I have a Laplace tranform in the form given below
$$\mathcal{L}(s)=\text{exp}(-As^{2/\alpha}-Bs^{3/\beta})$$
which is an multiplication of two stretched exponential decay function
where $A,B >0$
I need to find the inverse Laplace transform of this.
$\textit{I know that this cannot be solved symbolically using elementary functions.}$
Can some one suggest and/or refer me to some solution?
Here, $\alpha$ and $\beta$ can take values like $3,4,5...$
Any approximation is also welcome
Hint: Try to expand the exponential function as a power sum.
$$\exp(-As^{2/\alpha}-Bs^{3/\beta})=1+\frac{-As^{2/\alpha}-Bs^{3/\beta}}{1!}+\frac{(-As^{2/\alpha}-Bs^{3/\beta})^2}{2!}+\cdots$$
You will need the following formula for inverting the powers of s.
$$\mathcal{L}^{-1}\left[ s^\alpha\right]=\frac{t^{-\alpha-1}}{\Gamma(-\alpha)}$$
Where $\Gamma$ is the Gamma function.