Hello, I have read in many articles that the trivial zeros of the Riemann zeta function are only the negative even integers (-2, -4, -6, -8, -10, ...).(adsbygoogle = window.adsbygoogle || []).push({});

The reason why these are the only ones is that when substituting them in the functional equation, the function is 0 because sin([itex]\frac{x·\pi}{2}[/itex])=0.

My question is: why aren't positive even integers trivial zeros too?

The sinus of k·[itex]\pi[/itex] =0 with either k[itex]\in[/itex]Z positive or negative.

Remember that the functional equation is:

[itex]\zeta[/itex](x)=[itex]\zeta[/itex](1-x)·[itex]\Gamma[/itex] (1-x)·2[itex]^{x}[/itex]·[itex]\pi[/itex][itex]^{x-1}[/itex]·sin ([itex]\frac{x·\pi}{2}[/itex])

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# Trivial zeros in the Riemann Zeta function

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