1

If I plug in 4 for s into:

$$\begin{equation} \zeta(s)=2^s\pi^{s-1}\Gamma(1-s)\sin\left(\frac{\pi s}{2}\right)\zeta(1-s). \end{equation}$$

Doesn't $$\zeta(4) = 0 $$ because of the sine function?

Does this mean that the above functional equation has parameters for where it can be used?

zerosofthezeta
  • 5,646
  • 5
    Perhaps you missed the pole of $\Gamma(1-s)$ when $s = 4$. – Antonio Vargas Dec 14 '13 at 07:51
  • 2
    $\sum_n \frac{1}{n^4} = 0$? I hope not. – Mustafa Said Dec 14 '13 at 08:14
  • 1
    It's way more involved than what you think. As already commented, the gamma function has singularities at the negative integers, which together with the zero of of the sine function and all the rest gives you $;\zeta(4)=\frac{\pi^4}{90};$ ...but it is not trivial to get to this vale. – DonAntonio Dec 14 '13 at 09:33

0 Answers0