## 5 Laplacian Transforms and the Wright Functions of the First Kind
\begin{array}{l}\int_{0}^{\infty}e^{-st}W_{1,2}(-\lambda\,t)\,dt=\,\frac{1 }{\sqrt{\lambda}}\int_{0}^{\infty}e^{-st}t^{-1/2}J_{1}(2\sqrt{\lambda\,t})\,dt= \\ \sqrt{\frac{\pi}{\lambda s}}e^{-\,\lambda/2s}I_{1/2}(\frac{\lambda}{2s})=\frac {2}{A}e^{-\,\lambda/2s}\sinh(\frac{\lambda}{2s})=\frac{1}{\lambda}\,(1-e^{-\, \lambda/s})\\ \int_{0}^{\infty}e^{-st}W_{2,1}(-\lambda\,t)\,dt=\frac{1}{s}\,E_{1,2}(-\frac{ \lambda}{s})\\ \tau=\frac{\lambda}{s}\\ E_{1,2}(-\tau)=\,\frac{1}{\tau}(1-e^{-\,\tau})\end{array}  \tag{31}\]  
This immediately gives, by using (43),  
\[ \begin{array}{l}E_{1,3/2}(z)=\sqrt{\frac{\pi}{2}}\,e^{z}\,erf(\sqrt{z})\\ E_{1,5/2}(z)=\frac{e^{z}}{2^{3/2}}\,[\sqrt{\pi}\,e^{z}erf(\sqrt{z})-z\,e^{-z}]\\ E_{1,7/2}(z)=\frac{e^{z}}{2^{5/2}}\big{\{}\,2\,[\sqrt{\pi}\,e^{z}erf(\sqrt{z})- ze^{-z}]\,-z^{2}\,e^{-z}\big{\}}\end{array}  \tag{51}\]  
The Laplacian transforms of the Mainardi functions are  
\[ \begin{array}{l}L\Big{\{}\frac{1}{t}F_{\alpha}(\frac{\lambda}{P})\Big{\}}=L \Big{\{}\frac{\alpha\,\lambda}{\mu+1}M_{\alpha}(\frac{\lambda}{P})\Big{\}}=e^{ -\lambda\,s^{\alpha}}\\ 0<\alpha<1\quad;\quad\lambda>0\end{array}  \tag{52}\]  
and  
\[ \begin{array}{l}L\Big{\{}\frac{1}{\alpha}F_{\alpha}(\frac{\lambda}{P})\Big{\}} =L\Big{\{}\frac{\lambda}{P^{\alpha}}M_{\alpha}(\frac{\lambda}{P})\Big{\}}= \frac{\lambda}{s^{1-\alpha}}e^{-\lambda\,s^{\alpha}}\\ 0<\alpha<1\quad;\quad\lambda>0\end{array}  \tag{53}\]  
or the term of the Wright function:  
\[ \begin{array}{l}L\Big{\{}\frac{1}{\alpha}W_{-\,\alpha,0}(-\frac{\lambda}{P} )\Big{\}}=L\Big{\{}\frac{\lambda}{P^{\alpha}}W_{-\,\alpha,1-\,\alpha}(-\frac{ \lambda}{P})\Big{\}}=\frac{\lambda}{s^{1-\alpha}}e^{-\lambda\,s^{\alpha}}\\ 0<\alpha<1\quad;\quad\lambda>0\end{array}  \tag{54}\]  
The inverse Laplacian transforms are known only for \(\alpha=1/2\) and \(\alpha=1/3\). \[ \begin{array}{l}L\Big{\{}\frac{1}{t}F_{1/2}(\frac{\lambda}{t^{1/2}})\Big{\}} =L\Big{\{}\frac{\lambda}{2t^{3/2}}M_{1/2}(\frac{\lambda}{t^{1/2}})\Big{\}}=e^ {-\lambda\,s^{1/2}}\\ \lambda>0\\ L^{-1}\Big{\{}e^{-\lambda\,s^{1/2}}\Big{\}}=\frac{\lambda e^{-\lambda^{2}/4t} }{2\sqrt{\pi}\,t^{3/2}}\\ \frac{1}{t}F_{1/2}(\frac{\lambda}{t^{1/2}})=\frac{\lambda}{2t^{3/2}}M_{1/2}( \frac{\lambda}{t^{1/2}})=\frac{\lambda\,e^{-\lambda^{2}/4t}}{2\sqrt{\pi}\,t^{ 3/2}}\\ F_{1/2}(\lambda\,\tau)=\frac{\lambda\,\tau^{2}}{2}M_{1/2}(\lambda\,\tau)=\frac {\lambda\,\tau^{2}e^{-\lambda^{2}/4t}}{2\sqrt{\pi}\,t}\end{array}  \tag{55}\]  
The multiplication by \(t\) of the Mainardi function in (55) is equivalent to  
\[ \begin{array}{l}L\Big{\{}F_{1/2}(\frac{\lambda}{t^{1/2}})\Big{\}}=L\Big{\{} \frac{\lambda}{2t^{1/2}}M_{1/2}(\frac{\lambda}{t^{1/2}})\Big{\}}=-\frac{d}{ds }\Big{\{}e^{-\lambda\,s^{1/2}}\Big{\}}=\frac{\lambda}{2\,s^{1/2}}e^{-\lambda\, s^{1/2}}\\ L^{-1}\Big{\{}\frac{\lambda}{2\,s^{1/2}}e^{-\lambda\,s^{1/2}}\Big{\}}=\frac{ \lambda\,e^{-\lambda^{2}/4t}}{2\sqrt{\pi}\,t}\\ F_{1/2}(\frac{\lambda}{t^{1/2}})=\frac{\lambda}{2\,t^{1/2}}M_{1/2}(\frac{ \lambda}{t^{1/2}})=\frac{\lambda e^{-\lambda^{2}/4t}}{2\sqrt{\pi}\,t}\\ F_{1/2}(\lambda\,\tau)=\frac{\lambda\,\tau}{2}M_{1/2}(\lambda\,\tau)=\frac{ \lambda\,\tau e^{-\lambda^{2}/4t}}{2\sqrt{\pi}\,t}\end{array}  \tag{56}\]  
The same results, but in terms of the Wright functions, can be written as follows:  
\[ \begin{array}{l}L\Big{\{}2\ W_{-1/2,0}(-\frac{\lambda}{t^{1/2}})\Big{\}}=L \Big{\{}\frac{\lambda}{t^{1/2}}W_{-\,1/2,1\,/2}(-\frac{\lambda}{t^{1/2}})\Big{\}} =\frac{\lambda}{s^{1/2}}e^{-\lambda\,s^{1/2}}\\ L^{-1}\Big{\{}\frac{\lambda}{s^{1/2}}e^{-\lambda\,s^{1/2}}\Big{\}}=\frac{1}{ \sqrt{\pi}\,t}e^{-\lambda^{2}/4t}\\ 2\ W_{-\,1/2,0}(-\frac{\lambda}{t^{1/2}})=\frac{\lambda\,\tau}{2}W_{-\,1/2,1\,/2 }(-\frac{\lambda}{t^{1/2}})=\frac{\lambda\,\tau}{\sqrt{\pi}\,t}\,e^{-\lambda^{2 }\,\tau^{2}/4}\\ 2\ W_{-\,1/2,0}(-\lambda\,\tau)=\lambda\,\tau\,W_{-\,1/2,1\,/2}(-\lambda\,\tau)= \frac{\lambda\,\tau}{\sqrt{\pi}\,t}\,e^{-\lambda^{2}\,\tau^{2}/4}\end{array}