is not an entire function: branch cuts replace the discrete zeros of $D(\lambda)$).

The meromorphic continuation of $I_q(s,\lambda)$ is thus important at $s=0$. If (for $\lambda > - \inf V$ initially) we compute the expansion $$\label{BET}
(V(q) + \lambda)^{-s+1/2} \sim 
\sum_\sigma \beta_\sigma (s) q^{\sigma-Ns} \quad 
{\rm for\ } q \to +\infty \qquad
(\sigma={N \over 2},\ {N \over 2} -1, \cdots)$$ (the $\beta_\sigma$ of course also depend on $\lambda$ and on the potential), then $$\label{IA}
I_q(s,\lambda) \sim 
-\sum_\sigma \beta_\sigma(s) {q^{\sigma+1-Ns} \over \sigma+1-Ns} 
\qquad (q \to +\infty) ,$$ hence it satisfies $$\label{Tres}
\lim_{s \to 0} s I_q(s,\lambda) = \beta_{-1}(0)/N = -Z(0,\lambda)/2;$$ $\beta_{-1}(s)$ is actually independent of $\lambda$ except for $N=2$; the latter value for the residue directly comes from eq.(\[ZC\]), making another explicit statement of the trace identity (\[Tr0\]).

We denote here $$\label{FPI}
{\mathcal I}_q(\lambda) \defi 
\lim_{s \to 0} \Bigl\{ I_q(s,\lambda) - {\beta_{-1}(0) \over Ns} \Bigr\} \qquad
(\mbox{the {\sl finite part\/} of } I_q(s,\lambda) \mbox{ at } s=0) .$$ Now, some of our earlier statements revolving around this quantity [@V0; @V6] require corrections in the most general setting $\beta_{-1}(s) \not\equiv 0$. Finite parts are also to be extracted: first on the expansion (\[IA\]), giving $$\label{I0}
{\mathcal I}_q(\lambda) \sim 
-\sum_{\sigma \ne -1} \beta_\sigma(0) 