Laplace transform of : $t^{\gamma-1} F(\alpha,\beta,\delta,\frac{t}{d})$, where $F$ is the Gauss' hypergeometric function What is the Laplace transform of : $t^{\gamma-1} F(\alpha,\beta,\delta,\frac{t}{d})$, where $\gamma >0 $ and $F$ is the Gauss' hypergeometric function.
Note that I have the Laplace transform of : $t^{\gamma-1} F(\alpha,\beta,\delta,-t)$.
Thanks!

Since $F\left( {\alpha ,\beta ,\delta ;t} \right) = \sum {\frac{{\left( \alpha  \right)_n \left( \beta  \right)_n }}{{n!\left( \delta  \right)_n }}t^n } 
$. Assuming the uniform convergence of the series,  then term by term integration yields that 
\begin{align}
\int_0^\infty  {e^{ - st} } t^{\gamma  - 1} F\left( {\alpha ,\beta ,\delta ;t} \right)dt
\\
 &= \int_0^\infty  {e^{ - st} } t^{\gamma  - 1} \left( {\sum {\frac{{\left( \alpha  \right)_n \left( \beta  \right)_n }}{{n!\left( \delta  \right)_n }}t^n } } \right)dt
\\
&= \sum {\frac{{\left( \alpha  \right)_n \left( \beta  \right)_n }}{{n!\left( \delta  \right)_n }}\left( {\int_0^\infty  {e^{ - st} } t^{n + \gamma  - 1} dt} \right)} 
\\
&= \sum {\frac{{\left( \alpha  \right)_n \left( \beta  \right)_n }}{{n!\left( \delta  \right)_n }}\frac{{\Gamma \left( {n + \gamma } \right)}}{{s^{n + \gamma } }}} 
\end{align}
where $$
\int_0^\infty  {e^{ - st} } t^{n + \gamma  - 1} dt = \frac{{\Gamma \left( {n + \gamma } \right)}}{{s^{n + \gamma } }}
,$$ valid if $n+\gamma-1>-1$.  To write the answer in more general form, we may use the generalized Hypergeometric function as follows:$$
_p F_q \left( {a_1 ,a_2 , \ldots ,a_p ;b_1 ,b_2 , \ldots ,b_q ;t} \right) = \sum\limits_{n = 0}^\infty  {\frac{{\left( {a_1 } \right)_n \left( {a_2 } \right)_n  \cdots \left( {a_p } \right)_n }}{{\left( {b_1 } \right)_n \left( {b_2 } \right)_n  \cdots \left( {b_q } \right)_n }}\frac{{t^n }}{{n!}}} 
,$$ so in our case we have $
p = 3,q = 1,a_1  = \alpha ,a_2  = \beta ,a_3  = \gamma ,b_1  = \delta ,t = \frac{1}{s}
$ and since $
\left( \alpha  \right)_n  = \frac{{\Gamma \left( {\alpha  + n} \right)}}{{\Gamma \left( \alpha  \right)}}
$, then we may write the above series as:
\begin{align}
\sum {\frac{{\left( \alpha  \right)_n \left( \beta  \right)_n }}{{n!\left( \delta  \right)_n }}\frac{{\Gamma \left( {n + \gamma } \right)}}{{s^{n + \gamma } }}}  &= \frac{{\Gamma \left( \delta  \right)}}{{\Gamma \left( \alpha  \right)\Gamma \left( \beta  \right)}}
\sum {\frac{{\Gamma \left( {\alpha  + n} \right)\Gamma \left( {\beta  + n} \right)}}{{n!\Gamma \left( {\delta  + n} \right)}}\frac{{\Gamma \left( {n + \gamma } \right)}}{{\Gamma \left( \gamma  \right)}}\frac{{\Gamma \left( \gamma  \right)}}{{s^{n + \gamma } }}} 
\\
 & = 
\frac{{\Gamma \left( \delta  \right)\Gamma \left( \gamma  \right)}}{{\Gamma \left( \alpha  \right)\Gamma \left( \beta  \right)}}s^\gamma  \sum {\frac{{\Gamma \left( {\alpha  + n} \right)\Gamma \left( {\beta  + n} \right)}}{{n!\Gamma \left( {\delta  + n} \right)}}\frac{1}{{s^n }}} 
\\
&= \frac{{\Gamma \left( \delta  \right)\Gamma \left( \gamma  \right)}}{{\Gamma \left( \alpha  \right)\Gamma \left( \beta  \right)}}s^\gamma \cdot 
{}_3 F_1 \left( {\alpha ,\beta ,\gamma ;\delta ;\frac{1}{s}} \right)
\end{align}
