Is the functional equation for $\zeta (s) \left(1-\frac{1}{3^{s-1}}\right)$ known?

Question

It says in wikipedia that Hardy gave a simple proof of the functional equation for:

$$\eta(s)=\zeta (s) \left(1-\frac{1}{2^{s-1}}\right)$$

and that it is:

$$\eta(-s) = 2 \frac{1-2^{-s-1}}{1-2^{-s}} \pi^{-s-1} s \sin\left({\pi s \over 2}\right) \Gamma(s)\eta(s+1)$$

Trying to generalize this to the von Mangoldt function, I am wondering as a first step if the functional equation for:

$$\zeta (s) \left(1-\frac{1}{3^{s-1}}\right)$$

is known?

Answer 1

Assume $a$ is a real number such that $a^{s-1}\neq1$. If you set $$ z(s)=\zeta (s) \left(1-\frac{1}{a^{s-1}}\right), $$ then, using the functional equation for $\zeta$, you obtain $$ \begin{align} z(1-s)&=\zeta (1-s) \left(1-\frac{1}{a^{-s}}\right)\\\\ &=2\frac{\Gamma(s)}{(2\pi)^s} \cos\left(\frac{\pi}{2}s\right) \zeta(s)\times \left(1-\frac{1}{a^{-s}}\right)\\\\ &=2\frac{\Gamma(s)}{(2\pi)^s} \cos\left(\frac{\pi}{2}s\right) \frac{1-a^s}{1-\frac{1}{a^{s-1}}} \times \zeta(s)\times \left(1-\frac{1}{a^{s-1}}\right)\\\\ &=-2a^s\frac{\Gamma(s)}{(2\pi)^s} \cos\left(\frac{\pi}{2}s\right) \frac{1-a^{-s}}{1-a^{1-s}} \times z(s)\\\\ \end{align} $$ or, with $s \to s+1$,

$$ \color{blue}{z(-s)}=2\:\left(\frac{a}{2}\right)^{s+1}\frac{1-a^{-s-1}}{1-a^{-s}}\pi^{-s-1}s\sin\left(\frac{\pi s}{2}\right)\Gamma(s) \:\color{blue}{z(s+1)}. $$