Laplace transform of $f(t)=e^{-Kt^n} $

Question

Could someone help me find the Laplace transform of the time funtion $f(t)=e^{-Kt^n} $ where $K$ is real and positive and $0<n<1$ ? Thank you.

Answer 1

We have that : $$ \frac{\Gamma(\omega)}{m^{\omega}}=\int_{0}^{\infty}x^{\omega^{-1}}e^{-mn}\;\text{d}x $$ Now applying the Laplace Transform of $e^{-Kt^{n}}$, we get : $$ \mathcal{L}\left(e^{-Kt^{n}}\right)=\int_{0}^{\infty}e^{-Kt^{n}}e^{-st}\;\text{d}t \tag{1}$$ Since : $$ e^{-Kt^{n}}=\sum_{r=0}^{\infty}\frac{(-K)^{r}}{r!}t^{nr} \tag{2}$$ Then by substituting $(2)$ in $(1)$ we get : $$ \int_{0}^{\infty}\sum_{r=0}^{\infty}\frac{(-K)^{r}}{r!}t^{nr}e^{-st}\;\text{d}t $$ Hence, we get that : $$ \boxed{F(s)=\sum_{r=0}^{\infty}\frac{(-K)^{r}}{r!}\frac{\Gamma(1+nr)}{s^{1+nr}}} $$

[Note] : $1+nr>0$ to avoid $\Gamma$ diverging to $\pm\infty$