# Inverse Laplace Transform how to find the exact solution

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

How can I get the solution?

• 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] Mar 1, 2020 at 13:58
• That means Mathematica doesn't know the answer. Mar 1, 2020 at 14:07

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,

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