Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Assuming I have a Gamma distributed random Variable $x \sim Gamma( \alpha, \beta )$. Now I like to have the following two expectation values (integrals):

  1. $E \left[ x \ln x \right]$
  2. $E \left[ \ln \Gamma \left( x \right) \right]$ with $\Gamma \left( x \right)$ being the Gamma function

Many thanks in advance

share|cite|improve this question
At least at first glance, the second integral seems quite complicated, considering the expressions for the Log Gamma Function (see, for example, – Shai Covo Feb 10 '11 at 3:42

First integral. $X$ has density function $$ f(x;\alpha ,\beta ) = \frac{{\beta ^\alpha }}{{\Gamma (\alpha )}}x^{\alpha - 1} e^{ - \beta x},\;\; x > 0, $$ with $\alpha,\beta > 0$ fixed. It follows readily from $$ \Gamma '(\alpha ) = \int_0^\infty {x^{\alpha - 1} e^{ - x} \ln x\,{\rm d}x} $$ and $$ \Gamma '(\alpha ) = \Gamma (\alpha )\psi _0 (\alpha ), $$ where $\psi_0$ is the digamma function, that $$ {\rm E}[X\ln X] = \frac{{\beta ^\alpha }}{{\Gamma (\alpha )}}\int_0^\infty {(x\ln x)x^{\alpha - 1} e^{ - \beta x} dx} = \frac{\alpha }{\beta }[\psi_0 (\alpha + 1) - \ln \beta ]. $$

EDIT: Elaborating (in response to the OP's request).

First, you can find here the above formulas for $\Gamma'(\alpha)$. Now, using a change of variable, we have $$ \Gamma'(\alpha+1) = \int_0^\infty {\beta ^{\alpha+1} x^{\alpha} e^{ - \beta x} \ln (\beta x)\,{\rm d}x} = \ln \beta \int_0^\infty {\beta ^{\alpha+1} x^{\alpha} e^{ - \beta x} \,{\rm d}x} + \int_0^\infty { \beta ^{\alpha+1} x^{\alpha} e^{ - \beta x} \ln x\,{\rm d}x}. $$ The first integral on the right-hand side is equal to $\Gamma(\alpha+1)\ln \beta$, and the second integral to $\Gamma(\alpha)\beta {\rm E}[X\ln X]$. It thus follows that $$ \Gamma (\alpha + 1)\psi _0 (\alpha + 1) = \Gamma (\alpha + 1)\ln \beta + \Gamma (\alpha )\beta {\rm E}[X\ln X]. $$ Finally, by $\Gamma(\alpha+1)=\alpha \Gamma(\alpha)$, we get $$ {\rm E}[X\ln X] = \frac{\alpha }{\beta }[\psi_0 (\alpha + 1) - \ln \beta ]. $$

share|cite|improve this answer
Thank you very much! Can you give a hint, how you came to you solution? For me this is not clear... Do you have an idea about the other integral? – Matthias Feb 9 '11 at 14:30

1. E(XlnX)=(formula of expectation, group x) =E(lnY) with Y~gamma(alpha+1,beta)

E(lnY)=ψ(alpha+1)-ln(beta) where ψ(k) is the digamma function.

share|cite|improve this answer
wow. Really? Statistical manipulations are so inscrutable. How do you go about figuring out those answers? – Mitch Feb 9 '11 at 1:59
Hi! Thanks for your answer. But it looks like you are wrong. I sampled gamma distributed data and compared your answer, the answer of Shai Covo and the sampled expectation value. Your formula is quite far away from the sampled. However the one of Shai Covo is very close. – Matthias Feb 9 '11 at 14:26
yep, i was wrong, i forget to include de alpha/beta factor. – Sebastian Vallejo Feb 10 '11 at 5:15

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.