Inferring asymptotic behaviour from the dominant pole of the Laplace transform

Hi,

I am reposting the following question with the hope that a more detailed description will lead to a more descriptive response:

dominant pole in the laplace transform

I have a vector function $X(t)=(X_1(t),X_2(t),...,X_n(t))$ whose components satisfy a delay-difference equation and whose Laplace transform $\hat{X}(s)$ has a unique pole $x_0$ of largest real part with $x_0$ on the real axis.\

I am able to show that the poles of $\hat{X}(s)e^{st}$ are confined to the vertical strip $\{z \in \mathbb{C}:-R\leq \textrm{Re}(z) \leq x_0\}$, for some $R>0$, and that for each $t$, \begin{equation*} X_i(t) = \lim_{n\rightarrow\infty}\sum_{s_j \in \Gamma_n}\textrm{Res}(\hat{X}_i(s)e^{st};s_j) \end{equation*} where $(\Gamma_n)_{n=1,2,...}$ is a sequence of contours that expands to fill the region

$\{ z \in \mathbb{C}:\textrm{Re}(z) < x_0+\epsilon \}$

for some $\epsilon$, and $s_j \in \Gamma_n$ denotes a pole in the interior of the contour $\Gamma_n$. In any vertical strip just to the left of $x_0$, there are an infinite number of poles. If $x_0$ has order $m$ as a pole of $\hat{X}_i(s)e^{st}$, then $X_i(t)/(e^{x_0t}t^{m-1})$ should converge to a positive constant as $t\rightarrow\infty$, however I am having trouble showing that the contribution of the leading pole dominates the combined contribution of other poles. The way that I imagine doing this is in two parts:

1. Showing that the combined contribution of poles $s_j$ with $\textrm{Re}(s_j)\geq x_0-\delta$ are small

2. Showing that the combined contribution of the remaining poles is bounded by $Ce^{(x_0-\delta+\Delta)t}$, for some $C>0$, as $t\rightarrow\infty$. Here $\delta>0$ is a sufficiently small positive constant and $0 <\Delta<\delta$.

It appears that item 1 can be proved by bounding the density of poles near the line $\{z:\textrm{Re}(z)=x_0\}$. Item 2 seems more obvious. Is anyone aware of a general result from which item 2 follows? Thanks in advance.

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