# How to evaluate $\int_{0}^{\infty}\ln^2(x)\ln(1+x)\ln^2\left(1+\frac{1}{x}\right)\frac{dx}{x}$

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

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

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

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.

• 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. Jul 6, 2019 at 19:14
• Mathematica gives $$I=6(\zeta(3))^2+\frac{\pi^6}{60}\approx24.69279801\dots$$ Jul 6, 2019 at 19:21
• pheeew, I am glad there is a closed form!
– user569129
Jul 6, 2019 at 19:28
• 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. Jul 7, 2019 at 13:46


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

\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

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

• 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 :-) Jul 7, 2019 at 0:06
• @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. Jul 7, 2019 at 17:54
• You may also be interested in the generalization. Currently working out if this gives us some values of the Nielsen polylogarithm. Jul 7, 2019 at 17:57
• 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. Jul 9, 2019 at 17:00
• 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
– FDP
Aug 13, 2019 at 23:28

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.

• 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. Jul 6, 2019 at 21:46
• Oop well yes the third one is fairly obvious, but the other two... D: Jul 7, 2019 at 0:04
• 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. Jul 7, 2019 at 4:00
• Each of these parts have known solution. See here. May 21, 2020 at 5:04