A sine integral $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$

Here's another approach.

We have $$\begin{eqnarray*} \int_0^\infty dx\, \left(\frac{\sin x}{x}\right)^n &=& \lim_{\epsilon\to 0^+} \frac{1}{2} \int_{-\infty}^\infty dx\, \left(\frac{\sin x}{x-i\epsilon}\right)^n \\ &=& \lim_{\epsilon\to 0^+} \frac{1}{2} \int_{-\infty}^\infty dx\, \frac{1}{(x-i\epsilon)^n} \left(\frac{e^{i x}-e^{-i x}}{2i}\right)^n \\ &=& \lim_{\epsilon\to 0^+} \frac{1}{2} \frac{1}{(2i)^n} \int_{-\infty}^\infty dx\, \frac{1}{(x-i\epsilon)^n} \sum_{k=0}^n (-1)^k {n \choose k} e^{i x(n-2k)} \\ &=& \lim_{\epsilon\to 0^+} \frac{1}{2} \frac{1}{(2i)^n} \sum_{k=0}^n (-1)^k {n \choose k} \int_{-\infty}^\infty dx\, \frac{e^{i x(n-2k)}}{(x-i\epsilon)^n}. \end{eqnarray*}$$ If $n-2k \ge 0$ we close the contour in the upper half-plane and pick up the residue at $x=i\epsilon$. Otherwise we close the contour in the lower half-plane and pick up no residues. The upper limit of the sum is thus $\lfloor n/2\rfloor$. Therefore, using the Cauchy differentiation formula, we find $$\begin{eqnarray*} \int_0^\infty dx\, \left(\frac{\sin x}{x}\right)^n &=& \frac{1}{2} \frac{1}{(2i)^n} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} \frac{2\pi i}{(n-1)!} \left.\frac{d^{n-1}}{d x^{n-1}} e^{i x(n-2k)}\right|_{x=0} \\ &=& \frac{1}{2} \frac{1}{(2i)^n} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} \frac{2\pi i}{(n-1)!} (i(n-2k))^{n-1} \\ &=& \frac{\pi}{2^n (n-1)!} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1}. \end{eqnarray*}$$ The sum can be written in terms of the hypergeometric function but the result is not particularly enlightening.

Just to verify oen's post (since there is a post with a different answer), I will post the answer I got.

$|\sin(z)|\le e^{|\mathrm{Im}(z)|}$; therefore, on the strip $|\mathrm{Im}(z)|\le1$, we have $|\sin(z)|\le e$. Thus, $\left(\frac{\sin(z)}{z}\right)^n$ vanishes as $|z|\to\infty$ in that strip and therefore, $$ \int_{-\infty}^\infty\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z =\int_{-\infty-i}^{\infty-i}\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z\tag{1} $$ Next define two contours $\gamma^+$ and $\gamma^-$. $\gamma^+$ goes from $-R-i$ to $R-i$ then circles back through the upper half plane along $|z+i|=R$. $\gamma^-$ goes from $-R-i$ to $R-i$ then circles back through the lower half plane along $|z+i|=R$.

Using the binomial theorem, we get $$ \left(\frac{\sin(z)}{z}\right)^n=\frac1{(2iz)^n}\sum_{k=0}^n(-1)^k\binom{n}{k}e^{(n-2k)iz}\tag{2} $$ Integrate the terms where $n-2k\ge0$ along $\gamma^+$ and the others along $\gamma^-$. Since $\gamma^-$ doesn't enclose any singularities, we can ignore that integral. Therefore, $$ \begin{align} \int_0^\infty\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z &=\frac12\int_{\gamma^+}\frac1{(2iz)^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}e^{(n-2k)iz}\mathrm{d}z\\ &=\frac{\pi i}{(2i)^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}\mathrm{Res}\left(\frac{e^{(n-2k)iz}}{z^n},0\right)\\ &=\frac{\pi i}{(2i)^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}\frac{(n-2k)^{n-1}i^{n-1}}{(n-1)!}\\ &=\frac{\pi}{2^n(n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}(n-2k)^{n-1}\tag{3} \end{align} $$

I'll write $I = \int_{-\infty}^{\infty} \left(\frac{\sin z}{z} \right)^n dz$

First, to simplify matters let's take $n$ odd and $\geq 3$. Let $C_{\epsilon}^+$ be the contour along the real line that takes a semicircular detour into the upper half plane about the origin, and let $C_{\epsilon}^-$ be the same for the lower half plane. We use continuity of the integrand to argue that $$ I = \lim_{\epsilon \rightarrow 0} \int_{C_{\epsilon}^{\pm}} = \frac{1}{2} \lim_{\epsilon \rightarrow 0} \left( \int_{C_{\epsilon}^+} + \int_{C_{\epsilon}^-} \right) $$ Now think about $(\sin x)^n$: it's a sum of exponential terms of the form $e^{i l x}$ for $-n \leq l \leq n$ with some coefficients. You should convince yourself that any $l < 0$ term is killed by $\int_{C_{\epsilon}^-}$ and any $l > 0$ term is killed by $\int_{C_{\epsilon}^+}$. Moreover by completing these contours with large semicircles, you can derive ($l > 0$): $$ \int_{C_{\epsilon}^{\mp}} \frac{e^{\pm i l x}}{x^n} dx = \mp 2 \pi i \frac{(\pm i l)^{n-1}}{(n-1)!} $$ Summing everything up and noticing that there is no $\epsilon$ dependence, and keeping track of signs (which I failed to do on a first pass) we've shown that, $$ I = \frac{\pi }{2^{n-1} (n-1)!} \sum_{l = 0}^{(n-1)/2} (-1)^{n-1-l}\left(\begin{array}{c}n \\ l \end{array} \right) (n-2l)^{n-1} $$ I hope that wasn't too much (or too little).