Looking to solve an integral of the form $\int_1^\infty (y-1)^{n-1} y^{-n} e^{(\alpha -\alpha n)\frac{(y-1) }{y}} \; dy$ Looking for a solution the following integral. With $n \geq2$, $\alpha>1$,
$$z(n,\alpha)=\frac{\left(\alpha  (n-1)\right)^n}{\Gamma (n)-\Gamma (n,(n-1) \alpha )} \int_1^\infty  (y-1)^{n-1} y^{-n} e^{(\alpha -\alpha  n)\frac{(y-1) }{y}} \; dy $$
I am adding the numerical integration for different values of $\alpha$.

I am showing how the integral behaves (for $n=5$ and $\alpha = 3/2$ as one answer was that it does not converge: .

Integrand behaves as $\frac{C}{y}$ as $y\rightarrow\infty$; hence the integral diverges.
Indeed, 
$$ (y-1)^{n-1} y^{-n} \sim y^{-1}$$
as $y\to\infty$, and 
$$ e^{(\alpha -\alpha  n)\frac{(y-1) }{y}}\to e^{(\alpha -\alpha  n) } $$
