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

Question

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 1

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.

Answer 2

Hint:

$$I=\int^\infty_0\frac{t^{a-1}}{t-u}e^{-bt}dt=e^{-bu}\underbrace{\int^\infty_0\frac{t^{a-1}}{t-u}e^{-b(t-u)}dt}_{=J(b)}$$

$$J’=-\int^\infty_0t^{a-1}e^{-b(t-u)}dt=-\frac{e^{bu}}{b^a}\Gamma(a)$$

Answer 3

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.

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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}