Entropy of Weilbull Distribution

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I have a hard integral that I would like to compute. As the tittle suggests, I'm interested in the Weibull distribution. For $k,\lambda >0$ its distribution function is given by:

$$f(x)= \frac{k}{\lambda}\bigg(\frac{x}{\lambda}\bigg)^{k-1}\exp\bigg[-\bigg(\frac{x}{\lambda}\bigg)^k\bigg]$$

I'm interested in computing its entropy, i.e.

$$-\int_0^{\infty} \frac{k}{\lambda}\bigg(\frac{x}{\lambda}\bigg)^{k-1}\exp\bigg[-\bigg(\frac{x}{\lambda}\bigg)^k\bigg] \ln \bigg\{ \frac{k}{\lambda}\bigg(\frac{x}{\lambda}\bigg)^{k-1}\exp\bigg[-\bigg(\frac{x}{\lambda}\bigg)^k\bigg]\bigg\}dx$$

All the usual techniques seem to fail, like integration by parts or substitution. I tried for instance setting $z:=(x/k)^{k-1}$ but it does not seem to lead anywhere. What can be some useful trick for this task?

Thanks in advance.