calculus - 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!

Answer

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

$$\begin{align} \text{PV}\int_0^\infty \frac{x^\alpha e^{-\beta x}}{x-1}\,dx&=(-1)^{\alpha+1} \frac{d^\alpha}{d\beta^\alpha}\left(e^{-\beta}\text{Ei}(\beta)\right)\\\\ &=-e^{-\beta}\text{Ei}(\beta)+\sum_{m=1}^\alpha \frac{(m-1)!}{\beta^m} \end{align}$$