I would like to ask if a close form (or even approximations) of the following inverse Laplace transform is know,
$$ {\cal L}^{-1}\left[\frac{1}{(s+1)(s+a)-ae^{-sL}}\right] = \;? $$
One usual thing to do is to identify the poles of the function and the inverse Laplace integral $\int_{\gamma - i\infty}^{\gamma +i\infty}$ just give the sum of residues. However the function above has infinitely many poles and they are not elementary functions of $a$ and $L$.
Highly appreciate your help!
One possibility is the use of Padé approximant for $e^{-sL}$ as
$$ e^{-sL} \approx \frac{1-\frac{L s}{2}+\frac{L^2 s^2}{10}-\frac{L^3 s^3}{120}+\cdots }{1+\frac{L s}{2}+\frac{L^2 s^2}{10}+\frac{L^3 s^3}{120}+\cdots} $$
depending on the value of $L$ regarding $\{1, a\}$the expansion can be simply
$$ e^{-sL} \approx \frac{1-\frac{L s}{2} }{1+\frac{L s}{2}} $$