Looking to solve an integral of the form $\int_1^\infty (y-1)^{n-1} y^{-n} e^{(\alpha -\alpha n)\frac{(y-1) }{y}} \; dy$

Question

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: .

Answer 1

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 $$ e^{(\alpha -\alpha n)\frac{(y-1) }{y}}\to e^{(\alpha -\alpha n) } $$

Answer 2

\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.

Answer 3

$$\begin{align}\int_1^\infty dy\, (y-1)^{n-1} y^{-n} e^{(\alpha -\alpha n)\frac{(y-1) }{y}} &= e^{-\alpha (n-1)} \int_0^1 \frac{du}{u^2} \left (\frac1{u}-1 \right )^{n-1} u^n e^{\alpha (n-1) u} \\ &= e^{-\alpha (n-1)} \int_0^1 du (1-u)^{n-1} u^{-1} e^{\alpha (n-1) u} \end{align}$$

The integral diverges for all values of $n$.