Let $w(x) = x \log{x}$

$x \sim Gamma(\alpha = 3.7, \lambda = 1)$

Find $E[w(x)]$

I have set up the following integral:

$\int_0^{\infty} x\log{x} \frac{\lambda^{\alpha}}{\Gamma(\alpha)} x^{\alpha -1} e^{-(\lambda)x}dx$

Brute-forcing it doesn't seem to be working, but I can't find the "trick".

I know from simulation that it should be approximately 5.32, I just need to compare it to the analytical solution.

  • $\begingroup$ you are missing $x^{\alpha-1}$ in the Gamma density $\endgroup$
    – Lii
    Commented Mar 29, 2020 at 15:29
  • $\begingroup$ Thank you, unfortunately I have not made that typo on my page - it's still a baffling integral. $\endgroup$
    – jbpib27
    Commented Mar 29, 2020 at 15:37
  • $\begingroup$ The solutions at stats.stackexchange.com/questions/370880 apply with little change. $\endgroup$
    – whuber
    Commented Mar 29, 2020 at 18:56
  • $\begingroup$ I’ve used it in my post @whuber $\endgroup$
    – gunes
    Commented Mar 29, 2020 at 19:02
  • 1
    $\begingroup$ @gunes Yes, I noticed that after posting my comment (and upvoted your answer, btw). $\endgroup$
    – whuber
    Commented Mar 29, 2020 at 19:08

1 Answer 1


$$\begin{align}E[X\log X]&=\int_0^\infty \log x \frac{\lambda^\alpha}{\Gamma(\alpha)}x^\alpha e^{-\lambda x}dx\\&=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\lambda}\int_0^\infty\log x\frac{\lambda^{\alpha+1}}{\Gamma(\alpha+1)}x^{(\alpha+1)-1}e^{-\lambda x}dx\\&=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\lambda}E[\log Y]\end{align}$$ where $Y\sim \text{Gamma}(\alpha_y=\alpha+1, \lambda_y=\lambda)$. From here, the expected value of the logarithm of a gamma RV is: $$E[\log Y]=-\log\lambda+\psi(\alpha+1)$$ where $\psi$ stands for polygamma function.

So, the overall result is (put $\alpha=3.7, \lambda=1$): $$E[X\log X]=\underbrace{\frac{\Gamma(4.7)}{1\times\Gamma(3.7)}}_{3.7}(-\log 1 +\psi(4.7))\approx5.32$$

I've used Matlab's psi function to calculate the polygamma function.

  • 1
    $\begingroup$ Thanks, I have added your derivation here. $\endgroup$
    – Joram Soch
    Commented Nov 3, 2021 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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