MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Yes, you may need Digamma function. See – Anand Jan 11 '12 at 15:58

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 $n-2$.

share|cite|improve this answer

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?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.