closed form expression $\int_{0}^{\infty}[1-\epsilon e^{-\mu(z-\alpha)}]^n \lambda e^{-\lambda \alpha} d\alpha$ I am trying to solve the following integration (closed form solution)
$$
\int_{0}^{\infty}[1-\epsilon e^{-\mu(z-\alpha)}]^n \lambda e^{-\lambda \alpha} d\alpha
$$
I tried using substitution and then using beta function definition [$\beta(x,y)=\int_0^1 t^{x-1} (1-t)^{y-1}dt$]  to get the closed form solution as taking $y=\epsilon e^{-\mu(z-\alpha)}$ and the interval is (after substation).
$$
\int_{0}^{\infty}[1-y]^n y^{-1-\frac{\mu}{\lambda}} dy
$$
which is unsolvable or the integral goes to infinity (using matlab).
 A: With help of Mathematica I have:
$$\int_0^{\infty } (1-\epsilon  \exp (-\mu  (z-\alpha )))^n \lambda  \exp (-\lambda  \alpha ) \, d\alpha =\\\lambda  \left(\frac{e^{-\lambda  \left(z+\frac{i \pi }{\mu }\right)} \epsilon ^{\lambda /\mu }
   \Gamma \left(-n+\frac{\lambda }{\mu }\right) \Gamma \left(-\frac{\lambda }{\mu }\right)}{\mu  \Gamma (-n)}+\frac{\, _2F_1\left(-n,-\frac{\lambda }{\mu };1-\frac{\lambda }{\mu };e^{-z \mu } \epsilon
   \right)}{\lambda }\right)=\\\lambda  \left(\frac{e^{-\lambda  \left(z+\frac{i \pi }{\mu }\right)} \epsilon ^{\lambda /\mu } \Gamma \left(-n+\frac{\lambda }{\mu }\right) \Gamma \left(-\frac{\lambda
   }{\mu }\right)}{\mu  \Gamma (-n)}-\frac{e^{-z \lambda } \epsilon ^{\lambda /\mu } \lambda  B_{e^{-z \mu } \epsilon }\left(-\frac{\lambda }{\mu },1+n\right)}{\mu  \lambda }\right)$$
Mathematica code:
Integrate[(1 - \[Epsilon] Exp[-\[Mu] (z - \[Alpha])])^ n*\[Lambda] Exp[-\[Lambda] \[Alpha]], {\[Alpha], 0,  Infinity}] == \[Lambda] (( E^(-\[Lambda] (z + (I \[Pi])/\[Mu])) \[Epsilon]^(\[Lambda]/\[Mu]) Gamma[-n + \[Lambda]/\[Mu]] Gamma[-(\[Lambda]/\[Mu])])/(\[Mu] \ Gamma[-n]) +  Hypergeometric2F1[-n, -(\[Lambda]/\[Mu]), 1 - \[Lambda]/\[Mu],  E^(-z \[Mu]) \[Epsilon]]/\[Lambda]) == \[Lambda] (( E^(-\[Lambda] (z + (I \[Pi])/\[Mu])) \[Epsilon]^(\[Lambda]/\[Mu]) Gamma[-n + \[Lambda]/\[Mu]] Gamma[-(\[Lambda]/\[Mu])])/(\[Mu] \ Gamma[-n]) - ( E^(-z \[Lambda]) \[Epsilon]^(\[Lambda]/\[Mu]) \[Lambda] Beta[ E^(-z \[Mu]) \[Epsilon], -(\[Lambda]/\[Mu]),  1 + n])/(\[Mu]*\[Lambda]))
If: $\epsilon <e^{z \mu }\land n \mu <\lambda \land e^z\leq \epsilon ^{1/\mu}$
