consider a 2-parameter deformation $$I(a,b)=\int_{0}^{\infty}x^{b-1}e^{-ax}\left(e^{ix}-e^{-ix}\right)^m dx =\sum_{k=0}^m(-1)^{m-k}{m\choose k}\int_0^{\infty}x^{b-1}e^{-(a+i(m-2k))x}dx$$ with $a>0$ and $b>0$. Obviously, it is expressed in terms of gamma functions: \begin{align} I(a,b)=\sum_{k=0}^m (-1)^{m-k}\bigl(a+i(m-2k)\bigr)^{-b}{m\choose k}\Gamma(b)=\\=\frac{\pi}{\Gamma(1-b)}\sum_{k=0}^m \frac{(-1)^{m-k}\bigl(a+i(m-2k)\bigr)^{-b}}{\sin\pi b}{m\choose k} \end{align} Now we are interested in the limit $a\rightarrow 0$, $b\rightarrow 1-m$. The prefactor becomes $\displaystyle \frac{\pi}{(m-1)!}$. To take the limit of the sum, write it as \begin{align}\sum_{k=0}^{[m/2]}\left[\frac{(-1)^{m-k}\bigl(a+i(m-2k)\bigr)^{-b}}{\sin\pi b}+\frac{(-1)^{k}\bigl(a-i(m-2k)\bigr)^{-b}}{\sin\pi b}\right]{m\choose k}\substack{{a\rightarrow0}\\ \rightarrow}\\ \sum_{k=0}^{[m/2]}\frac{(-1)^{m-k}e^{-i\pi b/2}+(-1)^{k}e^{i\pi b/2}}{(m-2k)^b\sin\pi b}{m\choose k}=e^{-i\pi m/2}\sum_{k=0}^{[m/2]}\frac{2\cos\frac{\pi(b+m-2k)}{2}}{(m-2k)^b\sin\pi b}{m\choose k}. \end{align} Finally, the limit of $\displaystyle \frac{2\cos\frac{\pi(b+m-2k)}{2}}{\sin\pi b}$ as $b\rightarrow 1-m$ is computed by L'Hopital to be $(-1)^{k+m}$. Summarizing everything, we get for our integral the expression $$(2i)^{-m}I(0,1-m)=\frac{\pi\cdot 2^{-m}}{(m-1)!}\sum_{k=0}^{[m/2]}(-1)^k(m-2k)^{m-1}{m\choose k},$$ which coincides with the expression of achille hui.

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+1. This is nice as well! – user17762 May 1 '13 at 22:47
Is there an easy way to show that $I_m$ is a decreasing sequence? – user17762 May 1 '13 at 22:57
(+1) very nice. – Mhenni Benghorbal May 2 '13 at 13:04