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

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.
