A Bound based on Jensen's Inequality Consider:
$$X \sim \text{Gamma}(\alpha, \beta)$$
$$Y = \frac{1}{X+c}, \ c > 0$$
I am interested in $E(Y)$, which I'm pretty sure is intractable...
$$E(Y) = \frac{\beta^\alpha}{\Gamma(\alpha)}\int_0^\infty \frac{x^{\alpha-1}e^{-\beta x}}{x+c} dx $$
If an exact analytic answer is unavailable, I would like to approximate this in terms of $\alpha$, $\beta$ and $c$.

By Jensen's Inequality, we have:
$$E\left[\frac{1}{X+c}\right] \geq \frac{1}{E(X) + c}$$
Or equivalently, for some $\epsilon(\alpha, \beta, c) > 0$:
$$E(Y) = \frac{1}{E(X) + c} + \epsilon(\alpha, \beta, c)$$
Can we say anything about the magnitude of $\epsilon(\alpha, \beta, c)$. If not, is there a better way to approach this problem?
Thanks in advance!
 A: According to Rockower (Integral Identities for Random Variables, $\it{The \ American \ Statistician}$, February 1988, Vol. 42, No. 1), the expectation is given by $$E \left[ \frac{1}{X+c} \right] =\int_0^{\infty} e^{-cs} \cal{L}(s)ds,$$ where $\cal{L} (s)$ is the Laplace transform of the pdf of the random variable $X$.
From a table of Laplace transforms we find $$\cal{L} \left( x^{\alpha-1} e^{-x/{\beta}} \right)=\frac{\Gamma \left( \alpha \right)}{\left(s+\frac{1}{\beta} \right)^{\alpha} } $$ 
So for the Laplace transform of the pdf we get $$\cal{L} \left( \frac{x^{\alpha-1} e^{-x/{\beta}}}{\beta^{\alpha} \ \Gamma  \left( \alpha \right)  }\right)=\frac{1}{\beta^{\alpha} \left(s+\frac{1}{\beta} \right)^{\alpha} } $$
Then $$E \left[ \frac{1}{X+c} \right] =\int_0^{\infty} \frac{e^{-cs}}{\beta^{\alpha} \left( s + \frac{1}{\beta} \right)^{\alpha}} ds=\int_0^{\infty} \frac{e^{-cs}}{ \left( \beta s + 1 \right)^{\alpha}} ds$$
Now make the substitution $q=\beta s + 1.$ The result is
$$E \left[ \frac{1}{X+c} \right] =\int_1^{\infty} \frac{e^{-c \left( \frac{q-1}{\beta} \right)}}{\beta \ q ^{\alpha}} dq=\frac{1}{\beta}e^{c/{\beta}}\int_1^{\infty} \frac{e^ { - \left( {\frac{c}{\beta}} \right) q } }{  q^{\alpha}} dq$$
From the definition for the generalized exponential integral function given in the comment by wolfies, we have finally 
$$ E \left[ \frac{1}{X+c} \right]= \frac{e^{c/{\beta}}}{\beta} E_{\alpha} \left( \frac{c}{\beta} \right), $$
which agrees with his answer using the computer algebra system. 
