Expectation of $\frac{1}{1+X}$ for Gamma 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. 
 A: $\newcommand{\+}{^{\dagger}}%
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$\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.
A: 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
$$
I=\mathrm e^{c\beta}\beta^{\alpha}c^{1-\alpha}\int_{c\beta}^\infty s^{-\alpha}\mathrm e^{-s}\mathrm ds=\mathrm e^{c\beta}\beta^{\alpha}c^{1-\alpha}\Gamma(1-\alpha, \beta c). $$
