Find the residue of ζ(2s) and ζ(6s)

Question

I know that the residue of the Riemann zeta function is 1. I'm wondering how one would calculate the residue if you have ζ(2s) and ζ(6s) as the pole has shifted to 1/2 and 1/6 respectively. Also do the trivial zeros change? So ζ(2s) has trivial zeros at s=-1,-2,-3,...?

Answer 1

You are correct about the placement of the trivial zeros of $\zeta(2s).$ This could be seen, for example, from the functional equation (i.e. reflection formula) of $\zeta(2s)$:

$$\zeta(2s)=\zeta(1-2s)\Gamma(1-2s)\sin(πs)π^{2(s-1)}4^s$$ which follows from the functional equation of $\zeta$ after some algebraic massage. This thing goes to 0 precisely when $s=-n\quad\land\quad 1\le n\in\mathbb{N}.$ So yes, they do in fact change values.

The poles of $\zeta(2s)$ and $\zeta(6s)$ are, in fact, simple, and as you have stated are $\frac{1}{2}$ and $\frac{1}{6}$ respectively, so we have the residue of $f$ at a simple pole $x_0$: $\lim_{z\to x_0}(z-x_0)f(z)$ and as to your question, the residues could be found by:

$\lim_{z\to\frac{1}{2}}(z-\frac{1}{2})\zeta(2z)$

$\lim_{z\to\frac{1}{6}}(z-\frac{1}{6})\zeta(6z)$

respectively.