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 $$ J(\lambda, n) = \frac{\lambda^n}{\Gamma(n)} \int_0^\infty t^{n-1} e^{-\lambda t} \log(1+t)\ dt = \frac{1}{\Gamma(n)} \int_0^\infty s^{n-1} e^{-s} \log(1+s/\lambda) \ ds$$
Using Maple, I get
$$\Psi \left( n \right) -\ln \left( \lambda \right) +{\frac { {\mbox{$_2$F$_2$}(1,1;\,2,2-n;\lambda)}\lambda}{n-1}}+{\frac { \left( -1 \right) ^{-n}\pi }{\sin \left( \pi n \right) }}-{\frac { \left( -1 \right) ^{-n}\pi \Gamma \left( n,-\lambda \right) }{\sin \left( \pi n \right) \Gamma \left( n \right) }} $$
which seems to be correct when $n$ is a non-integer. 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 $n-2$.
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?