complex analysis - Solution for the Laplace Transform $mathscr{L}left{frac{t^{alpha-1}}{t-mu}right}(beta)$

I have been looking for an explicit solution to the following Laplace transform for $\alpha,\mu,\beta>0$

$$\begin{equation} \frac{\beta^\alpha}{\Gamma(\alpha)}\mathscr{L}\left\{\frac{t^{\alpha-1}}{t-\mu}\right\}(\beta) =\frac{\beta^\alpha}{\Gamma(\alpha)}\operatorname{PV\int_0^\infty}\frac{t^{\alpha-1}}{t-\mu}e^{-\beta t}\,\mathrm dt. \end{equation}$$
Notice that this is equivalent to $E((X-\mu)^{-1})$ for $X\sim\text{Gamma}(\alpha,\beta)$. I attacked this problem by substituting $x=t-\mu$ and then writing
$$\begin{equation} \tag{1} \lim_{\epsilon\to0^+} \frac{\beta^\alpha}{\Gamma(\alpha)}e^{-\beta\mu}\left(\int_{-\mu}^0+\int_0^\infty\right) x^{\epsilon-1}(x+\mu)^{\alpha-1}e^{-\beta x}\,\mathrm dx. \end{equation}$$
After evaluating these integrals, I took the limit and and then dropped the residue term $-f_X(0)\pi\mathrm i$ resulting from the pole when $\epsilon=0$. This took me about six pages to do and involves some nasty derivatives of hypergoemetric functions w.r.t. parameters, meijer $G$ functions, residue expansions, etc. The solution I got was
$$\begin{equation} \frac{\beta^\alpha}{\Gamma(\alpha)}\mathscr{L}\left\{\frac{t^{\alpha-1}}{t-\mu}\right\}(\beta) =\beta e^{-\beta\mu}\,\Re\left\{E_\alpha(-\beta\mu)\right\}, \end{equation}$$
where $E_\nu(z)$ is the generalized exponential integral. I have explicit solution for the real part which must be separated into cases for integer and non-integer $\alpha$. Could this solution have been easily reached with contour integrals? After ending up at such a simple solution, I feel like I solved this the hard way and there is a much easier way to do it.

Answer

This may be somewhat simpler. Let $0 < \alpha < 1, \operatorname{Im} \mu \neq 0, 0 < \beta$ (so it may be a bit misleading to refer to the integral as the Laplace transform). We have

$$\int_0^\infty \frac d {d\beta} \left( e^{\mu \beta} \frac {t^{\alpha - 1}} {t - \mu} e^{-\beta t} \right) dt = -\Gamma(\alpha) \beta^{-\alpha} e^{\mu \beta}, \\ \int_0^\infty \frac {t^{\alpha-1}} {t-\mu} e^{-\beta t} dt = \underbrace {\Gamma(\alpha) (-\mu)^{\alpha-1} e^{-\mu \beta} \Gamma(1-\alpha, -\mu \beta)}_{=I(\alpha, \mu, \beta)} + C(\alpha, \mu).$$
The integral and $I(\alpha, \mu, \beta)$ are continuous at $\beta=0$, and
$$\int_0^\infty \frac {t^{\alpha-1}} {t-\mu} dt = \pi (-\mu)^{\alpha-1} \csc \pi \alpha = I(\alpha, \mu, 0),$$
from which we can conclude that $C(\alpha, \mu) = 0$.

For positive $\mu$, the p.v. integral will be equal to $I(\alpha, \mu - i0, \beta)$ plus $i \pi$ times the residue of the integrand at $t=\mu$. $\Gamma(a, z)$ is continuous from above at negative $z$, thus $I(\alpha, \mu, \beta)$ is continuous from below: $I(\alpha, \mu - i0, \beta) = I(\alpha, \mu, \beta)$, therefore

$$\operatorname{v.\!p.} \int_0^\infty \frac {t^{\alpha-1}} {t-\mu} e^{-\beta t} dt = I(\alpha, \mu, \beta) + i \pi \mu^{\alpha-1} e^{-\mu \beta}.$$
The result extends to $1 \leq \alpha$ by analytic continuation.