Inverse Laplace transform of $\frac{1}{s-e^{-s}}$

by vyaman   Last Updated December 24, 2017 23:20 PM

I am trying to get Inverse Laplace Transform of $f(s) = \frac{1}{s-e^{-s}}$.

According to the book Integral Transforms and Their Applications of Debnath (page 174) in most of the cases $L^{-1}${f(s)} is equal to the sum of the residues of $e^{st}f(s)$ ($f(s)$ is the Laplace transformed function of $f(t)$).

I think $e^{st}f(s)$ has a simple pole at $s = ProductLog[1]$. So I find residue as $\lim_{s\to ProductLog[1]} (s-ProductLog[1]) \frac{e^{st}}{s-e^{-s}}$ which makes $f(t) = \frac{e^{t ProductLog[1]}}{1+ ProductLog[1]}$.

I am not sure if this is correct. Would appreciate your comments.

Answers 1

$$\frac{1}{s-e^{-s}}=\frac{1}{s}\cdot\frac{1}{1-\frac{1}{s e^s}}=\frac{1}{s}\left[1+\frac{1}{s e^s}+\frac{1}{s^2 e^{2s}}+\ldots\right] $$ and $$\mathcal{L}^{-1}\left(\frac{1}{s}\right)=1,\quad \mathcal{L}^{-1}\left(\frac{1}{s^{k+1} e^{sk}}\right)=\frac{(x-k)^k}{k!}\mathbb{1}_{\geq k}(x)$$ so, by the linearity of $\mathcal{L}^{-1}$, the inverse Laplace transform of $\frac{1}{s-e^{-s}}$ is a continuous and piecewise-smooth function, always greater than $1$ and close to $\frac{e^{x W(1)}}{1+W(1)}$ for $x\in(0,10)$. On the other hand $$ \lim_{x\to -\infty}\frac{e^{x W(1)}}{1+W(1)} = 0 \neq 1 = \lim_{x\to -\infty}\left(\mathcal{L}^{-1}\frac{1}{s-e^{-s}}\right)(x).$$ The explanation of this odd phenomenon has been outlined in the comments.

Jack D'Aurizio
Jack D'Aurizio
December 24, 2017 22:47 PM

Related Questions