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

Question

How to calculate the inverse Laplace transform of

$F(s) = \frac{1}{1+se^{s}}$

Answer 1

• 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 : $$F(s) = \frac{1}{1+s e^s} = \sum_{k=0}^\infty (-1)^{k} s^k e^{sk} = \int_{-\infty}^\infty e^{-st} \sum_{k=0}^\infty \delta^{(k)}(t+k) dt$$