0$, so that it falls into the error term of the expression \eqref{eqtheoaebde} we are going to obtain for $P_n$. Otherwise we may consider the maximal pair $(u,v)$ (with respect to lexicographic order) for which this function is not holomorphic; then \eqref{contribexcep} is equal to $$ \frac{ \omega'_{u,v}}{2 i \pi} \int_{C_R} \frac{(\rho-z)^{T}\log(\rho-z)^{v }}{z^{n+1}} \cdot (1+o(1))\dd z $$ for some $T\in\QQ$ and $ \omega'_{u,v} \in \Qbaretoile \omega_{u,v}\subset \Gamma(\QQ) \cdot \GG$ (using Assertion (iii) of Theorem~\ref{thintrocce}). We obtain finally the following formula for \eqref{contribexcep} (see \cite[p.\,387]{Bible}): $$ \begin{cases} \dfrac{ \omega'_{u,v}}{\Gamma(-T)}\,\rho^{T-n} n^{-T-1}\log(n)^{v} (1+o(1))& \textup{if } T\not\in\NN, \\[8pt] \displaystyle\omega'_{u,v} \rho^{T-n} n^{-T-1} \log(n)^{v -1} (1+o(1))&\textup{if $T\in\NN$ (so that $v\geq 1$).} \end{cases} $$ This contribution can either fall into the error term of \eqref{eqtheoaebde}, or give a term with $ \capa_1= \ldots = \capa_{d-1}= \te_1=\ldots = \te_{d-1}=0$. Let us study now the terms in \eqref{eqpnde} for which $ \alpha_k\neq 0$; since $E(z)$ is not a polynomial there is at least one such term. The function $$ \frac{e^{\alpha_k B(z)}}{z^{n+1}}\cdot A(z)B(z)^{u_k}\log(B(z))^{v_k} $$ is smooth on $C_R $ (except on the cuts of $\log(B(z))$) and the integral can be estimated as $n\to \infty$ by finding the critical points of $\alpha_k B(z)-n \log(z)$, \ie the solutions of $zB'(z)=n/\alpha_k$. For large $n$, any critical point $z$ must be close to $\rho$ (since $zB'(z)$ is bounded away from $\rho$ for $|z| \leq |\rho|$). Now in a neighborhood of $z=\rho$ we have $$ zB'(z) \equi -\frac{\rho \tau \mathfrak{B}}{\sigma} \cdot \frac{1}{(z-\rho)^{1+\tau/\sigma}}, $$ so that we have $\tau+\sigma$ critical points $z_{j,k}(n)$, for $j=0, \ldots, \sigma+\tau-1$, with $$ z_{j,k}(n) - \rho \equi e^{2 i \pi j\sigma/(\sigma+\tau)}\cdot \Bigl(-\frac{\sigma n}{\rho \mathfrak{B} \tau \alpha_k}\Bigr)^{-\sigma/(\sigma+\tau)}. $$ Using \eqref{eqasyab} and letting $\kappa=t/s \in \QQ$ we deduce that $$ A(z_{j,k}(n)) \equi \mathfrak{A} e^{2 i \pi j\sigma\kappa/(\sigma+\tau)}\cdot \Bigl(-\frac{\sigma n}{\rho \mathfrak{B} \tau \alpha_k} \Bigr)^{-\sigma\kappa /(\sigma+\tau)} \neq 0. $$ Moreover we have $$\alpha_k B(z_{j,k}(n)) \equi \frac{-\sigma}{\tau} ( z_{j,k}(n) - \rho) \alpha_k B'(z_{j,k}(n)) \equi \frac{-\sigma n}{\rho \tau} ( z_{j,k}(n) - \rho) \equi \ddjk n^{\tau/(\sigma+\tau)}$$ with\vspace*{-.3\baselineskip} \begin{equation} \label{eqdefdk} \ddjk = \Big( \alpha_k \mathfrak{B} e^{2i\pi j}\Big)^{\sigma/(\sigma+\tau)} \Big(\frac{-\sigma}{\rho \tau}\Big)^{ \tau / (\sigma+\tau)}\neq 0. \end{equation} To apply the saddle point method, we need to estimate the second derivative $\Delta_{j,k}(n)$ of $\alpha_kB(z)-n\log(z)$ at $z=z_{j,k}(n)$. We obtain\vspace*{-.3\baselineskip}\enlargethispage{.1\baselineskip}% \begin{multline*} \Delta_{j,k}(n)=\alpha_kB''(z_{j,k}(n))+\frac{n}{z_{j,k}(n)^2}\\[-3pt] \equi \frac{\tau(\sigma+\tau)}{\sigma^2} (\alpha_k \mathfrak{B})^{-\sigma/(\sigma+\tau)} e^{-2i\pi j\frac{2\sigma+\tau}{\sigma+\tau}} \Bigl(-\frac{\sigma} {\rho \tau } \Bigr)^{\frac{ 2\sigma+\tau}{ \sigma+ \tau}} n^{\frac{ 2\sigma+\tau}{ \sigma+ \tau}}. \end{multline*} Finally,\vspace*{-.5\baselineskip} $$ B(z_{j,k}(n))^{u_k}\equi (\ddjk / \alpha_k)^{u_k} n^{\tau u_k / (\sigma+\tau)}. $$ This enables us to apply the saddle point method. This yields a non-empty subset $J_k$ of $\{0,\ldots,\sigma+\tau-1\}$ such that the term corresponding to $\alpha_k$ in \eqref{eqpnde} is equal to $$ \sum_{j\in J_k} \frac{ \omega_k }{\sqrt{2\pi \Delta_{j,k}(n)}}\, \frac{e^{\alpha_k B(z_{j,k}(n))}}{z_{j,k}(n)^{n +1}}\, A(z_{j,k}(n))B(z_{j,k}(n))^{u_k} \log(B(z_{j,k}(n)))^{v_k} (1+o(1)) .$$ Now for any pair $(j,k)$, $\alpha_k B(z_{j,k}(n))$ is an algebraic function of $n$ so that it can be expanded as follows as $n\to\infty$: \begin{equation} \label{eqasydk} \alpha_k B(z_{j,k}(n)) = \sum_{\ell=0}^{d'} \capa_{j,k,\ell} n^{\ell / d} + o(1), \end{equation} with $\capa_{j,k,\ell} \in \Qbar$, $0