Laplace transform of a rational function Let $\alpha>1$ and $z>0$.  I am looking for a solution to the following Laplace transofrm:
\begin{equation}
\alpha\cdot\mathcal{L}\left\{{_{2}F_{1}(1-\alpha,1;2;-t)}\right\}(z)=\mathcal{L}\left\{\frac{(1+t)^\alpha-1}{t}\right\}(z)
=\int_{0}^{\infty}\frac{(1+t)^\alpha-1}{t}\,e^{-zt}\,\mathrm{d}t
\end{equation}
I found a solution to the Laplace transform of ${_{2}F_{1}}(-t)$ here but it does not seems to work.
I also found this (Gradshteyn & Ryzhik, ed. 8, #7.522.1):
\begin{equation}
\int_{0}^{\infty}e^{-\lambda x}x^{\gamma-1}{_{2}F_1}(\alpha,\beta;\delta;-x)\,\mathrm{d}x=\frac{\Gamma(\delta)\lambda^{-\gamma}}{\Gamma(\alpha)\Gamma(\beta)}E(\alpha\,;\beta:\gamma\,;\delta:\lambda),
\end{equation}
for $\Re \lambda>0$ and $\Re \gamma>0$ where $E(p\,;\alpha_{r}:q\,;\varrho_{s}:x)$ is the MacRobert E-Function. Apparently, the E-function canbe expressed as MeijerG; however, I am having trouble understanding how to do so. In the Table of Integral Transforms (vol. 1, pg. 384) we have
\begin{equation}
E(p\,;\alpha_{r}:q\,;\beta_{s}:x)=
G_{\,q+1,p}^{\,p,1}\left(x\,\middle\vert
\begin{array}{c}
1,\beta_1,\dots,\beta_q\\
\alpha_1,\dots,\alpha_p
\end{array}
\right).
\end{equation}
 A: You can convert the hypergeometric function to the Meijer G-function and use the formula for the integral of a product of two linear G-functions:
$$\alpha  \int_0^\infty {_2F_1}(1 - \alpha, 1; 2; -t) e^{-z t} dt = \\
\frac \alpha {\Gamma(1 - \alpha)} \int_0^\infty
 G_{2, 2}^{1, 2} \left(t \middle| {0, \alpha \atop 0, -1} \right)
 G_{0, 1}^{1, 0} \left(z t \middle| {- \atop 0} \right) dt = \\
\frac \alpha {\Gamma(1 - \alpha)}
 G_{2, 3}^{3, 1} \left(z \middle| {0, 1 \atop 0, 0, -\alpha} \right).$$
In terms of more elementary function, it'll be
$$\frac z {1 + \alpha} \,{_2F_2}(1, 1; 2, 2 + \alpha; z) +
 \Gamma(\alpha + 1) z^{-\alpha} (-z)^\alpha \Gamma(-\alpha, -z) + \\
 \ln z + \pi \csc (\pi \alpha) z^{-\alpha} (-z)^\alpha -
 \psi(\alpha) - \pi \cot \pi \alpha - \frac 1 \alpha,$$
$\alpha>0, \operatorname{Re} z > 0$. For integral $\alpha_0$, this is still valid as a limit, with the singular terms yielding
$$\lim_{\alpha \to \alpha_0}(
 \pi \csc (\pi \alpha) z^{-\alpha} (-z)^\alpha - \pi \cot \pi \alpha) =
\ln(-z) - \ln z.$$
A: Wolfram Language all the way:
LaplaceTransform[Hypergeometric2F1[1-a,1,2,-t],t,z]
produces
$$\frac{-\mathcal{L}_t\left[\frac{1}{t}\right](z)+\frac{z \,
   _2F_2(1,1;2,a+2;z)}{a+1}+\frac{\pi  z^{-a} \csc (\pi  a) \left((-1)^{a+1} z^a
   \Gamma (1-a,-z)+e^z\right)}{\Gamma (1-a)}+\pi  e^{i \pi  a} \csc (\pi  a)-\psi
   ^{(0)}(-a)-\gamma }{a}$$
Here $\gamma$ is the Euler constant, and $\psi^{(0)}(-\alpha)$ is the PolyGamma function. You still have an LT in here, unfortunately, but it seems reduced compared with the original.
