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

Solution 1:

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

Solution 2:

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.

Solution 3:

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

$$ L(m)=\frac{\pi}{2^{m}(m-1)!}\sum_{k=0}^{\left \lfloor{\frac{m}{2}}\right \rfloor }(-1)^{k}\binom{m}{k}(m-2k)^{m-1}. $$