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

Question

doing some work on a PDE system I have stumbled across a Laplace transform which I'm not sure how to invert: $$ F(s) = e^{\frac{1}{s}-s} $$ I can't find it in any table and the strong singular growth for $s=0$ makes me think that perhaps the inverse doesn't exist? Does it exist? And if it does, how can I find it?

Thanks for your help,

Maxi

Answer 1

Hint: Write $$e^{s-1/s}=\sum_{k\ge 0}\frac{e^{-s}}{k! s^k}$$ and use $$\mathcal{L}\{f(t-a)\}=e^{-sa}\mathcal{L}\{f(t)\}\\\mathcal{L}\{t^k H(t)\}=\frac{\Gamma(k+1)}{s^{k+1}}$$

Answer 2

The inverse most definitely exists. Write the integrand of the ILT in its Laurent expansion about $s=0$ as follows:

$$F(s) e^{s t} = e^{\frac1s} e^{(t-1) s} = \left (1+\frac1s +\frac1{2! s^2} + \frac1{3! s^3}+\cdots \right ) \left [1+(t-1) s+\frac1{2!} (t-1)^2 s^2 + \cdots \right ]$$

The ILT is simply the residue of the integrand at $s=0$, or namely the coefficient of $s^{-1}$ in the above expansion. This coefficient is

$$\sum_{k=0}^{\infty} \frac{(t-1)^{k}}{k!(k+1)!} H(t-1) = \frac{I_1\left (2 \sqrt{t-1} \right )}{\sqrt{t-1}} H(t-1)$$

where $H$ is the Heaviside function and $I_1$ is the modified Bessel function of the first kind of first order.