I am reading a paper which in part of that authors used to calculate the inverse Laplace transform in a way that I can not understand.
actually suppose we have $ \hat{F}(z)=\sum_{i=0}^{\infty} \frac{A^{i}}{z^{i+1}}$ where $A$ is square matrix with positive elements. For calculation of $F(t)$ the paper used this way :
First of all with my knowledge the formula for inverse Laplace transform is not $\int_{0}^{\infty} e^{zt}\hat{F}(z)dz$ and I don't know where this formula came from.
However if I consider this formula true, I still can't drive the final $e^{At}$.
I would be appreciated for any help.
First off, let's define the exponential of a matrix via its power series. i.e. $$e^{At} = \sum _{n=0}^\infty \frac{ A^n }{n!} t^n $$ Now, if you don't like the inverse Laplace, let's use that fact that the Laplace transform is injective. We see that \begin{align*} \hat F =\mathcal{L} ( e^{At} ) =& \int_0^\infty e^{- zt} e^{At} dt \quad \text{def'n of Laplace}\\ =& \sum_{n=0}^\infty A^n \int_{0}^\infty \frac{t^ne^{-zt} }{n!} dt \quad \text{def'n of matrix exp, with uniform convergence and linearity} \\ = & \sum_{n=0}^\infty A^n \frac{1}{z^{n+1} } \quad z > 0 \end{align*} Now since the transform is injective, we have that the inverse of $\hat F$ is actually $e^{At}$