What is $\sum_{k=1}^\infty \rm{sinc}^8(k)$ using the sine cardinal function? 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.
 A: Using Bernoulli polynomials, one can make a general formula:
$$S_n=\sum_{k=1}^{\infty}\frac{\sin^n k}{k^n}=-\frac{\pi^n}{2n!}\sum_{k=0}^{n}(-1)^k\binom{n}{k}B_n\left(\Big\{\frac{n-2k}{2\pi}\Big\}\right),$$
where $\{x\}=x-\lfloor x\rfloor$ denotes fractional part of $x$. Say, continuing the examples,
$$S_{10}=-\frac{1}{2}-\frac{1093\pi}{672}+\frac{5883\pi^2}{896}-\frac{2449\pi^3}{288}+\frac{563\pi^4}{96}\\-\frac{1423\pi^5}{576}+\frac{43\pi^6}{64}-\frac{103\pi^7}{864}+\frac{3\pi^8}{224}-\frac{\pi^9}{1152}+\frac{\pi^{10}}{40320}.$$
BTW, $n=7$ is the first with $n>2\pi$, which causes the complication.
A: This supplements metamorphy's accepted answer which allowed me to investigate higher $n$. Define,
$$I_n=\int_0^\infty \rm{sinc}^n(k)\,dk$$
$$F_n=\frac12-I_n+\sum_{k=1}^\infty \rm{sinc}^n(k)$$
We have $F_1 = F_2 = F_3 = F_4 = F_5 = F_6 = 0$. Then the simple evaluations,
$$I_7 = \frac{5887}{23040}\pi,\quad F_7 = \frac{\pi\, v^6}{6!},\quad v =\pi-\tfrac72$$
$$\;I_8 =\frac{151}{630}\pi,\quad F_8 = -\frac{\pi\, v^7}{7!},\quad v =\pi-\tfrac82$$
While the pattern for $I_n$ as a rational multiple of $\pi$ continues, the simple form of $F_n$ does not. 

The next $F_n$ are palindromic and near-palindromic,
$$F_9 = \frac{\pi}{2^5\,8!}\,P_0$$
$$P_0 = 1+10v+28v^2+70v^3+70v^4+70v^5+28v^6+10v^7+v^8$$
where $v= 2(\pi-4)$.
$$F_{10} = \frac{\pi}{9!}\big(1+3P_1\big)$$
$$P_1 = 3+30v+120v^2+280v^3+420v^4+420v^5+280v^6+120v^7+30v^8+3v^9$$ 
where $v = \pi-5$.
$$F_{11} = \frac{\pi}{10!}\big(11+15P_2\big)$$
$$P_2 = \small{3+36v+168v^2+432v^3+784v^4+\frac{4536}5v^5+784v^6+432v^7+168v^8+36v^9+3v^{10}}$$
and where $v = \pi-9/2$.
Note: Unfortunately, higher $n$ do not seem to have similar forms. The answer given by metamorphy does not immediately imply palindromic polynomials, so one can wonder why these appear.
