(TLDR: Need an asymptotic expansion for the square root which works at the origin, specifically. The classical Taylor series won't do: i need it to be exact at zero, outside of which it can have mediocre convergence. I have found an expression that seems to fulfill my needs, but can't justify it on rigorous grounds.)
I am working on the control of a system represented by the following expression in the Laplace Domain:
$$\exp\left(-\sqrt{s^2+2as}\right)$$
It has the obvious asymptote $\exp(-s-a)$ away from the origin. Unfortunately, however, it is exactly its behaviour around zero that interests me. I would like to address it with a Laurent series of some type, which should be exact there (like a Taylor series around that point would be).
The problem is clearly the square root, which is ill-behaved around zero. I am well aware that one might easily find the following series expansion:
$$\sqrt{s^2+2as} = (s+a)\sqrt{1 - \frac{a^2}{(s+a)^2}} = (s+a) \sum \limits_{k=0}^{\infty} {\frac{1}{2} \choose k}\frac{(-a^2)^{k}}{(s+a)^{2k}} $$
The issue: it converges extremely poorly for $s=0$. Do note that this is not a matter of choosing a good expansion point or whatnot. The branch cut at the origin basically means that any Taylor or Laurent-like expansion is bound not to decently represent the function there.
However, it should be possible to derive some asymptotic expansion for this square root which is exact at the origin, takes a small detour because of the awful derivatives there, and comes back on track. Hell, I've tried a few myself:
$$\sqrt{s^2+2as} \sim s+a - \frac{1}{N}\sum_{n=1}^N \frac{a^{n+1}}{(s+a)^{n}}$$
Which fits very well for N small. However, i have no clue on how to justify such a thing. I have tried most usual methods on series and a few methods on divergent asymptotic series, but nothing seems to work.
So. How would one go about finding an asymptotic expansion for this square root (or better yet, its exponential) which works at the origin?