Expectation of $\frac{1}{1+X}$ for Gamma

Question

I am trying to evaluate the following integral:

$$ \int_{0}^{\infty} \frac{1}{c+x} \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} dx $$

I have tried simple transformation and by-parts and nothing worked. Then I found this simple solution on internet at http://www.physicsforums.com/showthread.php?t=655393 :

$$ \int_{0}^{\infty} \frac{1}{c+x} \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} dx = \beta^{\alpha} c^{\alpha-1} e^{\beta c}\Gamma(1-\alpha, \beta c) $$

It would be great if someone could verify if this is indeed the answer. I don't know Maple or any other software that can evaluate integrals and hence there is no way to verify the correctness. Any reference (paper,book) would be great too.

I am a Bayesian Statistician and I need this for evaluating a posterior quantity.

Answer 1

Use $\displaystyle\frac1{c+x}=\int_0^\infty\mathrm e^{-(c+x)t}\mathrm dt$ and interchange the order of the integrals. This yields that the integral you wish to compute is $$ I=\int_0^\infty\mathrm e^{-ct}\int_{0}^{\infty}\frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1}\mathrm e^{-(\beta+t) x} \mathrm dx\mathrm dt=\int_0^\infty\mathrm e^{-ct}\frac{\beta^{\alpha}}{(\beta+t)^{\alpha}}\mathrm dt. $$ The change of variable $s=c(\beta+t)$ yields $$ I=\mathrm e^{c\beta}\beta^{\alpha}c^{1-\alpha}\int_{c\beta}^\infty s^{-\alpha}\mathrm e^{-s}\mathrm ds=\mathrm e^{c\beta}\beta^{\alpha}c^{1-\alpha}\Gamma(1-\alpha, \beta c). $$

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{{\cal J}\pars{\alpha, \beta, c} \equiv \int_{0}^{\infty}{1 \over c + x}\, {\beta^{\alpha} \over \Gamma\pars{\alpha}}\,x^{\alpha - 1}\expo{-\beta x}\,\dd x = {\beta^{\alpha} \over \Gamma\pars{\alpha}}\,{\cal K}\pars{\alpha, \beta, c}}\tag{1}$

\begin{align} &\mbox{where}\\ {\cal K}\pars{\alpha, \beta, c} & \equiv \int_{0}^{\infty}{x^{\alpha - 1}\expo{-\beta x} \over c + x}\,\dd x = \beta^{1 - \alpha}\int_{0}^{\infty}{\pars{\beta x}^{\alpha - 1}\expo{-\beta x} \over \beta x + \beta c}\,\dd\pars{\beta x} \\[3mm]&= \beta^{1 - \alpha}\int_{0}^{\infty}{x^{-\pars{1 - \alpha}}\expo{-x} \over x + \beta c}\,\dd x = \beta^{1 - \alpha}\, {\Gamma\pars{1 - \bracks{1 - \alpha}}\Gamma\pars{1 - \alpha,\beta c} \over \expo{-\beta c}\pars{\beta c}^{\alpha}} \\&\mbox{See G&R}\ {\bf 8.353}.3\,,\quad 7^{\underline{\rm a}}\ \mbox{ed. Page 900}. \\&\mbox{The result is valid whenever}\ \Re\alpha >0\ \mbox{and}\ \beta, c > 0. \\[3mm] {\cal K}\pars{a,b,c} &= \beta^{1 - 2\alpha}c^{-\alpha}\expo{\beta c} \Gamma\pars{\alpha}\Gamma\pars{1 - \alpha,\beta c} \end{align}

By replacing this result in $\pars{1}$, we found: $$ \color{#0000ff}{\large% \int_{0}^{\infty}{x^{\alpha - 1}\expo{-\beta x} \over c + x}\,\dd x = \beta^{1 - \alpha}c^{-\alpha}\expo{\beta c}\Gamma\pars{1 - \alpha,\beta c}} $$ Notice that it's slightly different of the OP proposed answer.