Inverse Laplace transform of $F(s) = \frac1{s+e^{-s\tau}}$ where $\tau\in\Bbb R_+$

Question

I'm looking for the inverse Laplace transform of $$F(s) = \frac{1}{s + e^{-s\tau}}$$ where $\tau$ is a positive real parameter.

I am trying to use general inverse formula of Laplace transformation to solve it. But then, I need to find the singularities of $F(s)$, that is, $s + e^{-s\tau} = 0$.

Transforming the equation, I can get $ \tau = \frac{\log(-s)}{(-s)}$. Seems that the number of singularities depends on the value of parameter $\tau$. Then, question comes to me, how to find the residue at those possible singularities? And then how to proceed the calculation for the general inverse formula?

Many thanks in advance for your advice.

Answer 1

We use the notation $u_+= 0$ for $u<0$ and $u_+=u$ for $u\geq 0.$

$$L(s)=\frac{1}{s\left(1+\frac{e^{-sa}}{s}\right)}=\sum_{n=0}^{\infty}(-1)^n\frac{e^{-nas}}{s^{n+1}}$$$$=\sum_{n=0}^{\infty}(-1)^n\int_0^{\infty}e^{-sy-sna}\frac{y^n}{n!} =\int_0^{\infty}e^{-sx}f(x)dx$$ with $$f(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}(x-na)_+^n.$$

Answer 2

$$\mathcal{L}_s^{-1}\left[\frac{1}{s+\exp (-s \tau )}\right](t)=\sum _{m=0}^{\infty } \frac{(-t+m \tau )^m \theta (t-m \tau )}{\Gamma (1+m)}$$

where: $\theta (t-m \tau )$ is HeavisideTheta function.

$$\mathcal{L}_s^{-1}\left[\frac{1}{s+\exp (-s \tau )}\right](t)=\\\mathcal{M}_q^{-1}\left[\mathcal{L}_s^{-1}\left[\mathcal{M}_A\left[\frac{1}{s+A \exp (-s \tau )}\right](q)\right](t)\right](1)=\\\mathcal{M}_q^{-1}\left[\mathcal{L}_s^{-1}\left[e^{q s \tau } \pi s^{-1+q} \csc (\pi q)\right](t)\right](1)=\\\mathcal{M}_q^{-1}\left[\frac{\pi (t+q \tau )^{-q} \csc (\pi q) \theta (t+q \tau )}{\Gamma (1-q)}\right](1)=\\\sum _{m=0}^{\infty } \frac{(-t+m \tau )^m \theta (t-m \tau )}{\Gamma (1+m)}$$

where: $\mathcal{M}_q^{-1}$ is Inverse Mellin Transform, and $\mathcal{M}_A$ is Mellin Transform.