hi,


I have a function $X(t)$ whose Laplace transform $\hat{X}(s)$ has a unique pole of largest real part $x\_0$, which is a real number. I am able to show that for each $t$, $X(t)$ is a convergent sum of the residues of $\hat{X}(s)e^{st}$, over its poles. If $x\_0$ is a pole of order $m$, is it true in general that $X(t)/(t^{m-1}e^{x\_0t})$ converges as $t\rightarrow\infty$?



dominant pole in the laplace transform
Yes, this is true. Just write the Laplace transform as a sum of $a/(x-x\_0)^m$ and the rest.
The inverse Laplace of the first summand is your asymptotic term, and the inverse Laplace transform
of the rest is small.
(I suppose that your expression "unique pole of the largest real part" means that the real parts
of the other poles are at most $x\_0-\epsilon$).


