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While solving an ODE using the Laplace transform, I ran into the following problem:

$$f(t) = \mathcal L^{-1}\Bigg( {se^{cs} \over (e^s+e^{-s})(k-s)^2(k+s)^2}\Bigg) \tag 1$$

where $c$ and $k$ are constants. I can't seem to solve this because of the $(e^s+e^{-s})$ term in the denominator. How can I find the inverse Laplace transform of my expression?

FriendlyNeighborhoodEngineer
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    Ultimately you may try a direct computation with the help of residues. – Abezhiko Nov 19 '23 at 15:06
  • But isn't the residue theorem applicable only if I have polynomials in the numerator and denominator, where the degree of the denominator is greater than the degree of the numerator? – FriendlyNeighborhoodEngineer Nov 19 '23 at 15:29
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    No, it applies to such integrands too; the residue formula only needs the integrand to possess some Laurent series around the singularities (which may not be the case with branch points for instance). – Abezhiko Nov 19 '23 at 15:50
  • I see, I have to refresh my complex analysis math. Thank you. I will try to solve it using the residue theorem. Would you please write an answer so I can accept it? – FriendlyNeighborhoodEngineer Nov 19 '23 at 15:54

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With Mathematica help:

$$\mathcal{L}_s^{-1}\left[\frac{s \exp (c s)}{(\exp (s)+\exp (-s)) (k-s)^2 (k+s)^2}\right](t)=\\\sum _{m=0}^{\infty } -\frac{(-1)^m e^{-k (-1+c-2 m+t)} \left(-1+e^{2 k (-1+c-2 m+t)}\right) (1-c+2 m-t) \theta (-1+c-2 m+t)}{4 k}$$

where: $\theta (c-2 m+t-1)$ is the Heaviside theta function.

Mathematica can compute this series,but using FullSimplify gives long expression with Floor function.

Mathematica code:

Sum[-(((-1)^m E^(-k (-1 + c - 2 m + t)) (-1 + E^(2 k (-1 + c - 2 m + t))) (1 -c + 2 m - t) HeavisideTheta[-1 + c - 2 m + t])/(4 k)), {m, 0, Infinity}]

Mariusz Iwaniuk
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