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I am currently working on finding a closed-form equation for the variance of the ordered data of the three-parameter Weibull distribution. One method has been proposed, but the published solution is using numerical integration. A much faster solution would need to simplify the expression, which lead me to this expression.

Is there a solution in closed-form for this equation in which $n \in \mathbb{N} >1$?

$$ \int_0^1 (1-u)^n (\log(-\log(1-u)))^2 \mathrm{d}u $$

If not, is there a good approximation?

Denis Cousineau
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1 Answers1

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By the substitution $x = \ln\left(\frac{1}{1-u} \right)$ your integral is equal to $$ \int_{0}^{\infty}e^{-x(n+1)} \ln^2(x) \, \mathrm{d}x $$ which by this answer gives you $$ \boxed{\int_0^1 (1-u)^n \ln^2\left(\ln\left(\frac{1}{1-u}\right)\right)\, \mathrm{d}u = \frac{1}{n+1}\left(\dfrac{\pi^2}{6} + \left(\gamma+ \ln(n+1)\right)^2\right)} $$

Robert Lee
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