# Thread: trivial zeros of the Riemann zeta function

1. ## trivial zeros of the Riemann zeta function

I have read that the negative even integers are zeros for the Riemann zeta function because the function satisfies the equation
and at s being a negative even integer, the sin(pi*s/2) part vanishes. This argument would mean that the positive even integers would also be zeros for the same reason, except that the gamma function is only defined for positive integers, because the following integral would (and here is my question) diverge for negative real z. Is this correct?
Thanks.

Yes, $\displaystyle \Gamma(z)$ has a pole of order $\displaystyle 1$ at the negative integers and $0$. and $\displaystyle \sin$ is going to have a zero of order $1$ at the same spots. So these will indeed cancel out.