Evaluating integral (comes from a bigger problem in statistics) Let $\alpha, \beta>0$ be parameters. I wish to compute
$$\int_0^\infty \frac{x^\alpha}{x-1} e^{-\beta x} dx.$$
I managed to reduce this problem when $\alpha$ is integer by using
$$\frac{x^\alpha}{x-1}=\frac{1}{x-1}+\sum_{j=1}^{\alpha} x^{j-1}.$$
So the question is how to compute
$$\int_0^\infty \frac{1}{x-1} e^{-\beta x} dx.$$
Any ideas? Thanks a lot!
 A: The integral of interest, $\int_0^\infty \frac{e^{-\beta x}}{x-1}\,dx $ diverges due to the singularity at $x=1$.  However, the Cauchy Principal Value of the integral exists and can be expressed as
$$\begin{align}
\text{PV}\int_0^\infty \frac{e^{-\beta x}}{x-1}\,dx &=\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon}\frac{e^{-\beta x}}{x-1}\,dx+\int_{1+\epsilon}^\infty \frac{e^{-\beta x}}{x-1}\,dx\right)\\\\
&=e^{-\beta }\lim_{\epsilon\to0^+}\left(\int_{-\beta}^{-\beta\epsilon}\frac{e^{- x}}{x}\,dx+\int_{\beta\epsilon}^\infty \frac{e^{- x}}{x}\,dx\right)\\\\
&=-e^{-\beta}\text{Ei}(\beta)
\end{align}$$
in terms of the Exponential Integral $\text{Ei}(x)\equiv -\text{PV}\int_{-x}^\infty \frac{e^{-x}}{x}\,dx$.
If $\alpha\in \mathbb{N}$ with $\alpha\ge 1$, then we have
$$\begin{align}
\text{PV}\int_0^\infty \frac{x^\alpha e^{-\beta x}}{x-1}\,dx&=(-1)^{\alpha+1} \frac{d^\alpha}{d\beta^\alpha}\left(e^{-\beta}\text{Ei}(\beta)\right)\\\\
&=-e^{-\beta}\text{Ei}(\beta)+\sum_{m=1}^\alpha \frac{(m-1)!}{\beta^m}
\end{align}$$
A: One method is to use differentiation and 
$$\int_{0}^{\infty} \frac{x^{\alpha} \, dx}{x-1} = \pi \csc(\pi (\alpha + 1)) \, (-1)^{\alpha}.$$
Let $I(\beta)$ be 
$$I(\beta) = \int_{0}^{\infty} \frac{e^{-\beta x} \, x^{\alpha} \, dx}{x-1}$$
which leads to
\begin{align}
- I' - I &= \int_{0}^{\infty} e^{-\beta x} \, x^{\alpha} \, dx \\
&= \frac{\Gamma(\alpha + 1)}{\beta^{\alpha + 1}}.
\end{align}
The solution of $$I' + I = - \frac{\Gamma(\alpha + 1)}{\beta^{\alpha + 1}}$$
is $$I(\beta) = c_{1} \, e^{-\beta} + (-1)^{\alpha} \, \Gamma(\alpha + 1) \, e^{-\beta} \, \Gamma(-\alpha, - \beta)$$
The difficulty in this solution is that if $\beta = 0$ then to be a finite solution $R(\alpha) < 0$ must be stated. Other conditions may apply. Continuing without concern of ranges the case would be
$$I(0) = \pi \csc(\pi (\alpha + 1)) \, (-1)^{\alpha} = c_{1} + (-1)^{\alpha} \, \Gamma(\alpha + 1) \, \Gamma(-\alpha, 0)$$
and
$$I(\beta) = (-1)^{\alpha} \, e^{- \beta} \, \left[ \pi \csc(\pi (\alpha + 1)) + \Gamma(\alpha + 1) \, (\Gamma(-\alpha, - \beta) - \Gamma(-\alpha, 0)) \right],$$
where $Re(\alpha) < 0$ and other conditions may apply.
