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Trivial zeros in the Riemann Zeta function

  1. Oct 8, 2011 #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](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])
  2. jcsd
  3. Oct 8, 2011 #2
    At the even integers, the simple poles of [itex]\Gamma(1-z)[/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 non-zero, that is, the singularity is a removable one. For example consider the limit:

    [tex]\lim_{x\to 4} \; \Gamma(1-x) \sin(\pi x/2)=\frac{\pi}{12}[/tex]
  4. Oct 25, 2011 #3
    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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