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: .
 A: $$\begin{align}\int_1^\infty  dy\, (y-1)^{n-1} y^{-n} e^{(\alpha -\alpha  n)\frac{(y-1) }{y}} &= e^{-\alpha (n-1)} \int_0^1 \frac{du}{u^2} \left (\frac1{u}-1 \right )^{n-1} u^n e^{\alpha (n-1) u} \\ &= e^{-\alpha (n-1)} \int_0^1 du (1-u)^{n-1} u^{-1} e^{\alpha (n-1) u} \end{align}$$
The integral diverges for all values of $n$.
A: \begin{align}
\int_1^\infty (y-1)^{n-1} y^{-n} e^{(\alpha -\alpha n)\frac{(y-1) }{y}} \; dy
\\=\int_1^\infty \frac{(y-1)^{n-1}}{y^{n-1}y} e^{(\alpha -\alpha n)\frac{(y-1) }{y}} \; dy
\\=\int_1^\infty {(1-\frac{1}{y})^{n-1}}{\frac{1}{y}} e^{(\alpha -\alpha n){(1-\frac{1}{y}})} \; dy
\end{align}
Let z = -ln(y) then dz = -dy/y. So $\frac{1}{y} = e^{z}$.
When z = 1, y = 1, and when z = $\infty$, y = 0.
\begin{align}
\\=\int_0^1 {(1- e^z)^{n-1}}e^{(\alpha -\alpha n){(1-e^z})} \; dz
\\=\int_0^1 {(1- e^z)^{n-1}}e^{(\alpha -\alpha n){(1-e^z})} \; dz
\end{align}
Please check the above for any error during substitution. 
A: 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 
$$ e^{(\alpha -\alpha  n)\frac{(y-1) }{y}}\to e^{(\alpha -\alpha  n) } $$
