What are the branches of the $p$-adic zeta function? I'm reading the book  $p$-adic Numbers, $p$-adic Analysis, and Zeta-Functions by Neal Koblitz. In it, Koblitz wants to iterpolate the Riemann Zeta function for the values $\zeta_p(1-k)$ with $k \in \mathbb{N}$, where $\zeta_p(s) = (1-p^s)\zeta(s)$. Doing this, he defines the next function:
$$ \zeta_{p,s_0}(s) = \frac{1}{\alpha^{-(s_0+(p-1)s)}-1} \int_{\mathbb{Z}_p^*}x^{s_0+(p-1)s-1}\mu_{1,\alpha},$$ with $s_0 \in \{0,1,2,\ldots,p-2\}$ and $s \in \mathbb{Z}_p$.
He then says that this $\zeta_{p,s_0}$ are 'branches' of $\zeta_p$. My question is: how does $\zeta_{p,s_0}$ define a branch ? $\zeta_{p,s_0}$ generally does not interpolate the values $\zeta_p(1-k)$ even for $k \equiv s_0 \bmod (p-1)$, since we have 
$$k =s_0+k_0(p-1) \Rightarrow \zeta_p(1-k) = \zeta_{p,s_0}(k_0)$$ and not $\zeta_p(1-k)  = \zeta_{p,s_0}(1-k)$. So my question is really: what are the real branches of the $p$-adic zeta function ?
 A: According to the idea of Tate's thesis, the Riemann zeta function $\zeta(s)$ should be thought of as a distribution on the idèles $\text{GL}_1(\mathbf A_\mathbf Q)$ of $\mathbf Q$. Integrating this distribution against the complex characters of $\text{GL}_1(\mathbf A_\mathbf Q)$ gives the values of the zeta function.
Iwasawa showed that the $p$-adic Riemann zeta function constructed by Kubota and Leopoldt admits a similar construction. The object which plays the role of the idèles is $\mathbf Z_p^\times$; the role of complex characters of the idèles is played by the $p$-adic characters in the weight space $W=\text{Hom}_{\text{cont}}(\mathbf Z_p^\times, \mathbf Q_p^\times)$ (continuous homomorphisms). This group is canonically isomorphic to $\mathbf Z/(p-1)\mathbf Z \times \mathbf Z_p$. The zeta function can therefore be seen as a $p$-adic valued function on $\mathbf Z/(p-1)\mathbf Z \times \mathbf Z_p$. The various restrictions to $s \in \mathbf Z/(p-1)\mathbf Z$ are the "branches" of the zeta function as you are seeing them in Kobliz's book.
Edit: I have changed this answer, so Sanchez's comment is not related. (What I had before was true but I don't think it was the best way to see this.)
