I have a solution in laplace images:
\begin{align} &p_f(x,s) = - \frac{1}{s}\frac{b}{a} \frac{1}{\sqrt{\frac{s}{a}+ \frac{b}{a} \frac{\sqrt{ s}}{1+c\sqrt{s}}}} e^{-x \sqrt{\frac{s}{a}+ \frac{b}{a} \frac{\sqrt{ s}}{1+c\sqrt{s}}}} \label{pressure_frac_sol:laplace2c}\\ &p_r(x,y,s) = - \frac{1}{s}\frac{b}{a} \frac{ e^{-\sqrt{s}y}}{1+c\sqrt{s}} \frac{1}{\sqrt{\frac{s}{a}+ \frac{b}{a} \frac{\sqrt{ s}}{1+c\sqrt{s}}}} e^{-x \sqrt{\frac{s}{a}+ \frac{b}{a} \frac{\sqrt{ s}}{1+c\sqrt{s}}}}. \label{pressure_res_sol:laplacec} \end{align} where is $s$ is Laplace variable on a complex plane (it is used in Laplace transform to get rid of time variable $t$: $L\left[p(x,y,t)\right] =p(x,y,s)= \int_0^{\infty}p(x,y,t) e^{-st}dt$).
Now I want to go back to the originals - to the time variable $t$. I tried to do this with the help of Mellin transform by the contour $$F(t)={\frac {1}{2\pi i}}\int \limits _{{\gamma-i \infty }}^{{\gamma+i \infty }}e^{{st}}F(s)\,ds. \label{melline_fml}$$
Problem is that following integral diverges: \begin{align} \lim_{|s| \to 0} \int \limits _{DE} p_f (x,y,s) e^{st} ds = -{\frac {1}{2\pi i}}\int \limits _{{\gamma-i \infty }}^{{\gamma+i \infty }}\frac{b}{a} \frac{1}{\sqrt{\frac{s}{a}+ \frac{b}{a} \frac{\sqrt{ s}}{1+c\sqrt{s}}}} e^{st-x \sqrt{\frac{s}{a}+ \frac{b}{a} \frac{\sqrt{ s}}{1+c\sqrt{s}}}} \,\frac{ds}{s} =\nonumber \\ = -\lim_{\epsilon \to 0} {\frac {1}{2\pi i}}\int \limits _{- \pi}^{\pi} \frac{b}{a} \frac{1}{\sqrt{\frac{\epsilon e^{i \phi}}{a}+ \frac{b}{a} \frac{\sqrt{\epsilon} e^{i \frac{\phi}{2}}}{1+c\sqrt{\epsilon} e^{i \frac{\phi}{2}}}}} e^{\epsilon e^{i \phi}t-x \sqrt{\frac{\epsilon e^{i \phi}}{a}+ \frac{b}{a} \frac{\sqrt{\epsilon} e^{i \frac{\phi}{2}}}{1+c\sqrt{\epsilon} e^{i \frac{\phi}{2}}}}} \,i d{\phi} = -\infty \end{align}
Maybe someone can tell me how to switch from images to originals in this particular case - for $p_f, p_r$?