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How can we prove that:

$$\int_0^1\frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx=\frac{\pi^2}{8}\zeta(3)-\frac{27}{16}\zeta(5) $$

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  • $\begingroup$ It's curious that ${\tt \mbox{Mathematica 10.0.0.0}}$, in a MacBook Pro, yields $-\ln\left(\, 4\,\right)$ which is clearly ${\large\it\mbox{wrong !!!}}$. $$\verb* Clear[x]; Integrate[Log[x]^2 Log[1 - x] Log[1 + x] /x, {x, 0, 1}]* $$ W&A evaluates it correctly. $\endgroup$ – Felix Marin Nov 29 '14 at 23:10
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Performing integration by parts by taking $u=\ln(1-x)\ln(1+x)$, then $$\begin{align} \int_0^1\frac{\ln^2x\ln(1-x)\ln(1+x)}{x}\ dx&=\frac{1}{3}\bigg[\int_0^1\frac{\ln(1+x)\ln^3x}{1-x}\ dx-\int_0^1\frac{\ln(1-x)\ln^3x}{1+x}\ dx\bigg]\\ &=\frac{1}{3}\bigg[I-J\bigg]\\ \end{align}$$ Evaluation of $I$ : $$\begin{align} I&=\int_0^1\frac{\ln^3{x}\big[\ln(1-x^2)-\ln(1-x)\big]}{1-x}\,dx\\ &=\int_0^1\frac{(1+x)\ln^3{x}\ln(1-x^2)}{(1+x)(1-x)}\,dx-\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx\\ &=\int_0^1\frac{\ln^3{x}\ln(1-x^2)}{1-x^2}\,dx+\int_0^1\frac{x\ln^3{x}\ln(1-x^2)}{1-x^2}\,dx-\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx\\ &=\frac{1}{16}\int_0^1\frac{x^{-\frac{1}{2}}\ln^3{x}\ln(1-x)}{1-x}\,dx-\frac{15}{16}\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx\\ &=\frac{1}{16}\int_0^1\frac{x^{-\frac{1}{2}}\ln^3{x}\ln(1-x)}{1-x}\,dx-\frac{15}{16}\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx\\ &=\frac{1}{16}\lim_{\mu \to \frac{1}{2}\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\int_0^1 x^{\mu-1}(1-x)^{\nu-1}\,dx-\frac{15}{16}\lim_{\mu \to 1\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\int_0^1 x^{\mu-1}(1-x)^{\nu-1}\,dx\\ &=\frac{1}{16}\lim_{\mu \to \frac{1}{2}\,,\, \nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\text{B}(\mu,\nu)-\frac{15}{16}\lim_{\mu \to 1\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\text{B}(\mu,\nu)\\ &=12\zeta(5)-\frac{3\pi^2}{8}\zeta(3)-\frac{\pi^4}{8}\ln{2} \end{align}$$ Evaluation of $J$ : \begin{align} J&=\int_0^1\frac{\ln^3{x}\big[\ln(1-x^2)-\ln(1+x)\big]}{1+x}\,dx\\ &=\int_0^1\frac{(1-x)\ln^3{x}\ln(1-x^2)}{(1-x)(1+x)}\,dx-\int_0^1\frac{\ln^3{x}\ln(1+x)}{1+x}\,dx\\ &=\int_0^1\frac{\ln^3{x}\ln(1-x^2)}{1-x^2}\,dx-\int_0^1\frac{x\ln^3{x}\ln(1-x^2)}{1-x^2}\,dx-K\\ &=-\frac{1}{16}\int_0^1\frac{x^{-\frac{1}{2}}\ln^3{x}\ln(1-x)}{1-x}\,dx+\frac{1}{16}\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx-K\\ &=-\frac{1}{16}\lim_{\mu \to \frac{1}{2}\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\int_0^1 x^{\mu-1}(1-x)^{\nu-1}\,dx+\frac{1}{16}\lim_{\mu \to 1\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\int_0^1 x^{\mu-1}(1-x)^{\nu-1}\,dx-K\\ &=-\frac{1}{16}\lim_{\mu \to \frac{1}{2}\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\text{B}(\mu,\nu)+\frac{1}{16}\lim_{\mu \to 1\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\text{B}(\mu,\nu)-K\\ &=\frac{45}{2}\zeta(5)-\frac{5\pi^2}{4}\zeta(3)-\frac{\pi^4}{8}\ln 2-K \end{align} where \begin{align} K&=\int_0^1\frac{\ln^3{x}\ln(1+x)}{1+x}\,dx\\ &=\int_0^1\sum_{k=1}^\infty (-1)^{k-1}H_{k}\,x^k\ln^3x\,dx\tag{1}\\ &=\sum_{k=1}^\infty (-1)^{k-1}H_{k}\int_0^1x^k\ln^3x\,dx\\ &=-6\sum_{k=1}^\infty (-1)^{k-1}\frac{H_{k}}{(k+1)^4}\tag{2}\\ &=-6\sum_{k=1}^\infty (-1)^{k-1}\left[\frac{H_{k+1}}{(k+1)^4}-\frac{1}{(k+1)^5}\right]\tag{3}\\ &=6\sum_{k=1}^\infty (-1)^{k-1}\left[\frac{H_{k}}{k^4}-\frac{1}{k^5}\right]\\ &=9\eta(5)-3\zeta(5)-6\eta(2)\zeta(3)\tag{4}\\ &=\frac{87}{16}\zeta(5)-\frac{\pi^2}{2}\zeta(3)\tag{5} \end{align}

Putting these together, we will get the desired result.


Explanation :

$(1)$ Use generating function $\displaystyle\sum_{k=1}^\infty (-1)^{k-1}H_{k}\,x^k=\frac{\ln(1+x)}{1+x}$

$(2)$ Use formula $\displaystyle\int_0^1 x^k \ln^n x\ dx=\frac{(-1)^n n!}{(k+1)^{n+1}}\quad,\ n\in\mathbb{Z}_{n\ge0}$

$(3)$ Use property $\displaystyle H_{k}=H_{k+1}-\frac{1}{k+1}$

$(4)$ Use formula $\displaystyle \sum_{k=1}^\infty (-1)^{k-1}\frac{H_{k}}{k^{2n}}=\frac{(2n+1)\eta(2n+1)}{2}-\sum_{k=0}^{n-1}\eta(2k)\zeta(2n+1-2k)\,,\,n\in\mathbb{Z}_{n\ge1}$

$(5)$ Use property of Dirichlet eta function $\displaystyle \eta(s)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^s}=\left(1-2^{1-s}\right)\zeta(s)$

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    $\begingroup$ @FelixMarin I added some explanations in my answer. Thanks $\endgroup$ – Venus Nov 30 '14 at 11:48
  • $\begingroup$ +1 I think you could have avoided a lot of trouble by directly expanding the integrand as a double sum without integrating by parts. $\endgroup$ – M.N.C.E. Dec 1 '14 at 6:09
  • $\begingroup$ @M.N.C.E. Thanks. Could you post your answer using that method? I want to see a different approach than mine? Thanks. $\endgroup$ – Venus Dec 1 '14 at 10:04
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    $\begingroup$ I can't post a solution now as my laptop is damaged. By applying the aforementioned approach, one can show in a few steps that the integral is \begin{align} \small{2\sum^\infty_{j=1}\frac{(-1)^jH_j}{j^4}- 2\sum^\infty_{j=1}\frac{(-1)^j\left(\zeta(2)-H_j^{(2)}\right)}{j^3}- 2\sum^\infty_{j=1}\frac{(-1)^j\left(\zeta(3)-H_j^{(3)}\right)}{j^2}}\end{align} These sums can be trivially evaluated. I actually prefer this to differentiating the Beta function four times and expanding everything to compute the limit. $\endgroup$ – M.N.C.E. Dec 1 '14 at 14:38
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    $\begingroup$ Very nice demonstration.thank $\endgroup$ – user178256 Dec 2 '14 at 21:58
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A generalization of the present integral is given in the book (Almost) Impossible Integrals, Sums, and Series (see page 6), $$\int_0^1\frac{\log(1-x)\log^{2n}(x)\log(1+x)}{x} \textrm{d}x =\frac{1}{2}(2n)!\left(1-\frac{1}{2^{2n+1}}\right)\sum_{k=1}^{2n} \zeta(k+1)\zeta(2n-k+2)$$

$$-(2n)!\sum_{k=1}^{n}\left(1-\frac{1}{2^{2k-1}}\right)\zeta(2k)\zeta(2n-2k+3) + \frac{1}{2^{2n+3}} (2n+3-2^{2n+3})(2n)!\zeta(2n+3),$$ where in his solution the author exploits a classical series representation of $\log(1-x)\log(1+x)$.

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