 # 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 ? wurpenxd 2022-09-11 Answered
Could someone help me find the Laplace transform of the time funtion $f\left(t\right)={e}^{-K{t}^{n}}$ where K is real and positive and 0<n<1 ?
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We have that :
$\frac{\mathrm{\Gamma }\left(\omega \right)}{{m}^{\omega }}={\int }_{0}^{\mathrm{\infty }}{x}^{{\omega }^{-1}}{e}^{-mn}\phantom{\rule{thickmathspace}{0ex}}\text{d}x$
Now applying the Laplace Transform of ${e}^{-K{t}^{n}}$, we get :
$\begin{array}{}\text{(1)}& \mathcal{L}\left({e}^{-K{t}^{n}}\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-K{t}^{n}}{e}^{-st}\phantom{\rule{thickmathspace}{0ex}}\text{d}t\end{array}$
Since :
$\begin{array}{}\text{(2)}& {e}^{-K{t}^{n}}=\sum _{r=0}^{\mathrm{\infty }}\frac{\left(-K{\right)}^{r}}{r!}{t}^{nr}\end{array}$
Then by substituting (2) in (1) we get :
${\int }_{0}^{\mathrm{\infty }}\sum _{r=0}^{\mathrm{\infty }}\frac{\left(-K{\right)}^{r}}{r!}{t}^{nr}{e}^{-st}\phantom{\rule{thickmathspace}{0ex}}\text{d}t$
Hence, we get that :
$\overline{)F\left(s\right)=\sum _{r=0}^{\mathrm{\infty }}\frac{\left(-K{\right)}^{r}}{r!}\frac{\mathrm{\Gamma }\left(1+nr\right)}{{s}^{1+nr}}}$
[Note] : $1+nr>0$ to avoid $\mathrm{\Gamma }$ diverging to $±\mathrm{\infty }$

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