I want to ask the inverse Laplace transform of the following function:
$$F(s) = \frac{1}{s \cdot (1 + a \cdot s)^{m} \cdot (1 + b \cdot s)^{m-k}} \cdot \Bigl[\exp{(\frac{- c \cdot s}{ 1 + b \cdot s } )}\Bigr]^{m-k}$$
where $a,b,c,m,k$ are all positive constants. $m$ and $k$ are integers with $m\ge k$.
It is possible to get an approximate solution using the Euler summation-based technique. But I want to have an exact solution for the inverse transform.
Some similar questions but in a much less simple form are also found on Stackexchane, i.e., Inverse Laplace Transform of e$^{-c \sqrt{s}}/(\sqrt{s}(a - s))$ or find the inverse Laplace transform of complex function . Especially I am wondering how to deal with the point $s = \frac{1}{b}$ since it is a pole of order $(m-k)$ but also appears in the exponential function and also I am not sure about the contour integral considering that $s$ exists in both the nominator and denominator of the argument of the exponential function.
Thanks a lot.
It might be easier to solve by series approach:
Hint:
$\mathcal{L}^{-1}\left\{\dfrac{\left(e^\frac{-cs}{1+bs}\right)^{m-k}}{s(1+as)^m(1+bs)^{m-k}}\right\}$
$=\mathcal{L}^{-1}\left\{\dfrac{e^\frac{c(k-m)s}{1+bs}}{s(1+as)^m(1+bs)^{m-k}}\right\}$
$=\mathcal{L}^{-1}\left\{\sum\limits_{n=0}^\infty\dfrac{c^n(k-m)^ns^{n-1}}{n!(1+as)^m(1+bs)^{m+n-k}}\right\}$