Integral: how to evaluate $$\int_0^1{\ln ^3\left( 1-x \right) \ln ^2\left( 1+x \right) \text{d}x}$$

Same context I'm not sure of the closed form of the integral, as I haven't evaluated it yet. However, after evaluating the integral $$\int_0^1 \ln (1+x) \ln (1-x) \, dx,$$ I thought about integrating $$\int_0^1 \ln^3(1-x) \ln^2(1+x) \, dx.$$

I am also interested in evaluation of $$\int_0^1 \ln^n(1-x) \ln^m(1+x) \, dx.$$ If possible

My work

replace $x\rightarrow\frac{1-x}{1+x}$, there is $$\int_0^1{\ln ^3\left( 1-x \right) \ln ^2\left( 1+x \right) \text{d}x}=2\underset{I}{\underbrace{\int_0^1{\frac{\ln ^2\left( \dfrac{2}{1+x} \right) \ln ^3\left( \dfrac{2x}{1+x} \right)}{\left( 1+x \right) ^2}\text{d}x}}}$$

Expand the integral

\begin{align*} &\ln ^2\left( \frac{2}{1+x} \right) \ln ^3\left( \frac{2x}{1+x} \right) =-\ln ^2\text{2}\ln ^3\left( 1+x \right) +\text{2}\ln\text{2}\ln ^4\left( 1+x \right) -\ln ^5\left( 1+x \right) \\ &+\text{3}\ln ^2\text{2}\ln ^2\left( 1+x \right) \ln \left( 2x \right) -\text{6}\ln\text{2}\ln ^3\left( 1+x \right) \ln \left( 2x \right) +\text{3}\ln ^4\left( 1+x \right) \ln \left( 2x \right) \\ &-\text{3}\ln ^2\text{2}\ln \left( 1+x \right) \ln ^2\left( 2x \right) +\text{6}\ln\text{2}\ln ^2\left( 1+x \right) \ln ^2\left( 2x \right) -\text{3}\ln ^3\left( 1+x \right) \ln ^2\left( 2x \right) \\ &+\ln ^2\text{2}\ln ^3\left( 2x \right) -\text{2}\ln\text{2}\ln \left( 1+x \right) \ln ^3\left( 2x \right) +\ln ^2\left( 1+x \right) \ln ^3\left( 2x \right) \end{align*}

Evaluation of $I_1$

$$I_1=-\ln ^22\int_0^1{\frac{\ln ^3\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=\frac{\ln ^52}{2}+\frac{\text{3}\ln ^42}{2}+\text{3}\ln ^32-\text{3}\ln ^22$$

Evaluation of $I_2$

$$I_2=\text{2}\ln 2\int_0^1{\frac{\ln ^4\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=-\ln ^52-\text{4}\ln ^42-\text{12}\ln ^32-\text{24}\ln ^22+\text{24}\ln 2$$

Evaluation of $I_3$

$$I_3=-\int_0^1{\frac{\ln ^5\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=-60+\frac{\ln ^52}{2}+\frac{\text{5}\ln ^42}{2}+\text{10}\ln ^32+\text{30}\ln ^22+\text{60}\ln 2$$

Evaluation of $I_4$

\begin{align*} I_4&=\text{3}\ln ^22\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln \left( 2x \right)}{\left( 1+x \right) ^2}}\text{d}x\\&=\text{3}\ln ^22\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x+\text{3}\ln ^32\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x \end{align*}

$$I_{41}=\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{4}\zeta \left( 3 \right) +\frac{\pi ^2}{6}-\frac{1}{3}\ln ^32-\ln ^22-\text{2}\ln 2$$

$$I_{42}=\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=1-\frac{1}{2}\ln ^22-\ln 2$$

Expanding the integral does not seem like a great way to evaluate it. I did not provide explanations for some smaller integrals due to the length of the post.

Closed form by Mathematica

\begin{align*} &-120 + 6 \pi^2 + \frac{\pi^4}{30} + 96 \log(2) - 6 \pi^2 \log(2) - \frac{1}{30} \pi^4 \log(2) \\ &\quad - 48 \log^2(2) + 2 \pi^2 \log^2(2) + 16 \log^3(2) - \frac{2}{3} \pi^2 \log^3(2) \\ &\quad - 3 \log^4(2) + \frac{3}{5} \log^5(2) + 24 \text{Li}_4 \left( \frac{1}{2} \right) + 24 \text{Li}_5 \left( \frac{1}{2} \right) \\ &\quad + 45 \zeta(3) - 2 \pi^2 \zeta(3) - 24 \log(2) \zeta(3) + 12 \log^2(2) \zeta(3) + \frac{3}{5} \zeta(5) \end{align*}

  • $\begingroup$ Answering this question might help in your problem $\endgroup$
    – Sam
    Commented May 15 at 2:17

4 Answers 4


If you're not shying away from some significant machinery, you can employ the developments and algebra of (harmonic) polylogarithms. For a starter see https://arxiv.org/abs/hep-ph/9905237 and https://www.physik.uzh.ch/data/HPL/ This won't give you directly what you're after, but you can use the package PolyLogTools Mathematica package https://arxiv.org/abs/1904.07279 and use even heavier machinery.

HPL[{1},x] is -log(1-x), and HPL[{-1},x] is log(1+x), and we turn this into the polylog notation of the PolyLogTools package

<< PolyLogTools`
In: int = (-HPL[{1}, x])^3 HPL[{-1}, x]^2 // HPLToG
Out: G[-1,x]^2 G[1,x]^3

You can use the shuffle algebra to reduce this to a sum of higher weight polylogarithms:

 In:   int // ShuffleG
Out: 12 G[-1, -1, 1, 1, 1, x] + 12 G[-1, 1, -1, 1, 1, x] + 
 12 G[-1, 1, 1, -1, 1, x] + 12 G[-1, 1, 1, 1, -1, x] + 
 12 G[1, -1, -1, 1, 1, x] + 12 G[1, -1, 1, -1, 1, x] + 
 12 G[1, -1, 1, 1, -1, x] + 12 G[1, 1, -1, -1, 1, x] + 
 12 G[1, 1, -1, 1, -1, x] + 12 G[1, 1, 1, -1, -1, x]

The function space is constructed to behave nicely under integration, so this is worked out algebraically:

In: GIntegrate[int, x]
Out: long list of polylogarithms

The expression vanishes for $x=0$ (check), so you get the desired integral by evaluating at $x=1$:

% /. x -> 1 // Simplify // Expand

$$ 24 \text{Li}_4\left(\frac{1}{2}\right)+24 \text{Li}_5\left(\frac{1}{2}\right)-2 \pi ^2 \zeta (3)+45 \zeta (3)+\frac{3 \zeta (5)}{4}+12 \zeta (3) \log ^2(2)-24 \zeta (3) \log (2)+\frac{\pi ^4}{30}+6 \pi ^2-120+\frac{3 \log ^5(2)}{5}-3 \log ^4(2)-\frac{2}{3} \pi ^2 \log ^3(2)+16 \log ^3(2)+2 \pi ^2 \log ^2(2)-48 \log ^2(2)-\frac{1}{30} \pi ^4 \log (2)+96 \log (2)-\frac{2}{3} \pi ^2 \log(512) $$

Things really blow up for larger (n,m), and you will get constants of higher transcendental weight, so I'm not sure about a closed formula for (n,m).


Too large for a comment: You may start with calculating $$\int_0^1\ln ^3\left( 1-x \right) \ln ^2\left( 1+x \right) \textrm{d}x+\color{blue}{\int_0^1\ln ^2\left( 1-x \right) \ln ^3\left( 1+x \right) \textrm{d}x},$$ by exploiting $$\int_{-1}^1 (1-t)^{x-1} (1+t)^{y-1} \textrm{d}t=2^{x+y-1} \operatorname{B}(x,y),$$ which is a result also presented and successfully exploited in (Almost) Impossible Integrals, Sums, and Series (2019), page 76. The integral in blue can be easily finished by exploiting known algebraic identities to reduce the calculations to manageable integrals.

End of story

  • $\begingroup$ I did not understand why + sign. $\endgroup$ Commented May 17 at 7:25
  • 1
    $\begingroup$ Because exploiting the result above you arrive at $$\int_{-1}^1 (\ln ^3\left( 1-x \right) \ln ^2\left( 1+x \right) +\ln ^2\left( 1-x \right) \ln ^3\left( 1+x \right)) \textrm{d}x,$$ and splitting at $x=0$ and letting the variable change $x \mapsto -x$ in the first integral, we get $$\int_{-1}^1 (\ln ^3\left( 1-x \right) \ln ^2\left( 1+x \right) +\ln ^2\left( 1-x \right) \ln ^3\left( 1+x \right)) \textrm{d}x$$ $$=2\int_0^1 (\ln ^3\left( 1-x \right) \ln ^2\left( 1+x \right) +\ln ^2\left( 1-x \right) \ln ^3\left( 1+x \right)) \textrm{d}x.$$ $\endgroup$ Commented May 17 at 7:44

I don't feel like substituting would help. Let's try a different approach. We could multiply and divided with $\ln(1+x)$. This gives us $I=\int_ {0}^{1} \frac{(\ln(1-x)\ln(1+x))^3}{ln(1+x)} dx$. Now we could apply the rule $\int_{} f(x)g(x)dx=F(x)g(x)-\int_ {}F(x)g'(x)dx$ (here $F(x) $is the antiderivative of $f(x)$). This hopefully simplifies the question to a greater extent. As I'm new here and not that great with integration, if I made a mistake anywhere please do let me know.


This idea have not leaded to the closed form. But it can be interesting.

$\color{brown}{\textbf{Taylor series.}}$

Since $$\ln\left(1-t\right)=-\sum\limits_{m=1}^\infty \frac{t^m}{m},\tag1$$ then $$\begin{align} &\ln^2\left(1-t\right)=\left(-\sum\limits_{m=1}^\infty \frac{t^m}{m} \right)^2 =\sum\limits_{m=1}^\infty \frac{t^{2m}}{m^2}+2\sum\limits_{k=2}^\infty \sum\limits_{j=1}^{k-1} \frac{t^{k+j}}{kj}\\[4pt] &=\sum_{m=1}^\infty \frac{t^{2m}}{m^2} +2\sum_{m=2}^\infty\left(\sum_{j=1}^{m-1} \frac{t^{2m}}{j(2m-j)} + \sum_{j=1}^{m-1} \frac{t^{2m-1}}{k(2m-1-k)}\right)\\[4pt] &=\sum_{m=1}^\infty \frac{t^{2m}}{m^2} +2\sum_{m=2}^\infty \frac{t^{2m}}{2m}\sum_{j=1}^{m-1}\left(\frac1j+\frac1{2m-j}\right) + 2\sum_{m=2}^\infty \frac{t^{2m-1}}{2m-1} \sum_{j=1}^{m-1} \left(\frac1j+\frac1{2m-1-j}\right)\\[4pt] &=t^2+\sum_{m=2}^\infty \frac{t^{2m}}{m^2} +\sum_{m=2}^\infty \frac{t^{2m}}{m}\left(-\frac1m+\sum_{j=1}^{2m-1} \frac1j\right) +2\sum_{m=2}^\infty \frac{t^{2m-1}}{2m-1} \sum_{j=1}^{2m-2}\frac1j\\[4pt] &=2\sum_{m=1}^\infty \frac{t^{2m}}{2m}\,\text H_{2m-1} +2\sum_{m=2}^\infty \frac{t^{2m-1}}{2m-1}\,\text H_{2m-2} =2\sum_{m=2}^\infty \frac{t^{m}}{m}\,\text H_{m-1}, \end{align}$$ $$\ln^2\left(1-t\right)=2\sum_{m=2}^\infty \frac{t^{m}}{m}\,\text H_{m-1}.\tag2$$ $\color{brown}{\textbf{The integral.}}$

Applying of the substitution $\;x=1-y\;$ leads to the integral $$I=\int\limits_0^1 \ln^3(1-x)\ln^2(1+x)\,\text dx =\int\limits_0^1 \ln^3 y\ln^2(2-y)\,\text dy =\int\limits_0^1 \ln^3 y\left(\ln2+\ln\left(1-\frac y2\right)\right)^2 \,\text dy$$ $$=\int\limits_0^1\left(\ln^2(2)+2\ln2\ln\left(1-\frac y2\right)+\ln^2\left(1-\frac y2\right)^2\right)\ln^3y\,\text dy.$$ Taking in account the substitution $\;y=e^{-z},\;$ easily to get $$J_k=\int\limits_0^1 y^k \ln^3y \,\text dy =-\frac1{(k+1)^4}\int\limits_0^\infty ((k+1)z)^3\,e^{-(k+1)z}\,\text d((k+1)z)=-\frac{3!}{(k+1)^4},$$ $$J_k=-\frac6{(k+1)^4}.\tag3$$ Taking in account $(1)-(3),$ we can present the given integral in the form of $$I=I_1+I_2+I_3,\tag4$$ where $$\begin{align} &I_1=\ln^2 2 \int\limits_0^1 \ln^3(y)\,\text dy = \ln^2(2)J_0 = -6\ln^22 \approx -2.88271\ 80835\ 09208\ 54800\ 262,\\[4pt] &I_2=-2\ln2\sum\limits_{m=1}^\infty \frac{2^{-2m}}{m}\, J_m =12\ln2\sum\limits_{m=1}^\infty \frac{2^{-2m}}{m(m+1)^4}\\[4pt] &=12\ln2\sum\limits_{m=1}^\infty 2^{-2m}\left(\frac1m-\frac 1{m+1}-\frac 1{(m+1)^2}-\frac 1{(m+1)^3}-\frac 1{(m+1)^4}\right)\\[4pt] &=\big(12\ln^2(2)\big)-\big(12\ln2(\ln4-1)\big)-\big(2\ln(2)(\pi^2-6(1+\ln^2(2)))\big)\\[4pt] &+\big(\ln2(-12+4\ln^3(2)-\pi^2\ln4+21\zeta(3))\big)+\left(12\ln2\left(2\operatorname{Li_4}\left(\frac12\right)-1\right)\right)\\[4pt] &=\Big(48-2\pi^2(1-\ln2)+2\ln4(-3-\ln^2(2)+\ln8) -24\operatorname{Li_4}\left(\frac12\right)-21\zeta(3)\Big)\ln2\\[4pt] &\approx 0.27438\ 02814\ 41802\ 15724\ 137,\\[4pt] &I_3=2\sum_{m=2}^\infty \frac{2^{-m}}{m}\,\text H_{m-1}\,J_m =-12\sum_{m=2}^\infty \frac{2^{-m}}{m(m+1)^4}\,\text H_{m-1}\\[4pt] &=-12\sum\limits_{m=1}^\infty \frac{\text H_{m-1}}{2^{m}}\left(\frac1m-\frac1{m+1}-\frac 1{(m+1)^2}-\frac 1{(m+1)^3}-\frac 1{(m+1)^4}\right)\\[4pt] &=\big(-12\ln^2(2)\big)+\big(24(-1+\ln2+\ln^2(2))\big)+\big(4\pi^2+24\ln2 - 24\ln^2(2)-8\ln^3(2)+6(-8+\zeta(3))\big)\\[4pt] &+12\sum_{m=2}^\infty \frac{\text H_{m-1}}{{2^m}}\left(\frac 1{(m+1)^3}+\frac 1{(m+1)^4}\right)\\[4pt] &=72-4\pi^2-48\ln2+12\ln^2(2)+8\ln^3(2)-6\zeta(3) +12\sum_{m=2}^\infty \frac{\text H_{m-1}}{{2^m}}\frac{m+2}{(m+1)^4}\\[4pt] &\approx -0.23390\ 48463\ 68738\ 94699\ 6262 +0.21174\ 49825\ 13423\ 3855\\[4pt] &\approx -0.02215\ 98638\ 55315\ 561. \end{align}$$ $$I=\big(-6\ln^22\big)+\ln2\bigg(48-2\pi^2-12\ln2+2\pi^2\ln2+12\ln^22 - 4\ln^32$$ $$-24\operatorname{Li}_4\left(\frac12\right)-21\zeta(3)\bigg) +I_3\approx -2.6304976659227219522575.$$


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