Trivial zeros in the Riemann Zeta function

[msariols]

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($\frac{x·\pi}{2}$)=0.

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

The sinus of k·$\pi$ =0 with either k$\in$Z positive or negative.Remember that the functional equation is:

$\zeta$(x)=$\zeta$(1-x)·$\Gamma$ (1-x)·2$^{x}$·$\pi$$^{x-1}$·sin ($\frac{x·\pi}{2}$)

 

[jackmell]

At the even integers, the simple poles of $\Gamma(1-z)$ are canceled by the simple zeros of $\sin(\pi z/2)$ 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:

$$\lim_{x\to 4} \; \Gamma(1-x) \sin(\pi x/2)=\frac{\pi}{12}$$

 

[camilus]

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.