TITLE: What is $\sum_{k=1}^\infty \rm{sinc}^8(k)$ using the sine cardinal function?
QUESTION [8 upvotes]: Given the sine cardinal function,
$$\rm{sinc}(x) = \frac{\sin x}x$$
for $x\neq0$. We have the nice evaluations,
$$\sum_{k=1}^\infty \rm{sinc}(k) = \sum_{k=1}^\infty \rm{sinc}^2(k)=-\tfrac12+\tfrac12\pi$$
$$\sum_{k=1}^\infty \rm{sinc}^3(k)=-\tfrac12+\tfrac38\pi$$
$$\sum_{k=1}^\infty \rm{sinc}^4(k)=-\tfrac12+\tfrac13\pi$$
$$\sum_{k=1}^\infty \rm{sinc}^5(k)=-\tfrac12+\tfrac{115}{384}\pi$$
$$\sum_{k=1}^\infty \rm{sinc}^6(k)=-\tfrac12+\tfrac{11}{40}\pi$$
then the not-so-nice,
$$\sum_{k=1}^\infty \rm{sinc}^7(k)=-\tfrac12+\quad\\ \tfrac{1}{46080}(129423\pi-201684\pi^2+144060\pi^3-54880\pi^4+11760\pi^5-1344\pi^6+64\pi^7)$$
However, I found this can be prettified as,
$$\sum_{k=1}^\infty \rm{sinc}^7(k)=-\frac12+\frac{7\cdot29^2\,\pi}{2^5\,6!}+\frac{\pi\big(\pi-\tfrac72\big)^6}{6!}$$


Questions:


Why is the closed-form for $n=7$ far more complicated than $n<7$? (And a good lesson that "patterns" may be illusory.)
What is $n=8$ in terms of $\pi$? (Maybe also for $n=9$?)


Update: Courtesy of Oliver Oloa's comment, for $n=8$, after some tweaking is,
$$\sum_{k=1}^\infty \rm{sinc}^8(k)=-\frac12+\frac{151\pi}{630}-\frac{\pi\big(\pi-\tfrac82\big)^7}{7!}$$
but $n=9$ is more complicated. See second answer below.

REPLY [4 votes]: Fourier Analytic Approach
The Fourier Transform of $\frac{\sin(x)}x$ is
$$
f(x)=\pi\!\left[-\tfrac1{2\pi}\le\xi\le\tfrac1{2\pi}\right]\tag1
$$
This would mean that the Fourier Transform of $\frac{\sin^n(x)}{x^n}$ is $f_n(\xi)=\left(\ast^n\right)\!f(\xi)$, which is the convolution of $n$ copies of $f$.
The Poisson Summation Formula says that
$$
\sum_{k\in\mathbb{Z}}\frac{\sin^n(k)}{k^n}=\sum_{k\in\mathbb{Z}}f_n(k)\tag2
$$
The support of $f$ is $\left[-\frac1{2\pi},\frac1{2\pi}\right]$; therefore, the support of $f_n$ is $\left[-\frac{n}{2\pi},\frac{n}{2\pi}\right]$. Furthermore, since $f$ is even, $f_n$ is also. Thus,
$$
\sum_{k=1}^\infty\frac{\sin^n(k)}{k^n}=\frac{f_n(0)-1}2+\sum_{k=1}^{\left\lfloor\frac{n}{2\pi}\right\rfloor}f_n(k)\tag3
$$
For $n\le6$, the right side of $(3)$ is $\frac{f_n(0)-1}2$. For $7\le n\le12$, the right side of $(3)$ is $\frac{f_n(0)-1}2+f_n(1)$. For $13\le n\le18$, the right side of $(3)$ is $\frac{f_n(0)-1}2+f_n(1)+f_n(2)$. And so on.

Contour Integration
We can use contour integration to get
$$
\begin{align}
f_n(\xi)
&=\int_{-\infty}^\infty\frac{\sin^n(x)}{x^n}e^{-2\pi ix\xi}\,\mathrm{d}x\\
&=\int_{-\infty-i}^{\infty-i}\frac{\left(e^{ix}-e^{-ix}\right)^n}{(2ix)^n}e^{-2\pi ix\xi}\,\mathrm{d}x\\
&=\sum_{k=0}^n(-1)^k\binom{n}{k}\int_{-\infty-i}^{\infty-i}\frac{e^{i(n-2k-2\pi\xi)x}}{(2ix)^n}\,\mathrm{d}x\\
&=\sum_{k=0}^{\lfloor n/2-\pi\xi\rfloor}(-1)^k\binom{n}{k}2\pi\frac{(n-2k-2\pi\xi)^{n-1}}{2^n(n-1)!}\\
&=\frac{\pi}{2^{n-1}(n-1)!}\sum_{k=0}^{\lfloor n/2-\pi\xi\rfloor}(-1)^k\binom{n}{k}(n-2k-2\pi\xi)^{n-1}\tag4
\end{align}
$$

Computation
Applying $(4)$ to $(3)$, we can compute $\sum\limits_{k=1}^\infty\frac{\sin^n(k)}{k^n}$ for any $n$:
$$
\begin{array}{l|l}
n&\sum\limits_{k=1}^\infty\frac{\sin^n(k)}{k^n}\\\hline
1&\frac{\pi-1}2\\
2&\frac{\pi-1}2\\
3&\frac{3\pi-4}8\\
4&\frac{2\pi-3}6\\
5&\frac{115\pi-192}{384}\\
6&\frac{11\pi-20}{40}\\
7&\frac{5887\pi-11520}{23040}+\frac{\pi(7-2\pi)^6}{46080}\\
8&\frac{151\pi-315}{630}+\frac{\pi(4-\pi)^7}{5040}\\
9&\frac{259723\pi-573440}{1146880}+\frac{\pi(9-2\pi)^8}{10321920}-\frac{\pi(7-2\pi)^8}{1146880}\\
10&\frac{15619\pi-36288}{72576}+\frac{\pi(5-\pi)^9}{362880}-\frac{\pi(4-\pi)^9}{36288}\\
11&\frac{381773117\pi-928972800}{1857945600}+\frac{\pi(11-2\pi)^{10}}{3715891200}-\frac{11\pi(9-2\pi)^{10}}{3715891200}+\frac{11\pi(7-2\pi)^{10}}{743178240}\\
12&\frac{655177\pi-1663200}{3326400}+\frac{\pi(6-\pi)^{11}}{39916800}-\frac{\pi(5-\pi)^{11}}{3326400}+\frac{\pi(4-\pi)^{11}}{604800}
\end{array}
$$