# Euler's Basel problem continued… $\zeta(2n)$ expressed in terms of $sinc$?

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? $$\zeta(u)=(2\pi)^{\frac{2u}{\pi}} \; \frac{u}{\pi} sinc(u)\;\Gamma(1-\frac{2u}{\pi}) \zeta(1-\frac{2u}{\pi})$$

• Related if not duplicate. math.stackexchange.com/questions/240993 – user17762 May 28 '13 at 6:49
• The link is only far related as not focussing really on sinc. My background is that sinc is a function that connects many different areas around number theory and physics. And a description of Zeta even in terms of sinc interesting. – al-Hwarizmi May 28 '13 at 7:25
• It does focus on $\operatorname{sinc}$; I don't see how it could focus on it any more than that. – joriki May 29 '13 at 11:00
• Joriki - I can not see this in the equation (*) – al-Hwarizmi May 30 '13 at 20:54