Sunday, 29 January 2017

Trivial zeros of the Riemann Zeta function



A question that has been puzzling me for quite some time now:




Why is the value of the Riemann Zeta function equal to 0 for every even negative number?



I assume that even negative refers to the real part of the number, while its imaginary part is 0.



So consider 2 for example:



f(2)=n=11n2=112+122+132+=12+22+32+=



What am I missing here?


Answer



The Zeta function is defined as ζ(s)=n1ns only for sC with (s)>1!



The function on the whole complex plane (except a few poles) is the analytic continuation of that function.



On the Wikipedia page, you can find the formula:
ζ(s)=2s1s12s0sin(sarctant)(1+t2)s2(eπt+1)dt



for s1. Maybe working on this integral for s a negative integer will give you the result.


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