How can we evaluate this definite integral $$I=\int_0^{1/\phi}x\log(x)\log(1+x)\log(1-x)\,dx,$$ where $\displaystyle\phi=\frac{1+\sqrt5}2$ is the golden ratio?
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epimorphic
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Laila Podlesny
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2what have u tried? what are your thoughts? – tired Aug 14 '15 at 19:48
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Differtitating w.r.t to a parameter seems to be a good idea. ugy trilogs will be involved... – tired Aug 14 '15 at 20:04
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2@tired Trilogs are beautiful! – Vladimir Reshetnikov Aug 14 '15 at 21:38
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@VladimirReshetnikov Tastes are different, luckily :) – tired Aug 15 '15 at 11:03
2 Answers
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How can we evaluate ?
Integrating by parts, we get \begin{align} \mathcal{I}(x)&=\int x \ln x\ln(1+x)\ln(1-x)\,dx=\\ &=-\frac12\int \ln x\ln(1+x)\ln(1-x)\,d(1-x^2)=\\ &=-\frac12 (1-x^2)\ln x\ln(1+x)\ln(1-x)\\ &\quad+\frac12\int\frac{\ln (1-x)\ln(1+x)}{x}dx\\&\quad-\frac12\underbrace{\int x\ln (1-x)\ln(1+x)\,dx}_{\text{elementary}}\\ &\quad +\underbrace{\frac12\int(1-x)\ln x\ln(1-x)dx-\frac12\int(1+x)\ln x\ln(1+x)dx}_{-\frac18\operatorname{Li}_2(x^2)+\text{elementary}}. \end{align} The second term antiderivative is the most complicated as naively it involves trilogarithms. However it can be simplified using the same trick that I used answering your other question: \begin{align*} \int\frac{\ln (1-x)\ln(1+x)}{x}dx=\frac12\int\frac{\ln^2(1-x^2)}{x}dx- \frac12\int\frac{\ln^2\frac{1+x}{1-x}}{x}dx \end{align*} Calculating these two antiderivatives separately, Mathematica simplifies them to a much more compact (one-line) expression, still involving trilogarithms.
It may well happen that due to special integration bounds the tri- (and?) dilogarithms may be lifted using various polylogarithm identities (see equation (2) here and equations (12)-(19) here for a few examples). However I find such calculations rather boring, so for that you should wait for an answer of You-Know-Who.
Start wearing purple
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(+1) nice, to know that other persons here also dislike this kind of calculations ;)! – tired Aug 15 '15 at 11:06
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6After simplifications, the result is $$\left(\frac{31}{24}-\phi\right)\ln^3\phi+\left(\frac{5\phi}4-\frac{13}8\right)\ln^2\phi+\left(\frac{\pi^2}{60}+\frac{11\phi}4-\frac{15}4\right)\ln\phi\-\frac14 \operatorname{Li}_3(1-\phi)-\frac3{10}\zeta(3)-\frac{\pi^2}{120}+\frac{3\phi}8- \frac34.$$ – Vladimir Reshetnikov Aug 15 '15 at 18:01
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@VladimirReshetnikov Great! Do you think that $\operatorname{Li}_3(1-\phi)$ is non-simplifiable? (I thought that $1-\phi=\phi^{-2}-1$ could lead to something). – Start wearing purple Aug 15 '15 at 18:11
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Just an idea (for now) $$ \lim_{a,b,c\to0}\partial_{abc}\int_0^{1/\phi}x^{a+1}(1+x)^b(1-x)^cdx $$ changing variables to $x\phi = y$ $$ \frac{1}{\phi^{a+2}}\int_0^{1}y^{a+1}(1+\frac{1}{\phi}y)^b(1-\frac{1}{\phi}y)^cdy $$ using the Appel Hyper geometric we find $$ \int_0^1x^{\alpha-1}(1-a_1x)^{-\beta}(1-b_1x)^{-\beta'}dx = \frac{\Gamma(\alpha)\Gamma(1)}{\Gamma(\alpha+1)}F_1\left(\alpha,\beta,\beta';\gamma;a_1,b_1\right) $$ where $\gamma-\alpha = 1$. comparing terms we find $$ a+1 = \alpha-1\to \alpha = a+2\\ \beta = -b\\ \beta' = -c\\ a_1 = \frac{1}{\phi}\\ b_1 = -\frac{1}{\phi} $$ thus the integral is $$ \lim_{a,b,c\to0}\partial_{abc}\frac{1}{\phi^{a+2}}\frac{\Gamma(a+2)\Gamma(1)}{\Gamma(a+3)}F_1\left(a+1,-b,-c;a+3;\frac{1}{\phi},-\frac{1}{\phi}\right)=\lim_{a,b,c\to0}\partial_{abc}\frac{1}{\phi^{a+2}}\frac{1}{a+2}F_1\left(a+1,-b,-c;a+3;\frac{1}{\phi},-\frac{1}{\phi}\right) $$
Chinny84
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