# A challenging logarithmic integral $\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx$

While playing around with Mathematica, I found that

$$\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx = \frac{1}{3}\log^3(2)-\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$

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Let $u=\log(1+x)$, $du=1/(1+x)\,dx$ and $1-x=2-e^u$ transforms our integral into $$\int_0^{\log2}u\log(2-e^u)\,\mathrm{d}u$$... not sure what to do from there, however. –  oldrinb Jun 1 '13 at 7:36
you can see:math.stackexchange.com/questions/405356/… –  math110 Jun 1 '13 at 7:47
@math110: I think I just understood what you are trying to say. We can relate it to the Euler Sum in your link. $$\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(H_n)^2$$ –  Integrals and Series Jun 1 '13 at 8:16
For a more generalized form of this integral, where the integral is over the range 0 to z rather than 0 to 1, see here -->> [mathhelpboards.com/questions-other-sites-52/… –  user98087 Oct 1 '13 at 19:32

Use your favorite program to compute the indefinite integral in terms of polylogarithms $$\int\frac{\ln(1+x)\ln(1-x)\,dx}{1+x}=\frac{\ln2}{2}\ln^2(1+x)-\ln(1+x)\,\mathrm{Li}_2\left(\frac{1+x}{2}\right)+\mathrm{Li}_3\left(\frac{1+x}{2}\right).$$ [This can be verified by straightforward differentiation].

To compute the definite integral, it suffices to know $\mathrm{Li}_{2,3}\left(\frac12\right)$ and $\mathrm{Li}_{2,3}(1)$. However, the definition of polylogarithm immediately implies $\mathrm{Li}_s(1)=\zeta(s)$. Also, the values $\mathrm{Li}_{2,3}\left(\frac12\right)$ can be found here (formulas (16), (17)).

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I can verify that the anti-derivative can be computed through some tedious integration by parts. –  Potato Jun 1 '13 at 8:46
@Potato It is maybe not easy to guess the form of the antiderivative. But once you have a correct guess, to check it is a one-line calculation using that $\mathrm{Li}_2'(z)=-\ln(1-z)/z$ and $\mathrm{Li}_3'(z)=\mathrm{Li}_2(z)/z$. –  O.L. Jun 1 '13 at 8:50
@Potato: I think I understood what you mean. One integration by parts gives roughly $\mathrm{Li}_2(smth)$ instead of $\frac{\ln(1-z)}{1+z}$ and $1/(1+z)$ instead of $\ln(1+z)$. Subsequent integration gives $\mathrm{Li}_3$. –  O.L. Jun 1 '13 at 8:56
Thank O.L.! I did not realize that the indefinite integral would be so easy. –  Integrals and Series Jun 1 '13 at 9:40
It's clear that the OP likes an analytically derivation. Otherwise, the OP can use a symbolic software to get the solution. –  Felix Marin Aug 13 at 19:45


With $\ds{0 < \mu < 1}$: \begin{align}&\color{#c00000}{\int_{0}^{\mu}% {\ln\pars{1 + x}\ln\pars{1 - x} \over 1+x}\,\dd x} \\[3mm]&=\half\,\ln^{2}\pars{1 + \mu}\ln\pars{1 - \mu} +\color{#00f}{\half\int_{0}^{\mu}{\ln^{2}\pars{1 + x} \over 1 - x}\,\dd x} \tag{1} \end{align}

\begin{align}&\color{#00f}{\half\int_{0}^{\mu}% {\ln^{2}\pars{1 + x} \over 1 - x}\,\dd x} =\half\int_{1}^{1 + \mu}{\ln^{2}\pars{x} \over 2 - x}\,\dd x =\half\int_{1/2}^{\pars{1 + \mu}/2}{\ln^{2}\pars{2x} \over 1 - x}\,\dd x \\[3mm]&=-\,\half\ln\pars{1 - {1 + \mu \over 2}}\ln^{2}\pars{1 + \mu} +\half\int_{1/2}^{\pars{1 + \mu}/2}\ln\pars{1 - x}\,{2\ln\pars{2x} \over x}\,\dd x \\[3mm]&=-\,\half\ln\pars{1 - \mu \over 2}\ln^{2}\pars{1 + \mu} -\int_{1/2}^{\pars{1 + \mu}/2}{\rm Li}_{2}'\pars{x}\ln\pars{2x}\,\dd x \\[3mm]&=-\,\half\,\ln^{2}\pars{1 + \mu}\ln\pars{1 - \mu} +\half\,\ln\pars{2}\ln^{2}\pars{1 + \mu} -{\rm Li}_{2}\pars{1 + \mu \over 2}\ln\pars{1 + \mu} \\[3mm]&+\int_{1/2}^{\pars{1 + \mu}/2}{\rm Li}_{2}\pars{x}\,{1 \over x}\,\dd x \\[3mm]&=-\,\half\,\ln^{2}\pars{1 + \mu}\ln\pars{1 - \mu} +\half\,\ln\pars{2}\ln^{2}\pars{1 + \mu} - {\rm Li}_{2}\pars{1 + \mu \over 2}\ln\pars{1 + \mu} \\[3mm]&+\int_{1/2}^{\pars{1 + \mu}/2}{\rm Li}_{3}'\pars{x}\,\dd x \end{align}

\begin{align} &\color{#00f}{\half\int_{0}^{\mu}% {\ln^{2}\pars{1 + x} \over 1 - x}\,\dd x} \\[3mm]&=-\,\half\,\ln^{2}\pars{1 + \mu}\ln\pars{1 - \mu} +\half\,\ln\pars{2}\ln^{2}\pars{1 + \mu} -{\rm Li}_{2}\pars{1 + \mu \over 2}\ln\pars{1 + \mu} \\[3mm]&+{\rm Li}_{3}\pars{1 + \mu \over 2} - {\rm Li}_{3}\pars{\half} \end{align}

Replacing in $\pars{1}$ and taking the limit $\ds{\mu \to 1^{-}}$: \begin{align}&\color{#c00000}{\int_{0}^{1}% {\ln\pars{1 + x}\ln\pars{1 - x} \over 1+x}\,\dd x} = \half\,\ln^{3}\pars{2} - {\rm Li}_{2}\pars{1}\ln\pars{2} +{\rm Li}_{3}\pars{1} - {\rm Li}_{3}\pars{\half} \end{align}

With the values: $${\rm Li}_{2}\pars{1} = {\pi^{2} \over 6}\,,\quad {\rm Li}_{3}\pars{1} = \zeta\pars{3}\,,\quad {\rm Li}_{3}\pars{\half} = {\ln^{3}\pars{2} \over 6} -{\ln\pars{2} \over 12}\,\pi^{2} + {7 \over 8}\,\zeta\pars{3}$$ we find

$$\color{#66f}{\large\int_{0}^{1}% {\ln\pars{1 + x}\ln\pars{1 - x} \over 1+x}\,\dd x = {\ln^{3}\pars{2} \over 3} - {\ln\pars{2} \over 12}\,\pi^{2} +{1 \over 8}\,\zeta\pars{3}} \approx {\tt -0.3088}$$

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This is a nice answer! –  Olivier Oloa Aug 13 at 21:42
@OlivierOloa Thanks. It's nice to rely on PolyLogarithms... –  Felix Marin Aug 16 at 21:18