How to prove that, the integral $$I_{a,b}:=\int_{1}^{+\infty}e^{-at}(1-t^{-1})^b\,dt ; \, a,b>0$$ is given by $\Gamma(b+1)$ times a Meijer G-function, i.e., $$I_{a,b}:=\Gamma(b+1) \times G^{m,n}_{p,q}(a_1,...,a_p; b_1,...,b_q/ z),$$ where $m,n,p,q,a_i,b_i$ and $z$ to be determined ?
Thank you in advance
$\int_1^\infty e^{-at}(1-t^{-1})^b~dt$
$=\int_1^\infty t^{-b}(t-1)^be^{-at}~dt$
$=\int_0^\infty(t+1)^{-b}t^be^{-a(t+1)}~d(t+1)$
$=e^{-a}\int_0^\infty t^b(t+1)^{-b}e^{-at}~dt$
$=e^{-a}\Gamma(b+1)U(b+1,2,a)$ (according to http://en.wikipedia.org/wiki/Confluent_hypergeometric_function#Integral_representations)
$=\dfrac{e^{-a}}{\Gamma(b)}G_{1,2}^{2,1}\left(a\,\middle|\begin{array}~~-b\\0,-1\end{array}\right)$ (according to http://functions.wolfram.com/HypergeometricFunctions/HypergeometricU/26/02/01)