What completes this analogy?
This: $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)\;\;\;\;\;\;\;\;\;\;(1)$$
is to:
$$\chi(s)=\pi ^{-\frac{s}{2}} \Gamma \left(\frac{s}{2}\right) \zeta (s)\;\;\;\;\;\;\;\;\;\;(2)$$
as:
$$\left(1-\frac{1}{a^{s-1}}\right)=-\frac{a^{-s} (-a + a^s)}{(-1 + a^s)} \times \left(1 - \frac{1}{a^{1-s-1}}\right)\;\;\;(3)$$
is to what? $(4)$
Repeat: $(1)$ is to $(2)$ as $(3)$ is to $(4)$.
In equality $(3)$ and $(4)$ $$p(s)=\left(1-\frac{1}{a^{s-1}}\right)$$ should remain a separate factor, so equality (3) has the abbreviated form:
$$p(s)=-\frac{a^{-s} (-a + a^s)}{(-1 + a^s)} \times p(1 - s)\;\;\;(3)$$
The important criteria is that the property $$\chi(s)=\chi(1-s)$$ satisfied by $(2)$, also should hold for $(4)$, the what part.
(*Mathematica 8 for equality (3)*)
s = 1.2345 - 5.4321*I;
a = 3
N[-a^-s (-a + a^s)/(-1 + a^s) 2^s \[Pi]^(s - 1)
Sin[(\[Pi] s)/2] Gamma[1 - s] Zeta[1 - s] (1 - 1/a^(1 - s - 1))]
N[Zeta[s]*(1 - 1/a^((s - 1)))]
N[-a^-s (-a + a^s)/(-1 + a^s) (1 - 1/a^(1 - s - 1))]
N[(1 - 1/a^((s - 1)))]