Evaluating $\int_0^1 (\ln(x)\ln(1-x))^n dx$

Question

Interested in evaluating:

$\displaystyle\int_0^1 (\ln(x)\ln(1-x))^n \mathrm{d}x,$ where $n \in \Bbb Z^+.$

I don't really know how to tackle this problem for $n > 1.$

I was able to get to this step: $\displaystyle\int_0^1 (\ln(x)\ln(1-x))^n \mathrm{d}x= \int_0^1\left(\sum_{n=1}^\infty\frac{x^n}n\ln x\right)^n \mathrm{d}x$.

This question is related to: Evaluate $ \int_{0}^{1} \ln(x)\ln(1-x)\,dx $

u-substitution and integration by parts – John Zimmerman Feb 11 '19 at 02:27 — John Zimmerman, Feb 11 '19 at 02:27
Why don't you use the beta function? (take some derivates). – Zacky Feb 11 '19 at 03:01 — Zacky, Feb 11 '19 at 03:01

score 9 · Accepted Answer · edited Feb 11 '19 at 14:02

9

$$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}dt$$ Where $B(x,y)$ is the Beta Function. Thus $$\int_0^1 \left(\ln(t)\ln(1-t)\right)^ndt=\left(\frac{d}{dy}\frac{d}{dx}\right)^nB(x,y)|_{(x,y)=(1,1)}$$

edited Feb 11 '19 at 14:02

Barry Cipra

79,832

answered Feb 11 '19 at 03:26

aleden

4,007

I love this method. Nice work (+1) – clathratus Feb 11 '19 at 04:50
1

Note that for actual computation, one should look at the log derivative. – Simply Beautiful Art Feb 11 '19 at 04:54
Came here to post this! you beat me to it! :-) – Feb 11 '19 at 06:55
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@SimplyBeautifulArt - Won't that fall apart with it being the nth derivative? It seems it will be a nightmarish mess no matter what approach is taken. – Feb 11 '19 at 07:04
@aleden - Not to critique your work - but for anyone reading, the ability to use this identity relies on the Beta Function's Integrand complying with both the Dominated Convergence Theorem and Leibniz's Integral Rule. [ – Feb 11 '19 at 08:21
@DavidG seems to be the nicest place to start, but yeah, it'll still be messy. – Simply Beautiful Art Feb 12 '19 at 02:56

Evaluating $\int_0^1 (\ln(x)\ln(1-x))^n dx$

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