# The inverse Laplace transform of a function (probably numerically)

I originally asked this question on MathOverflow but it was regarded as not being "research level". I repost the question here (hopefully it falls within forum's category this time) and will really appreciate if someone can shine light on it.

I need to do an inverse Laplace transform of a function with essential singularities for a specific problem. I find it is very similar to an equation J. Noolandi worked out in one of his papers in 1977 (Phys. Rev. B 16, 4466 (1977)). I have taken a snapshot of the Appendix page where the math problem and the solution were given (view it here). But I don't quite get some part of the solving process which I need help on deciphering.

If I can understand what he was doing, my specific problem can be solved. So now let's take the problem in the Appendix of Noolandi's paper for example. The function is

$\displaystyle \tilde I(s) = \frac{1-\exp(-a(s)t_0)}{a(s)}$

where $\displaystyle a(s)t_0 = \Sigma_{i=1}^n \frac{s\omega_i}{s+r_i} t_0 = \Sigma_{i=1}^n \frac{sM_i}{s+r_i}$. "$s$" is the variable. $\omega_i$, $r_i$, $t_0$, and $M_i$ are all real numbers greater than zero.

The question is how to get inverse Laplace transform $L^{-1}[\tilde I(s)]$. It looks like the author used Residue Theorem to evaluate the contour integral $\int_C \tilde I(s) \exp(st) ds$, but the hard part is the singularities are essential singularities, thus the residues can not be got conventionally.

However, it seems that the author constructed a contour and solved it numerically. I don't get Equation A5-A8 in the Appendix, and don't understand how to determine the parameter $R$ and $R_1$ he introduced in (probably something related to numerics).

Could anyone please help point out what he was doing? Thanks very much!

By the way, the author also said the inverse Laplace transform could also be evaluated using Laplace transform tables. And the result involved a convolution of the modified Bessel function of the first order. I still haven't worked it out either...

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