How to calculate the inverse Laplace transform of $F(s) = \frac{1}{1+se^{s}}$? How to calculate the inverse Laplace transform of 
$F(s) = \frac{1}{1+se^{s}}$
 A: *

*For $|s e^{s}| > 1 $  : $$F(s) = \frac{1}{1+s e^s} =\frac{\frac{ e^{-s}}{s}}{1+\frac{ e^{-s}}{s}} =  -\sum_{k=1}^\infty (-1)^{k} s^{-k} e^{-sk} = -\sum_{k=1}^\infty(-1)^k \int_{0}^\infty e^{-st}  \frac{(t-k)^{k-1}1_{t > k}}{(k-1)!} dt$$
a proof of $\int_0^\infty e^{-st} t^{k-1} dt = s^{-k} (k-1)!$ being there, and $$\int_0^\infty e^{-st} (t-k)^{k-1} 1_{t > k} dt= \int_0^\infty e^{-s(t+k)} t^{k-1} dt = e^{-sk} s^{-k}(k-1)!$$
Let $f(t) = -\sum_{k=1}^\infty(-1)^k  \frac{(t-k)^{k-1}1_{t > k}}{(k-1)!}$ then
$$|f(t)| <  1_{t > 0} \sum_{k=1}^\infty  \frac{t^{k-1}}{(k-1)!} = e^t  1_{t > 0}$$  thus for $Re(s) > 1$: $ \int_0^\infty e^{-st} f(t) dt$ converges, and since $|s e^s| > 1$ it is equal to $$-\sum_{k=1}^\infty (-1)^{k} s^{-k} e^{-sk} = F(s)$$.  Altogether it means $F(s) = \mathcal{L}[f](s), \ \  Re(s) > 1$ and $f(t) =  \mathcal{L}^{-1}[F(s)](t)$.

*For $|s e^{s}| < 1$ it becomes the Laplace transform of a (non tempered) distribution : $$F(s) = \frac{1}{1+s e^s}  = \sum_{k=0}^\infty (-1)^{k} s^k e^{sk} = \int_{-\infty}^\infty e^{-st} \sum_{k=0}^\infty \delta^{(k)}(t+k) dt$$
