In the Basel problem, the $sinc$ function is considered at the Wikipedia page.
Let me try to make an alternative function definition:
$$f(x) = \prod_{n=1}^\infty \left ( 1+ \frac{x^3}{n^3} \right ) = \prod_{n=1}^\infty \left ( 1+ \frac{x}{n} \right ) \left ( 1+ \frac{x}{n}\omega \right )\left ( 1+ \frac{x}{n} \omega^2 \right )$$ where $\omega = \exp(\frac{2 \pi i}{3})$ which after expanding to power series is equal to :
$$f(x) = \sum_{k=0}^\infty x^{3k}\zeta_k(3)$$
where we have defined:
$$\zeta_k(3) = \sum_{1\le n_1 <n_2 \cdots <n_k} \left(\frac{1}{n_1 n_2 \cdots n_k} \right) ^3$$
which can be seen as an infinite elementary symmetric polynomial in $1/1^k,1/2^k,1/3^k,\cdots$.
After looking at the properties of the sinc function on Wikipedia, I saw that it satisfies the Euler reflection formula.
My guess and question is, that the here defined $f(x)$ function satisfies a similar formula:
$$f(x) = \frac{1}{\Gamma(1+x)\Gamma(1+\omega x) \Gamma(1+\omega^2 x)}$$
I have done a series expansion of this last conjectured equality and I get in Sagemath:
$$x^{3} \zeta(3)$$
nothing new for $\zeta(3)$ at this step
$$\zeta_2(3) x^6 = -\frac{1}{1890} \, {\left(\pi^{6} - 945 \, \zeta(3)^{2}\right)} x^{6}$$
hence :
$$\zeta_2(3) = -\frac{1}{1890} \, {\left(\pi^{6} - 945 \, \zeta(3)^{2}\right)}$$
from which it follows by solving for $\zeta(3)$ that:
$$\zeta(3) = \sqrt{\frac{1890 \zeta_2(3)+\pi^6}{945}}$$
and similar:
$$\zeta_3(3) = -\frac{1}{1890} \, {\left(\pi^{6} \zeta(3) - 315 \, \zeta(3)^{3} - 630 \, \zeta(9)\right)}$$
$$\zeta_4(3) = -\frac{1}{5108103000} \, {\left(667 \, \pi^{12} + 1351350 \, \pi^{6} \zeta(3)^{2} - 212837625 \, \zeta(3)^{4} - 1702701000 \, \zeta(9) \zeta(3)\right)}$$
$$\zeta_5(3) = -\frac{1}{5108103000} \, {\left(667 \, \pi^{12} \zeta(3) + 450450 \, \pi^{6} \zeta(3)^{3} + 900900 \, \pi^{6} \zeta(9) - 42567525 \, \zeta(3)^{5} - 851350500 \, \zeta(9) \zeta(3)^{2} - 1021620600 \, \zeta(15)\right)}$$
etc.
I have four questions:
- Is the conjectured "reflection" equality true?
- Is there a "closed formula" for the $\zeta_k(3)$ or are the coefficients known numbers?
- Can this procedure be generalized to say $(1-\frac{x^l}{n^l})$?
- Is there anything known about the function $f(x)$?
Edit: By the comment of @TerryTao one can compute in a similar fashion:
If one starts with the Weierstrass factorisation $$ \Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty (1 + \frac{z}{k})^{-1} e^{z/k}$$ of the Gamma function, applied to $z = x, \omega x, \omega^2 x$ (where $\omega = e^{2\pi i/3}$ is the cube root of unity), and multiplies the three together, one obtains $$ \Gamma(x) \Gamma(\omega x) \Gamma(\omega^2 x) = \frac{1}{x^3} \prod_{k=1}^\infty (1+\frac{x^3}{k^3})^{-1}$$ and hence $$\prod_{k=1}^\infty (1+\frac{x^3}{k^3}) = \frac{1}{x^3 \Gamma(x) \Gamma(\omega x) \Gamma(\omega^2 x)}.$$
But using the formula $\Gamma(1+x) = x\Gamma(x)$ we obtain for $x=z,\omega z, \omega^2 z$:
$$\Gamma(1+z) = z \Gamma(z)$$ $$\Gamma(1+\omega z) = \omega z \Gamma(\omega z)$$ $$\Gamma(1+\omega^2 z) = \omega^2 z \Gamma( \omega^2 z)$$
and multiplying these we get:
$$ \frac{1}{\Gamma(1+x)\Gamma(1+\omega x) \Gamma(1+\omega^2 x)} = \frac{1}{x^3 \Gamma(x) \Gamma(\omega x) \Gamma(\omega^2 x) } = \prod_{k=1}^\infty (1+\frac{x^3}{k^3}) = f(x)$$
which proves the "reflection" formula and shows how to do this in the "general" case
$$f_l(x) = \prod_{n=1}^\infty (1+\left ( \frac{x}{n} \right )^l),l\ge 3$$
The problem to deduce the Taylor coefficients from the "reflection formula" is however not clear to me how to solve it. If someone would shed some light into this problem, that would be very nice. Thanks for your help!
