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$$I=\int_{0}^{\infty}\ln^2(x)\ln(1+x)\ln^2\left(1+\frac{1}{x}\right)\frac{\mathrm dx}{x}$$

$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$

$$\int_{0}^{\infty}\left(1-\frac{x}{2}+\frac{x^2}{3}+\cdots\right)\left[\ln(x)\ln\left(1+\frac{1}{x}\right)\right]^2 \mathrm dx$$

This integral takes the form of $$J=\int_{0}^{\infty}x^n\left[\ln(x)\ln\left(1+\frac{1}{x}\right)\right]^2 \mathrm dx, n\ge0$$

$$u=\left[\ln(x)\ln\left(1+\frac{1}{x}\right)\right]^2$$

$$u^{'}=\frac{2\ln(x)\ln(1+1/x)\left[(1+x)\ln(1+1/x)-\ln(x)\right]}{x(1+x)}$$

$$v=\frac{x^{n+1}}{n+1}$$

$$J=\frac{x^{n+1}}{n+1}\left[\ln(x)\ln\left(1+\frac{1}{x}\right)\right]^2-\frac{2}{n+1}\int_{0}^{\infty}x^{n+1}\cdot\frac{\ln(x)\ln(1+1/x)\left[(1+x)\ln(1+1/x)-\ln(x)\right]}{x(1+x)}\mathrm dx$$

Wow... this is getting too tough I am totally lose, any help.

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  • $\begingroup$ Firstly this expansion of $\ln{(1+x)}$ is only valid for $-1\lt x\le1$ which does not cover the given integral bounds. Then the integrand of $J$ does not converge to zero for $n\ge2$ so this approach does not seem like it would work anyway. $\endgroup$ – Peter Foreman Jul 6 '19 at 19:14
  • $\begingroup$ Mathematica gives $$I=6(\zeta(3))^2+\frac{\pi^6}{60}\approx24.69279801\dots$$ $\endgroup$ – Peter Foreman Jul 6 '19 at 19:21
  • $\begingroup$ pheeew, I am glad there is a closed form! $\endgroup$ – user569129 Jul 6 '19 at 19:28
  • $\begingroup$ The generalization of this problem is likewise solvable using the same approach as Felix Marin's answer and from my answer here we get some identities for Nielsen polylogarithm. $\endgroup$ – Simply Beautiful Art Jul 7 '19 at 13:46
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{I \equiv \int_{0}^{\infty}\ln^{2}\pars{x} \ln\pars{1 + x}\ln^{2}\pars{1 + {1 \over x}} \,{\dd x \over x} = {\pi^{6} \over 60} + 6\,\zeta^{2}\pars{3}:\ {\LARGE ?}}$.

\begin{align} I & \equiv \bbox[10px,#ffd]{\int_{0}^{\infty}\ln^{2}\pars{x} \ln\pars{1 + x}\ln^{2}\pars{1 + {1 \over x}} \,{\dd x \over x}} \\[5mm] & = \int_{0}^{\infty}\ln^{2}\pars{x} \ln\pars{1 + x} \bracks{\ln\pars{1 + x} - \ln\pars{x}}^{\, 2} \,{\dd x \over x} \\[5mm] & = \int_{1}^{\infty}\ln^{2}\pars{x - 1} \ln\pars{x} \bracks{\ln\pars{x} - \ln\pars{x - 1}}^{\, 2} \,{\dd x \over x - 1} \\[5mm] & = \int_{1}^{0}\ln^{2}\pars{{1 \over x} - 1} \ln\pars{1 \over x} \bracks{\ln\pars{1 \over x} - \ln\pars{{1 \over x} - 1}}^{\, 2}\ \,{-\dd x/x^{2} \over 1/x - 1} \\[5mm] & = -\int_{0}^{1} {\bracks{\ln\pars{1 - x} - \ln\pars{x}}^{\, 2} \ln\pars{x}\ln^{2}\pars{1 - x} \over x\pars{1 - x}}\,\dd x \\[8mm] & = -\int_{0}^{1} {\ln\pars{x}\ln^{4}\pars{1 - x} \over x\pars{1 - x}} \,\dd x + 2\int_{0}^{1} {\ln^{2}\pars{x}\ln^{3}\pars{1 - x} \over x\pars{1 - x}} \,\dd x \\[2mm] & - \int_{0}^{1}{\ln^{3}\pars{x}\ln^{2}\pars{1 - x} \over x\pars{1 - x}}\,\dd x \\[8mm] & = -\int_{0}^{1} {\ln\pars{x}\ln^{4}\pars{1 - x} \over x}\,\dd x -\int_{0}^{1} {\ln^{4}\pars{x}\ln\pars{1 - x} \over x}\,\dd x \\[2mm] & + 2\int_{0}^{1} {\ln^{2}\pars{x}\ln^{3}\pars{1 - x} \over x}\,\dd x + 2\int_{0}^{1} {\ln^{3}\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x \\[2mm] & - \int_{0}^{1}{\ln^{3}\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x - \int_{0}^{1}{\ln^{2}\pars{x}\ln^{3}\pars{1 - x} \over x}\,\dd x \\[8mm] & = -\int_{0}^{1} {\ln^{4}\pars{x}\ln\pars{1 - x} \over x}\,\dd x + \int_{0}^{1} {\ln^{3}\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x \\[2mm] & + \int_{0}^{1} {\ln^{2}\pars{x}\ln^{3}\pars{1 - x} \over x}\,\dd x -\int_{0}^{1} {\ln\pars{x}\ln^{4}\pars{1 - x} \over x}\,\dd x \end{align}

The above integrals are related to derivatives, respect $\ds{\mu}$ and $\ds{\nu}$ with $\ds{\pars{\mu,\nu} \to \pars{0^{+},0}}$, of

\begin{align} \mc{I}\pars{\mu,\nu} & \equiv \int_{0}^{1}{x^{\mu}\bracks{\pars{1 - x}^{\nu} - 1} \over x}\,\dd x \\[5mm] & = \int_{0}^{1}x^{\mu - 1}\pars{1 - x}^{\nu}\,\dd x - \int_{0}^{1}x^{\mu - 1}\,\dd x = {\Gamma\pars{\mu}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 1}} - {1 \over \mu} \\[5mm] & = {1 \over \mu}\bracks{{\Gamma\pars{\mu + 1}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 1}} - 1} \end{align} as $$ \int_{0}^{1}{\ln^{m}\pars{x}\ln^{n}\pars{1 - x} \over x} \, \dd x = \lim_{{\large\mu \to 0^{+}} \atop {\large\nu \to 0}}{\partial^{m + n}\mc{I}\pars{\mu,\nu} \over \partial\mu^{m}\,\partial\nu^{n}} $$


$$ \left\{\begin{array}{rcl} \ds{-\int_{0}^{1} {\ln^{4}\pars{x}\ln\pars{1 - x} \over x}\,\dd x} & \ds{=} & \ds{\phantom{-}{8\pi^{6} \over 315}} \\[2mm] \ds{\int_{0}^{1} {\ln^{3}\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x} & \ds{=} & \ds{-\,{\pi^{6} \over 105} + 6\,\zeta^{2}\pars{3}} \\[2mm] \ds{\int_{0}^{1} {\ln^{2}\pars{x}\ln^{3}\pars{1 - x} \over x}\,\dd x} & \ds{=} & \ds{-\,{23\pi^{6} \over 1260} + 12\,\zeta^{2}\pars{3}} \\[2mm] \ds{-\int_{0}^{1} {\ln\pars{x}\ln^{4}\pars{1 - x} \over x}\,\dd x} & \ds{=} & \ds{\phantom{-}{2\pi^{6} \over 105} - 12\,\zeta^{2}\pars{3}} \end{array}\right. $$

Note that



$\ds{{8\pi^{6} \over 315} + \bracks{-\,{\pi^{6} \over 105} + 6\,\zeta^{2}\pars{3}} + \bracks{-\,{23\pi^{6} \over 1260} + 12\,\zeta^{2}\pars{3}} + \bracks{{2\pi^{6} \over 105} - 12\,\zeta^{2}\pars{3}} = \bbx{{\pi^{6} \over 60} + 6\,\zeta^{2}\pars{3}}}$

which is the final answer.

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  • $\begingroup$ Nice, I was wondering how to get the bounds to $(0,1)$ and rewrite the integrand with $\ln(x)$ and $\ln(1-x)$, but just couldn't do it :-) $\endgroup$ – Simply Beautiful Art Jul 7 '19 at 0:06
  • $\begingroup$ @SimplyBeautifulArt Whenever we see a $\displaystyle 1 + x$ factor, we swithch to $\displaystyle 1 + x \mapsto x$. Later on, $\displaystyle x \mapsto 1/x$ yields intervals $\displaystyle\left(0,1\right)$ with factors $\displaystyle 1 - x$ everywhere. Thanks. $\endgroup$ – Felix Marin Jul 7 '19 at 17:54
  • $\begingroup$ You may also be interested in the generalization. Currently working out if this gives us some values of the Nielsen polylogarithm. $\endgroup$ – Simply Beautiful Art Jul 7 '19 at 17:57
  • $\begingroup$ Is there any way to get WA to evaluate $\mathcal I$? I'm trying to take the limit of 8th order partial derivatives, and the best I could do is get WA to tell me the derivatives, but not the limit. $\endgroup$ – Simply Beautiful Art Jul 9 '19 at 17:00
  • $\begingroup$ Maybe the change of variable $y=\dfrac{x}{1+x}$ ($x=\dfrac{y}{1-y},dx=\frac{1}{(1-y)^2}dy$) makes the things a little simpler $\endgroup$ – FDP Aug 13 '19 at 23:28
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As noted in the comments, the Taylor expansions of $\ln$ do not converge on the entirety of $(0,\infty)$. Instead, let $x\mapsto1/x$ on $(1,\infty)$ to get two integrals on $(0,1)$:

$$I_1=\int_0^1\ln^2(x)\ln(1+x)\ln^2\left(1+\frac1x\right)~\frac{\mathrm dx}x$$

$$I_2=\int_0^1\ln^2(x)\ln\left(1+\frac1x\right)\ln^2(1+x)~\frac{\mathrm dx}x$$

Since $\ln(1+1/x)=\ln(1+x)-\ln(x)$ we can multiply them out as follows:

$$I_1=\int_0^1\ln^2(x)\ln^3(1+x)-2\ln^3(x)\ln^2(1+x)+\ln^4(x)\ln(1+x)~\frac{\mathrm dx}x$$

$$I_2=\int_0^1\ln^2(x)\ln^3(1+x)-\ln^3(x)\ln^2(1+x)~\frac{\mathrm dx}x$$

$$I=I_1+I_2=\int_0^12\ln^2(x)\ln^3(1+x)-3\ln^3(x)\ln^2(1+x)+\ln^4(x)\ln(1+x)~\frac{\mathrm dx}x$$

None of these parts have known solution, so either I made a mistake, this is the wrong approach, or it is possible to continue with this specific combination of coefficients, or perhaps Mathematica's closed form is wrong, though I wouldn't know.

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  • $\begingroup$ The third one is pretty easy using the power expansion of $\ln(1+x)$. $$\int_0^1 \frac{\ln^4 x\ln(1+x)}{x}dx=\frac{93}{4}\zeta(6)$$ The first one is also known around here: math.stackexchange.com/q/972775/515527. But there's some simplification going on between the first and the second integral. I guess one might prefer to evaluate them togheter. $\endgroup$ – Zacky Jul 6 '19 at 21:46
  • $\begingroup$ Oop well yes the third one is fairly obvious, but the other two... D: $\endgroup$ – Simply Beautiful Art Jul 7 '19 at 0:04
  • $\begingroup$ Your approach is correct, and Mathematica's closed-form is also correct. Expressed as Nielsen polylogs, your answer is, $$I = -24S_{3,3}(-1)+36S_{4,2}(-1)-24S_{5,1}(-1)$$ with the third one as, $$I = -24S_{3,3}(-1)+36S_{4,2}(-1)+93/4\,\zeta(6)$$ Since $$I = 6\zeta^2(3)+\pi^6/60$$ This implies $$-24S_{3,3}(-1)+36S_{4,2}(-1) =6\zeta^3(3)-\pi^6/126$$ but neither of those two Nielsen polylogs (separately) have a known closed-form. $\endgroup$ – Tito Piezas III Jul 7 '19 at 4:00

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