# The sine cardinal function and $F_1 = F_2 = F_3 = F_4 = F_5 = F_6 = 0$

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,

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

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

and,

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

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

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.