The function in question
$fs1=exp(-a [Gamma(1-b)]^{1/2} s^{b/2}) $
$ fs2=exp(-a [Gamma(1-b)]^{1/2} s^{b/2})/(s^{b/2})$
$fs3=exp(-a [Gamma(1-b)]^{1/2} s^{b/2})/(s^b)$
$$\mathcal{L}^{-1} [e^{-a [Gamma(1-b)]^{1/2} s^{b}} {s^{-c}}] $$
where $a>0$ , $0<b<1$ and $c>0$.
It need transform s to t.
How to find the Inverse Laplace ?
Thank you very much.
Well, first of all we can use the '[convolution theorem][1]' of the [Laplace transform][2]:
$$\mathcal{I}\left(t\right):=\mathscr{L}_\text{s}^{-1}\left[\exp\left(\text{n}\cdot\text{s}^\text{m}\right)\cdot\text{s}^\text{p}\right]_{\left(t\right)}=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\exp\left(\text{n}\cdot\text{s}^\text{m}\right)\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\text{s}^\text{p}\right]_{\left(t-\tau\right)}\space\text{d}\tau\tag1$$
Using:
- The '[time shifting][1]' property of the [Laplace transform][2]:
$$\mathscr{L}_\text{s}^{-1}\left[\exp\left(\text{n}\cdot\text{s}^\text{m}\right)\right]_{\left(x\right)}=\theta\left(x+\text{n}\right)\cdot\mathscr{L}_\text{s}^{-1}\left[\exp\left(\text{s}^\text{m}\right)\right]_{\left(x+\text{n}\right)}\tag2$$
- Using the '[Table of selected Laplace transforms][3]':
$$\mathscr{L}_\text{s}^{-1}\left[\text{s}^\text{p}\right]_{\left(x\right)}=\frac{1}{\Gamma\left(-\text{p}\right)}\cdot\frac{1}{x^{1+\text{p}}}\tag3$$
So, we get:
$$\mathcal{I}\left(t\right)=\int_0^t\theta\left(t+\text{n}\right)\cdot\mathscr{L}_\text{s}^{-1}\left[\exp\left(\text{s}^\text{m}\right)\right]_{\left(t+\text{n}\right)}\cdot\frac{1}{\Gamma\left(-\text{p}\right)}\cdot\frac{1}{\left(t-\tau\right)^{1+\text{p}}}\space\text{d}\tau=$$
$$\int_0^t\mathscr{L}_\text{s}^{-1}\left[\exp\left(\text{s}^\text{m}\right)\right]_{\left(t+\text{n}\right)}\cdot\frac{1}{\Gamma\left(-\text{p}\right)}\cdot\frac{\theta\left(t+\text{n}\right)}{\left(t-\tau\right)^{1+\text{p}}}\space\text{d}\tau\tag4$$
Now, using the series for $\exp$ we can write:
$$\mathscr{L}_\text{s}^{-1}\left[\exp\left(\text{s}^\text{m}\right)\right]_{\left(t+\text{n}\right)}=\mathscr{L}_\text{s}^{-1}\left[\sum_{\beta=0}^\infty\frac{\left(\text{s}^\text{m}\right)^\beta}{\beta!}\right]_{\left(t+\text{n}\right)}=\sum_{\beta=0}^\infty\frac{1}{\beta!}\cdot\mathscr{L}_\text{s}^{-1}\left[\text{s}^{\beta\text{m}}\right]_{\left(t+\text{n}\right)}\tag5$$
Using the '[Table of selected Laplace transforms][3]':
$$\mathscr{L}_\text{s}^{-1}\left[\text{s}^{\beta\text{m}}\right]_{\left(t+\text{n}\right)}=\frac{1}{\Gamma\left(-\beta\text{m}\right)}\cdot\frac{1}{\left(t+\text{n}\right)^{1+\beta\text{m}}}\tag6$$
So, we get:
$$\mathcal{I}\left(t\right)=\int_0^t\left\{\sum_{\beta=0}^\infty\frac{1}{\beta!}\cdot\frac{1}{\Gamma\left(-\beta\text{m}\right)}\cdot\frac{1}{\left(t+\text{n}\right)^{1+\beta\text{m}}}\right\}\cdot\frac{1}{\Gamma\left(-\text{p}\right)}\cdot\frac{\theta\left(t+\text{n}\right)}{\left(t-\tau\right)^{1+\text{p}}}\space\text{d}\tau=$$
$$\sum_{\beta=0}^\infty\frac{1}{\beta!}\cdot\frac{1}{\Gamma\left(-\beta\text{m}\right)}\cdot\frac{1}{\Gamma\left(-\text{p}\right)}\int_0^t\frac{1}{\left(t+\text{n}\right)^{1+\beta\text{m}}}\cdot\frac{\theta\left(t+\text{n}\right)}{\left(t-\tau\right)^{1+\text{p}}}\space\text{d}\tau\tag7$$
[1]: https://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems
[2]: https://en.wikipedia.org/wiki/Laplace_transform
[3]: https://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms