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.
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).
You are asking for the inverse Laplace transform $g_\beta(t)$ of the stretched exponential function, $f_\beta(s)=\exp[-(s/s_0)^\beta]$. For $\beta=1$ this is a Dirac delta function, $g_1(t)=\delta(t-1/s_0)$, and for $\beta=1/2$ it is the Lévy distribution, $$g_{1/2}(t)=\exp\left(-\frac{1}{4 s_0 t}\right)\frac{1}{2t \sqrt{\pi s_0 t}}.$$ There are no expressions in terms of elementary functions for other values of $\beta$. A convenient integral representation, suitable for numerical evaluation, is given here, equation 29.
