How to find inverse Laplace of $\exp{[-(\sqrt{s^\alpha/k})\,x]}$?

Question

The function in question is

$f(x, s)=\exp{[-(\sqrt{s^\alpha/k})\,x]}$

where, $0<\alpha<1$, $k>0$ and $x\geq0$.

How to find the inverse Laplace?

Any suggestion?

Thanks

Answer 1

Expanding with infinite sum:

$$\mathcal{L}_s^{-1}\left[\exp \left(-\sqrt{\frac{s^{\alpha }}{k}} x\right)\right](t)=\mathcal{L}_s^{-1}\left[\sum _{n=0}^{\infty } \frac{\left(-\frac{x s^{\alpha /2}}{\sqrt{k}}\right)^n}{n!}\right](t)=\sum _{n=0}^{\infty } \mathcal{L}_s^{-1}\left[\frac{(-1)^n k^{-\frac{n}{2}} s^{\frac{n \alpha }{2}} x^n}{n!}\right](t)=\sum _{n=0}^{\infty } \frac{(-1)^n k^{-\frac{n}{2}} t^{-1-\frac{n \alpha }{2}} x^n}{n! \Gamma \left(-\frac{1}{2} (n \alpha )\right)}$$

case for $\alpha= 1/2$:

$-\frac{x \, _0F_2\left(;\frac{1}{2},\frac{3}{4};-\frac{x^4}{256 k^2 t}\right)}{\sqrt{k} t^{5/4} \Gamma \left(-\frac{1}{4}\right)}-\frac{x^2 \, _0F_2\left(;\frac{3}{4},\frac{5}{4};-\frac{x^4}{256 k^2 t}\right)}{4 k \sqrt{\pi } t^{3/2}}-\frac{x^3 \, _0F_2\left(;\frac{5}{4},\frac{3}{2};-\frac{x^4}{256 k^2 t}\right)}{6 k^{3/2} t^{7/4} \Gamma \left(-\frac{3}{4}\right)}$

Mathematica code:

Sum[((-1)^n k^(-n/2) t^(-1 - (n \[Alpha])/2) x^n)/(n! Gamma[-((n \[Alpha])/2)]) /. \[Alpha] -> 1/2, {n, 0, Infinity}, Assumptions -> {k > 0, x > 0, t > 0}] // Expand

(*-((x HypergeometricPFQ[{}, {1/2, 3/4}, -(x^4/(256 k^2 t))])/(
Sqrt[k] t^(5/4) Gamma[-(1/4)])) - (
x^2 HypergeometricPFQ[{}, {3/4, 5/4}, -(x^4/(256 k^2 t))])/(
4 k Sqrt[\[Pi]] t^(3/2)) - (
x^3 HypergeometricPFQ[{}, {5/4, 3/2}, -(x^4/(256 k^2 t))])/(
6 k^(3/2) t^(7/4) Gamma[-(3/4)])*)

EDITED:

case for $\alpha= 1$:

$\mathcal{L}_s^{-1}\left[e^{-\sqrt{\frac{s}{k}} x}\right](t)=\frac{e^{-\frac{x^2}{4 k t}} x}{2 \sqrt{k} \sqrt{\pi } t^{3/2}}\to \lim_{x\to 0} \, \frac{e^{-\frac{x^2}{4 k t}} x}{2 \sqrt{k} \sqrt{\pi } t^{3/2}}=0$