I would like to know if there is a closed form expression for the expectation of log(1+x) when x is a gamma random variable.
Thank you.
I would like to know if there is a closed form expression for the expectation of log(1+x) when x is a gamma random variable. Thank you. 


If $X$ has the gamma distribution with rate $\lambda$ and shape parameter $n$, you're asking for $$ J(\lambda, n) = \frac{\lambda^n}{\Gamma(n)} \int_0^\infty t^{n1} e^{\lambda t} \log(1+t)\ dt = \frac{1}{\Gamma(n)} \int_0^\infty s^{n1} e^{s} \log(1+s/\lambda) \ ds$$ Using Maple, I get $$\Psi \left( n \right) \ln \left( \lambda \right) +{\frac { {\mbox{$_2$F$_2$}(1,1;\,2,2n;\lambda)}\lambda}{n1}}+{\frac { \left( 1 \right) ^{n}\pi }{\sin \left( \pi n \right) }}{\frac { \left( 1 \right) ^{n}\pi \Gamma \left( n,\lambda \right) }{\sin \left( \pi n \right) \Gamma \left( n \right) }} $$ which seems to be correct when $n$ is a noninteger. For integer values of $n$, the result seems to be $\frac{\Gamma(n,\lambda)}{\Gamma(n)} Ei(1,\lambda)$ plus a polynomial in $\lambda$ of degree $n2$. 


I may be mistaken, but if you are making the change of variable $s = \lambda t$, shouldn't there be an extra factor of $\lambda$ outside the integral? 

