What is $\sum_{k=1}^\infty \rm{sinc}^8(k)$ using the sine cardinal function?

Question

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,

$$\sum_{k=1}^\infty \rm{sinc}^7(k)=-\frac12+\frac{7\cdot29^2\,\pi}{2^5\,6!}+\frac{\pi\big(\pi-\tfrac72\big)^6}{6!}$$

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Questions:

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

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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.

Answer 1

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.

Answer 2

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.

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Contour Integration

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

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Computation

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

Answer 3

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.

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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.