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This problem is similar to the one solved here except for an additional term in $u$:

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

in which $0 \le i \le n$ and both $i$ and $n \in \mathbb{N}$. In the specific case where $i=0$ we are back to the solution given in the above link. However, is there a solution in this more general problem?

This equation shows up when trying to solve an estimator for the Weibull distribution.

Denis Cousineau
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  • Seems related to: $$\int_{0}^{\infty}e^{-kx}\log^2 x,dx$$ for $k\in\mathbb N.$ Specifically, if you substitute $u=1-e^{-x}.$ But I don’t know how to do this integral, either. It just looks a smidge easier. – Thomas Andrews Nov 27 '21 at 22:49
  • The title say $n-1$ where the body says $n-i.$ – Thomas Andrews Nov 27 '21 at 22:50
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    Actually, you can get it into your original form by just expanding $$u^i=(1-(1-u))^i=\sum_{j=0}^{i}\binom ij (-1)^j(1-u)^j$$ – Thomas Andrews Nov 27 '21 at 23:03

1 Answers1

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As @ThomasAndrews mentions in his comment, you can obtain this case from the linked answer as follows: $$ \int_0^1 (1-u)^{(n-i)} u^i \ln^2(-\ln(1-u)) \mathrm{d}u =\sum_{j=0}^{i}\binom{i}{j} (-1)^j \int_{0}^{1} (1-u)^{n-i+j} \ln^2(-\ln(1-u)) \mathrm{d}u $$ and after some direct simplifications, this results in $$ \boxed{\int_0^1 (1-u)^{(n-i)} u^i \ln^2(-\ln(1-u)) \mathrm{d}u = \frac{\pi^2}{6} \frac{i! (n-i)!}{(n+1)!} +\sum_{k=1}^{i+1} \binom{i}{k-1} (-1)^{k+1} \frac{\left(\gamma + \ln(n-i+k) \right)^2}{(n-i+k)}} $$ which I doubt can be simplified further.

Robert Lee
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