residue theorem over the simples zeros of riemann zeta

Asked Feb 26 '17 at 20:22

Active Feb 26 '17 at 20:22

Viewed 56 times

given the circle integral

$$ \oint \frac{\zeta (2s)}{\zeta (s)} $$

if taken only over the reidues due to the simple zeros -2,-3-6 etc

i get the sum $$ \sum_{n=1}^{\infty} \frac{\zeta (-4n)}{\zeta '( -2n)} 2\pi i $$

which is 0 but what did i make wrong in the evaluation ?

i have omitted the term over the nontrivial zeros because i know how to evaluate it

asked Feb 26 '17 at 20:22

Jose Garcia

2

The singularities over the negative integers should be removable singularities, so no problem here... – Simply Beautiful Art Feb 26 '17 at 20:39
$\zeta(2s)/\zeta(s)$ is holomorphic on $Re(s) < 0$ so $\int_\gamma \frac{\zeta(2s)}{\zeta(s)} ds = 0$ for any Jordan curve $\gamma$ – reuns Feb 26 '17 at 20:50
does the same happens to $ \frac{\zeta (s) \zeta (s-1) \zeta (s-2)}{\zeta (2s-2)} $ – Jose Garcia Mar 02 '17 at 12:21

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