The integral is: $$\frac{1}{\Gamma{(1-\alpha)}}\int^\infty_0(1-e^{-\lambda y})e^{-y}y^{-(1+\alpha)} dy$$
where $\alpha\in(0,1)$ and $y,\lambda>0$.
I've tried using IBP, in which I get: $$\frac{1}{\alpha\Gamma(1-\alpha)}[-e^{-y}y^{-\alpha}|^\infty_0 - \Gamma(1-\alpha)]$$
where I don't think its possible to evaluate the first term.
I've also just tried using the integral definition of Gamma, which I got a nice answer with but is invalid as the integral for $\Gamma(-\alpha) = \int^\infty_0e^{-y}y^{-\alpha-1}dy$ is not defined for negative numbers.
Where should I go from here?