
#1
Oct811, 11:46 AM

P: 1

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, ...).
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](1x)·[itex]\Gamma[/itex] (1x)·2[itex]^{x}[/itex]·[itex]\pi[/itex][itex]^{x1}[/itex]·sin ([itex]\frac{x·\pi}{2}[/itex]) 



#2
Oct811, 01:06 PM

P: 1,666

At the even integers, the simple poles of [itex]\Gamma(1z)[/itex] are canceled by the simple zeros of [itex]\sin(\pi z/2)[/itex] and since the poles and zeros are of the same order (simple), this cancelation is nonzero, that is, the singularity is a removable one. For example consider the limit:
[tex]\lim_{x\to 4} \; \Gamma(1x) \sin(\pi x/2)=\frac{\pi}{12}[/tex] 



#3
Oct2511, 04:34 PM

P: 150

also because at the positive even integers, the zeta function is defined the Dirichlet series 1+1/2^s+1/3^s+1/4^s+... which converges for all positive even numbers.



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