I have a Laplace tranform in the form given below


$\mathcal{L}\_I(s)=\text{exp}(-\pi\lambda \Gamma(1+\frac{2}{\alpha})\Gamma(1-\frac{2}{\alpha})P^{2/\alpha}s^{2/\alpha})$


Can some one help me to find the inverse Laplace transform of it?


Here, $\alpha$ can take values like 1,2,3,4,5...


$P$ and $\lambda$ are constants.



How to find the Inverse Laplace Transform of the following?
In addition to Carlo Beenakker's answer. <http://journal.austms.org.au/ojs/index.php/ANZIAMJ/article/view/924/735> (The Kohlrausch function: properties and applications, by R.S. Anderssen, S.A. Husain and R.J. Loy),
besides the Doetsch's result for $g\_{1/2}(t)$ cited by Carlo, cites the following results ($s\_0=1$ is assumed):
$$g\_{1/3}(t)=\frac{x}{\pi}\sin{\left(\frac{\pi}{3}\right)}K\_{1/3}(x),\;\;x=2\left(\frac{1}{3t^{1/3}}\right )^{3/2},\;\;\mathrm{(Montroll\; and\; Bendler)}.$$
$$g\_{2/3}(t)=-\frac{1}{2t\sqrt{3\pi}}\exp{\left(-\frac{2}{27t^2}\right)}
W\_{-1/2,-1/6}\left(-\frac{4}{27t^2}\right),$$ which follows from Humbert
's result $$g\_{\beta}(t)=-\frac{1}{\pi}\sum\limits\_{k=0}^{\infty}\frac{(-1)^k}{k!}\sin{(\pi\beta k)}\frac{\Gamma(1+\beta k)}{t^{1+\beta k}}.$$
See also <http://arxiv.org/abs/0804.2702> (A relaxation function encompassing the stretched exponential and the compressed hyperbola, by M. Berberan-Santos) and for numerical calculation <http://nvlpubs.nist.gov/nistpubs/jres/095/jresv95n4p433_A1b.pdf> (Tables of the Inverse Laplace Transform of the Function $e^{-s^\beta}$, by M. Dishon, J.T. Bendler and G.H. Weiss).


