I have to make a brief intro before comming to my question.
To approach the famous Basel problem Euler starts with the $sinc$ function
\begin{align}\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots=\left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots\end{align}
He then multiplies out and collects the $x^2$ of the $sinc$ function
$$-\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots \right) =-\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}$$
landing
$$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.$$
This is the birthday of the $\zeta$ function for evens, with the general form by Euler
$$\zeta(2n)=\frac{(2\pi)^{2n}(-1)^{n+1}B_{2n}}{2\cdot(2n)!} \;(*)$$
Unfortunatley seems Euler never returns to the $sinc$ function, that was the origin.
How could we express $\zeta(2n)$ (equation *) elegantly in terms of $sinc$? Or is this not possible?
Thanks
PS: possibly helpful? $$\zeta(u)=(2\pi)^{\frac{2u}{\pi}} \; \frac{u}{\pi} sinc(u)\;\Gamma(1-\frac{2u}{\pi}) \zeta(1-\frac{2u}{\pi})$$
