## 5 Laplacian Transforms and the Wright Functions of the First Kind  
The Laplacian transforms of the Wright functions are expressed in terms of the two-parameter Mittag-Leffler functions [3,4]:  
\[L\big{\{}W_{\,\alpha,\beta}(\pm\lambda\,t)\big{\}}=\frac{1}{s}\,E_{\alpha,\beta }\bigg{(}\pm\frac{\lambda}{s}\bigg{)}\quad;\quad\alpha>0\quad;\quad\lambda>0 \tag{20}\]  
Applying operational rules of the Laplacian transformation, we have [16, 17, 18]:  
\[L\big{\{}e^{\pm\,\rho}W_{\,\alpha,\beta}(\lambda\,t)\big{\}}=\frac{1}{s\mp\rho} \,E_{\alpha,\beta}\bigg{(}\frac{\lambda}{s\mp\rho}\bigg{)}\quad;\quad\lambda, \rho>0 \tag{21}\]  
and this permits to obtain  
\[ \begin{array}{l}L\big{\{}\sinh(\rho\,t)\,\,W_{\,\alpha,\beta}(\lambda\,t) \big{\}}=\frac{1}{2}\big{\{}\frac{1}{s-\rho}\,E_{\alpha,\beta}\bigg{(}\frac{ \lambda}{s-\rho}\bigg{)}-\frac{1}{s+\rho}\,E_{\alpha,\beta}\bigg{(}\frac{ \lambda}{s+\rho}\bigg{)}\bigg{\}}\\ L\big{\{}\cosh(\rho\,t)\,\,W_{\,\alpha,\beta}(\lambda\,t)\big{\}}=\frac{1}{2} \big{\{}\frac{1}{s-\rho}\,E_{\alpha,\beta}\bigg{(}\frac{\lambda}{s-\rho} \bigg{)}+\frac{1}{s+\rho}\,E_{\alpha,\beta}\bigg{(}\frac{\lambda}{s+\rho} \bigg{)}\bigg{\}}\end{array}  \tag{22}\]Multiplication (20) by \(t\) gives  
\[L\big{\{}t\,W_{\alpha,\beta}(\lambda\,t)\big{\}}=-\frac{d}{ds}\bigg{\{}\frac{1}{s} \,E_{\alpha,\beta}\bigg{(}\frac{\lambda}{s}\bigg{)}\bigg{\}}=-\bigg{\{}-\frac{1 }{s^{2}}E_{\alpha,\beta}\bigg{(}\frac{\lambda}{s}\bigg{)}+\frac{1}{s}\frac{d}{ ds}E_{\alpha,\beta}\bigg{(}\frac{\lambda}{s}\bigg{)}\bigg{\}} \tag{23}\]  
The derivative of the Mittag-Leffler function is  
\[\frac{d}{ds}E_{\alpha,\beta}\bigg{(}\frac{\lambda}{s}\bigg{)}=-\frac{\lambda}{ s^{2}}\left\{\frac{E_{\alpha,\beta-1}\big{(}\frac{\lambda}{s}\big{)}-(\beta-1)E_{ \alpha,\beta}\bigg{(}\frac{\lambda}{s}\bigg{)}}{\alpha\big{(}\frac{\lambda}{s} \bigg{)}}\right\} \tag{24}\]  
Lastly, we have  
\[L\big{\{}t\,W_{\alpha,\beta}(\lambda\,t)\big{\}}=\frac{1}{s^{2}}\left\{\frac{ (\alpha\,\lambda-\beta+1)E_{\alpha,\beta}\Big{(}\frac{\lambda}{s}\Big{)}+E_{ \alpha,\beta-1}\Big{(}\frac{\lambda}{s}\Big{)}}{\alpha\,\lambda}\right\} \tag{25}\]  
In the case when the Wright functions are expressed as the Bessel functions (see (2.18)), the Laplacian transforms are known for \(\beta=0,1,2\)[15]:  
\[ \begin{array}{l}\int_{0}^{\infty}e^{-st}W_{1,1}(-\frac{\lambda^{2}t^{2}}{4 })\,dt=\int_{0}^{\infty}e^{-st}\,J_{0}(\lambda\,t)\,dt=\frac{1}{\sqrt{s^{2}+ \lambda^{2}}}\\ \int_{0}^{\infty}e^{-st}W_{1,2}(-\frac{\lambda^{2}t^{2}}{4})\,dt=\frac{2}{ \lambda}\,\int_{0}^{\infty}e^{-st}\frac{I_{1}(\lambda\,t)}{t}\,dt=\frac{2}{[s +\sqrt{s^{2}+\lambda^{2}}]}\\ \int_{0}^{\infty}e^{-st}W_{1,3}(-\frac{\lambda^{2}t^{2}}{4})\,dt=\frac{4}{ \lambda^{2}}\,\int_{0}^{\infty}e^{-st}\,\frac{I_{2}(\lambda\,t)}{t^{2}}\,dt= \\ \frac{1}{\lambda}\bigg{\{}\frac{\lambda}{[s+\sqrt{s^{2}+\lambda^{2}}]}+\frac{1 }{3}\bigg{[}\frac{\lambda}{[s+\sqrt{s^{2}+\lambda^{2}}]}\bigg{]}^{3}\bigg{\}} \end{array}  \tag{26}\]  
From (A1), it follows that  
\[\int_{0}^{\infty}e^{-st}W_{1,\beta+1}(-\lambda\,t)\,dt=\frac{1}{\lambda^{\beta /2}}\,\int_{0}^{\infty}e^{-st}t^{-\beta/2}J_{\beta}(2\sqrt{\lambda\,t})\,dt \tag{27}\]  
and this integral equality is useful in deriving the explicit forms of the Mittag-Leffler functions. Starting with \(\beta=0\), we have [19]  
\[\int_{0}^{\infty}e^{-st}W_{1,1}(-\lambda\,t)\,dt=\,\int_{0}^{\infty}e^{-st}J_ {0}(2\sqrt{\lambda\,t})\,dt=\frac{1}{s}\,e^{-\,\lambda/s}\quad;\quad Res>0 \tag{28}\]  
however, from (20),  
\[L\big{\{}W_{1,1}(-\lambda\,t)\big{\}}=\frac{1}{s}\,E_{1,1}(-\frac{\lambda}{s}) \tag{29}\]  
Therefore, via comparison, the expected result is reached:  
\[ \begin{array}{l}E_{1,1}(-\frac{\lambda}{s})=e^{-\,\lambda/s}\\ \tau=\frac{\lambda}{s}\\ E_{1,1}(-\tau)=e^{-\,\tau}\end{array}  \tag{30}\]  
Introducing \(\beta=1\) into (27), we have [19]  
\[