I've got the following problem:
Let's imagine we've got a laplace transform
$$\hat f(\lambda ) = {{(1 - \beta ){{1 - \hat \varphi (\lambda )} \over \lambda }} \over {1 - (1 - \beta )\hat \varphi (\lambda )}}$$ where ${\hat \varphi (\lambda )}$ is some function.
I want to find asymptotic expression for $f(t)$ for $t \to + \infty $ if I know asymptotic of $\hat \varphi (\lambda )$ for $\lambda \to 0$.
For example, if we have
$$\varphi (t) = \mu {e^{ - \mu t}}$$
then,
$$\hat \varphi (\lambda ) = {\mu \over {\lambda + \mu }}$$
and,
$$\hat f(\lambda ) = {{1 - \beta } \over {\lambda + \beta \mu }}$$$$\Rightarrow f(t) = (1 - \beta ){e^{ - \mu \beta t}}$$
But if I try to use expansion:
$$\hat \varphi (\lambda ) = {\mu \over {\lambda + \mu }} \approx 1 - {\mu ^{ - 1}}\lambda + ...$$
I get
$$\hat f(\lambda ) = {{1 - \beta } \over {(1 - \beta )\lambda + \beta \mu }}$$$$\Rightarrow f(t) = {e^{ - {{\mu \beta } \over {1 - \beta }}t}}$$
with a wrong exponentioal multiplier.
Of coure I understand that we can't get the full right expression, but in this case we haven't even got the right asymptotic at $t \to + \infty $.
Is there a right way to get it?
Knowing the behavior of $\hat \varphi$ at zero is not sufficient. If the Bromwich integral can be evaluated by applying the residue theorem, the asymptotic behavior of $f$ at infinity will be determined by the rightmost singularity of $\hat f$. The poles of $\hat \varphi$ won't be singularities of $\hat f$, but the points where $\hat \varphi = 1/(1 - \beta)$ will be simple poles. In your example there is a single pole at $\lambda = -\beta \mu$.