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So I want to obtain a inverse Laplace transform from mathematica but I get this:

enter image description here

How can I get the solution?

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  • $\begingroup$ I'm solving a heat transfer problem, so there is a solution for that inverse, I think. I only wrote this: InverseLaplaceTransform[Sinh[a sqrt(s)]/(s^2 Sinh[b sqrt(s)), s, t] $\endgroup$ Mar 1, 2020 at 13:58
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    $\begingroup$ That means Mathematica doesn't know the answer. $\endgroup$ Mar 1, 2020 at 14:07

1 Answer 1

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Using:

$$\frac{1}{\sinh \left(b \sqrt{s}\right)}=\text{csch}\left(b \sqrt{s}\right)=\sum _{k=0}^{\infty } 2 \exp \left(-(2 k+1) b \sqrt{s}\right)$$ then we have:

func = InverseLaplaceTransform[(2 E^(b (-1 - 2 k) Sqrt[s]) Sinh[a Sqrt[s]])/s^2, s, t]
Sum[func[[1]], {k, 0, Infinity}]

Solution only by Infinite Sum:

HoldForm[InverseLaplaceTransform[
Sinh[a Sqrt[s]]/(s^2*Sinh[b Sqrt[s]]), s, t] == 
Sum[1/(2 Sqrt[\[Pi]]) (a^2 Sqrt[\[Pi]] - 2 a b Sqrt[\[Pi]] + 
   b^2 Sqrt[\[Pi]] - 4 a b k Sqrt[\[Pi]] + 4 b^2 k Sqrt[\[Pi]] + 
   4 b^2 k^2 Sqrt[\[Pi]] + 
   2 a E^(-((-a + b + 2 b k)^2/(4 t))) Sqrt[t] - 
   2 b E^(-((-a + b + 2 b k)^2/(4 t))) Sqrt[t] + 
   2 a E^(-((a + b + 2 b k)^2/(4 t))) Sqrt[t] + 
   2 b E^(-((a + b + 2 b k)^2/(4 t))) Sqrt[t] - 
   4 b E^(-((-a + b + 2 b k)^2/(4 t))) k Sqrt[t] + 
   4 b E^(-((a + b + 2 b k)^2/(4 t))) k Sqrt[t] + 
   2 Sqrt[\[Pi]] t + 
   Sqrt[\[Pi]] (a^2 - 2 a (b + 2 b k) + (b + 2 b k)^2 + 2 t) Erf[(
     a - b (1 + 2 k))/(2 Sqrt[t])] - 
   Sqrt[\[Pi]] (a^2 + 2 a (b + 2 b k) + (b + 2 b k)^2 + 
      2 t) Erfc[(a + b + 2 b k)/(2 Sqrt[t])]), {k, 0, 
  Infinity}]] // TraditionalForm

I doubt there's a closed form for the Inverse Laplace Transform or Series.

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