Z(0) + \sum_{1 \le n \le \mu} \Res_{s=n} Z(s) \ {(-\lambda)^n \over n} .$$

### Spectral determinants [@V5]

Since $Z(s,\lambda)$ is regular at $s=0$, a [*spectral determinant*]{} can be defined by zeta-regularization, as $$\label{Det}
D(\lambda) \equiv \det(\hat H + \lambda) \defi 
\exp [- \partial_sZ(s,\lambda)]_{s=0} \ ;$$ it is an entire function of order $\mu$ with $\{-E_k\}$ as its set of zeros. Moreover, amidst all such functions, $D(\lambda)$ can be precisely picked out in at least two ways.

$\bullet$ On one side, eq.(\[tas\]) implies a [*canonical semiclassical*]{} behaviour for $D(\lambda)$, $$\begin{aligned}
\label{das}
-\log D(\lambda) &\sim& \sum_\rho c_\rho \Gamma_\rho(\lambda) 
\mbox{ for } \lambda \to +\infty, \qquad 
\Gamma_\rho(\lambda) \defi 
\partial_s \Bigl[{\G(s+\rho) \over \G(s)} \lambda^{-(s+\rho)} \Bigr]_{s=0}
\nonumber \\
\mbox{i.e.,} \quad \Gamma_\rho(\lambda) &=& \Bigl\{ \matrix{ 
\G(\rho) \lambda^{-\rho} & \mbox{if }-\rho \notin  \Bbb N \cr
(-(-\lambda)^m / m!) \bigl(\log\lambda-\sum_{r=1}^m 1/r \bigr)
& \mbox{if }-\rho =m \in \Bbb N } \Bigr.\end{aligned}$$ are the [*only*]{} terms allowed; no other type, including additive constants ($\propto \lambda^0$), can enter this expansion.

$\bullet$ Independently, $D(\lambda)$ is also fully specified by expansions around $\lambda=0$: firstly, in reference to the [*Fredholm determinant*]{} $\Delta(\lambda)$ (built as a Weierstrass infinite product): $$\begin{aligned}
\label{DTay}
D(\lambda) &\equiv& 
\exp \Bigl[ -Z'(0) -\sum_{1 \le n \le \mu}{{\tilde Z}(n) \over 