# In the functional equation for Riemann's zeta function, what happens at 2n, n natural

• Sep 27th 2011, 08:11 PM
In the functional equation for Riemann's zeta function, what happens at 2n, n natural
I am a bit confused on the functional equation for Riemann's zeta function,
$\displaystyle \zeta$(s) = $\displaystyle 2^{s}\pi^{s-1}sin(\frac{\pi s}{2})\Gamma (1-s) \zeta(1-s)$

for when s is a non-negative even integer. For example, when s = 2, then since
sin($\displaystyle \frac{\pi 2}{2}$) = 0 , or for s=0, sin(0) = 0
it would seem that
$\displaystyle \zeta$(s) = 0
which apparently is not true, for two reasons:
(a) $\displaystyle \zeta$(s) is supposed to agree, for s> 1, with the series
$\displaystyle \Sigma (n^{-s})$ from n = 1 to infinity,
but this latter sum will always be greater than zero for s>1 (to continue with the example of s=2, $\displaystyle \pi^{2}$/6 $\displaystyle \neq$ 0)
(b) non-negative even integers are not among the trivial zeros of the Riemann zeta function.

Another way to ask my question is why the functional equation =0 for negative even integers but not for non-negative even integers.