Define the function,

$$F_n=\frac12-\int_0^\infty \frac{\sin^n x}{x^n}\,dx+\sum_{x=1}^\infty \frac{\sin^n x}{x^n}\tag1$$

where $\rm{sinc}^n(x)=\frac{\sin^n x}{x^n}$ is the sine cardinal function. We have

$$F_1 = F_2 = F_3 = F_4 = F_5 = F_6 = 0$$

Then suddenly,

$$\begin{align*} F_7 &= \frac{1}{46080}\Bigl(117649\pi-201684\pi^2+144060\pi^3\\ &\qquad\qquad\qquad-54880\pi^4+11760\pi^5-1344\pi^6+64\pi^7\Bigr)\tag2 \end{align*}$$

Fortunately, this and the next can be simplified as,

$$F_7 = \frac{\pi\big(\tfrac72-\pi\big)^6}{6!}$$

$$F_8 = \frac{\pi\big(\tfrac82-\pi\big)^7}{7!}$$

Courtesy of an insight from robjohn's answer, it turns out the rest have beautifully consistent forms,

$$F_9 = \frac{\pi\big(\tfrac92-\pi\big)^8}{8!}-\frac{9\pi\big(\tfrac72-\pi\big)^8}{8!}$$

$$F_{10} = \frac{\pi\big(\tfrac{10}2-\pi\big)^9}{9!}-\frac{10\pi\big(\tfrac82-\pi\big)^9}{9!}$$


$$F_{11} = \frac{\pi\big(\tfrac{11}2-\pi\big)^{10}}{10!} -\frac{11\pi\big(\tfrac92-\pi\big)^{10}}{10!} + \frac{11\pi\big(\tfrac72-\pi\big)^{10}}{2\times9!}$$

$$F_{12} = \frac{\pi\big(\tfrac{12}2-\pi\big)^{11}}{11!} -\frac{12\pi\big(\tfrac{10}2-\pi\big)^{11}}{11!} + \frac{12\pi\big(\tfrac82-\pi\big)^{11}}{2\times10!}$$

and so on.

Q: Are there other functions $G_n$ similar to $(1)$ such that $G_n = 0$ for the first few $n$, then for higher $n$ is suddenly a polynomial in $\pi$ (or some well-known constant)?

P.S. The reason I ask is the polynomials of $G_n$ might have their own consistent forms similar to the one given by robjohn. I faintly remember a family, but can't explicitly recall it for now.


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