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.
 A: 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.
A: Hint:
$$I=\int^\infty_0\frac{t^{a-1}}{t-u}e^{-bt}dt=e^{-bu}\underbrace{\int^\infty_0\frac{t^{a-1}}{t-u}e^{-b(t-u)}dt}_{=J(b)}$$
$$J’=-\int^\infty_0t^{a-1}e^{-b(t-u)}dt=-\frac{e^{bu}}{b^a}\Gamma(a)$$
A: Here is my attempt at the solution.  I cannot seem to solve for the constant $c_1$ that comes from the solution of the differential equation.  I have solved this problem using an entirely different approach and get the same answer as here if I set $c_1=0$....Just not sure how to get it.

We wish to evaluate the following principal value integral
\begin{equation}
\Lambda=\frac{\beta^\alpha}{\Gamma(\alpha)}\operatorname{PV}\int_{0}^{\infty}\frac{x^{\alpha-1}}{x-\mu}e^{-\beta x}\,\mathrm{d}x.
\end{equation}
Begin by considering the contour integral
\begin{equation}
I=\oint_C\frac{z^{\alpha-1}}{z-\mu}e^{-\beta z}\,\mathrm{d}z,
\end{equation}
where $C$ is a wedge of angle $\pi/4$ of radius $R$ in the upper right quadrant, with a semicircular indentation of radius $\varepsilon$ into the upper half plane at $z=\mu$. The contour integral is equal to
\begin{multline}
I=%
\int_0^{\mu-\varepsilon}\frac{x^{\alpha-1}}{x-\mu}e^{-\beta x}\,\mathrm{d}x%
+\mathrm i\int_\pi^{0}(\mu+\varepsilon e^{\mathrm i\varphi})^{\alpha-1}e^{-\beta(\mu+\varepsilon e^{\mathrm i\varphi})}\,\mathrm{d}\varphi%
+\int_{\mu+\varepsilon}^{R}\frac{x^{\alpha-1}}{x-\mu}e^{-\beta x}\,\mathrm{d}x\\%
+\mathrm i\int_0^{\pi/4}\frac{(R e^{\mathrm i\varphi})^{\alpha}}{Re^{\mathrm i\varphi}-\mu}e^{-\beta R e^{\mathrm i\varphi}}\,\mathrm{d}\varphi%
+(e^{\mathrm i\pi/4})^\alpha\int_R^0\frac{t^{\alpha-1}}{e^{\mathrm i\pi/4}t-\mu}e^{-\beta e^{\mathrm i\pi/4}t}\,\mathrm{d}t%
\end{multline}
As $\varepsilon\to0$ and $R\to\infty$ the sum of the first and third integrals approaches the desired principal value and the fourth integral vanishes.  Since no poles lie inside the contour, the integral $I$ is equal to zero and we find
\begin{equation}
\operatorname{PV}\int_0^{\infty}\frac{x^{\alpha-1}}{x-\mu}e^{-\beta x}\,\mathrm{d}x=%
\mu^{\alpha-1}e^{-\beta\mu}\pi\mathrm i%
+(e^{\mathrm i\pi/4})^\alpha\int_0^\infty\frac{t^{\alpha-1}}{e^{\mathrm i\pi/4}t-\mu}e^{-\beta e^{\mathrm i\pi/4}t}\,\mathrm{d}t.
\end{equation}
For the remaining integral we define
\begin{equation}
J(\beta)=%
(e^{\mathrm i\pi/4})^\alpha e^{-\beta\mu}\int_0^\infty\frac{t^{\alpha-1}}{e^{\mathrm i\pi/4}t-\mu}e^{-(e^{\mathrm i\pi/4}t-\mu)\beta}\,\mathrm{d}t.
\end{equation}
Differentiating $J$ w.r.t. $\beta$ yields
\begin{equation}
J^\prime(\beta)=-\mu J(\beta)%
-(e^{\mathrm i\pi/4})^\alpha \int_0^\infty t^{\alpha-1}e^{-\beta e^{\mathrm i\pi/4}t}\,\mathrm{d}t.
\end{equation}
Integrating and then rearranging terms yields the first order equation
\begin{equation}
J^\prime(\beta)+\mu J(\beta)=-\frac{\Gamma(\alpha)}{\beta^\alpha}.
\end{equation}
Solving for $J$ one finds
\begin{equation}
J(\beta)=%
-e^{-\beta\mu}\int\frac{e^{\beta\mu}}{\beta^\alpha}\,\mathrm d\beta+c_1e^{-\beta\mu},
\end{equation}
where the constant $c_1$ needs to be determined. The integral is evaluated with formula $2.33.10$ in Gradshteyn & Ryzhik to get
\begin{equation}
J(\beta)=%
e^{-\beta\mu}(-\mu)^{\alpha-1}\Gamma(1-\alpha,-\beta\mu)+c_1e^{-\beta\mu}.
\end{equation}
Using the definition of the generalized exponential function $E_\nu(z)= z^{\nu-1}\Gamma(1-\nu,z)$ we write
\begin{equation}
J(\beta)=%
\Gamma(\alpha)\beta^{1-\alpha}e^{-\beta\mu} E_\alpha(-\beta\mu)+c_1e^{-\beta\mu}.
\end{equation}

suppose $c_1=0$, then

\begin{equation}
\operatorname{PV}\int_0^{\infty}\frac{x^{\alpha-1}}{x-\mu}e^{-\beta x}\,\mathrm{d}x=%
\mu^{\alpha-1}e^{-\beta\mu}\pi\mathrm i%
+\Gamma(\alpha)\beta^{1-\alpha}e^{-\beta\mu} E_\alpha(-\beta\mu).
\end{equation}
It is then obvious that
\begin{equation}
\Lambda=\beta e^{-\beta\mu}\left(\frac{(\beta\mu)^{\alpha-1}}{\Gamma(\alpha)}\pi\mathrm i%
+E_\alpha(-\beta\mu)\right).
\end{equation}
Using the series representation of the generalized exponential integral one can show
\begin{equation}
\Im\{E_\alpha(-\beta\mu)\}=-\frac{(\beta\mu)^{\alpha-1}}{\Gamma(\alpha)}\pi.
\end{equation}
This leads to the more compact expression
\begin{equation}
\Lambda=\beta e^{-\beta\mu}\Re\{E_\alpha(-\beta\mu)\}.
\end{equation}
