How to solve the integral:
$I = \int_0^T (1-\alpha \beta(t)) \beta(t)^{n-1} dt $
where $\beta(t) = \frac{1-e^{-rt}}{1-\alpha e^{-rt}}$ with the assumptions that $0<\alpha<1$, $n\in \Bbb{Z}$ and $n \ge1$?
I've tried using the substitution $u = e^{rt}$ to give:
$I= \frac{1-\alpha}{r} \int_1^{e^{rt}} \frac{(u-1)^{n-1}}{(u-\alpha)^n}du$
But I'm not sure how to proceed from here. Thanks.
The change of variables $u = (U - 1) t + 1$ gives $$\int_1^U \frac {(u - 1)^{n - 1}} {(u - \alpha)^n} du = \left( \frac {U - 1} {1 - \alpha} \right)^{\!n} \int_0^1 t^{n - 1} \left( 1 + \frac {U - 1} {1 - \alpha} t \right)^{\!-n} dt = \\ \frac 1 n \left( \frac {U - 1} {1 - \alpha} \right)^{\!n} {_2F_1} \!\left( n, n; n + 1; -\frac {U - 1} {1 - \alpha} \right).$$