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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.

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Yes, you may need Digamma function. See en.wikipedia.org/wiki/Polygamma_function –  Anand Jan 11 '12 at 15:58
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2 Answers 2

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 $n-2$.

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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?

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