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I'm after the numerical inverse Laplace transform of a function, so I type

f[t_?NumericQ] :=   InverseLaplaceTransform[1/(1 + s + ArcTanh[1/(s - 1)]), s, t];

This function may possibly suck, but it looks like its inverse Laplace transform should exist, albeit not in closed form (for instance, by the initial value theorem we know that f[t->0+]=1). However, when confronted with this task Mathematica simply replicates my command without giving a numerical solution.

Of course ArcTanh is a multivalued function in the complex plane, but shouldn't the algorithm used for Laplace inversion know how to handle this?

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    $\begingroup$ Support for numeric InverseLaplaceTransform was introduced in version 12.2. $\endgroup$ Jan 19, 2022 at 16:03
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    $\begingroup$ Your function works fine in Mathematica 13.0. $\endgroup$
    – John Doty
    Jan 19, 2022 at 17:39

1 Answer 1

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We can expand function 1/(1 + s + ArcTanh[1/(s - 1)]) by series:

f[s_] := 1/(1 + s + ArcTanh[1/(s - 1)]);

g[t_, M_] := InverseLaplaceTransform[Series[f[s], {s, Infinity, M}] // Normal, s, t]; 
Plot[{g[t, 50]}, {t, 0, 3}]

N[g[2, 50], 15](*for t=2*)
(*-1.39795847070074*)

InverseLaplaceTransform[f[s], s, 2.0, WorkingPrecision -> 14]
(*for t=2. Works only in version 12.2 or above.*)
(*-1.39795847070073 + 0.*10^-17 I*)

Probably does not have finite closed-form expression in terms of very large class of special functions.

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