A challenging integral $ -\int_0^1 \frac{\ln(1-x)}{1+x} \operatorname{Li}_2(x) \, \mathrm{d}x $

I'm interested in evaluating the following integral $ \DeclareMathOperator{\Li}{Li}$

$$ \mathcal{A} = -\int_0^1 \frac{\ln(1-x)}{1+x} \Li_2(x) \, \mathrm{d}x $$

My most successful attempt thus far went like this:

First, converting the dilogarithm to its integral form yields

$$ \mathcal{A} = \int_0^1 \int_0^1 \frac{\ln(1-x) \ln(1-xt) }{t(1+x)} \, \mathrm{d}t \, \mathrm{d}x $$

Interchanging the bounds of integration yields

$$ \mathcal{A} = \int_0^1 \int_0^1 \frac{\ln(1-x) \ln(1-xt) }{t(1+x)} \, \mathrm{d}x \,\mathrm{d}t $$

For the inner integral, we have

$$ \mathfrak{J}(t) = \int_0^1 \frac{ \ln(1-x)\ln(1-xt) }{(1+x)} \, \mathrm{d}x $$

Differentiating under the integral with respect to $t$ and then applying partial fractions yields

$$ \mathfrak{J}'(t) = \frac{1}{1+t} \int_{0}^{1} \frac{\ln(1-x) }{tx-1} \, \mathrm{d}x +\frac{1}{1+t} \int_0^1 \frac{\ln(1-x)}{1+x} \, \mathrm{d}x $$

This evaluates (not) very nicely to

$$ \mathfrak{J}'(t) = \frac{1}{t(1+t) } \Li_2\left(\frac{t}{1-t} \right) +\frac{1}{1+t} \left( \frac{\ln^2(2)}{2} -\frac{\pi^2}{12} \right) $$

This means that our original integral is equivalent to solving

$$ \mathcal{A} = \int_0^1 \int_0^t \frac{1}{at(1+a)} \Li_2 \left(\frac{a}{1-a} \right) \, \mathrm{d}a \, \mathrm{d}t + \left(\frac{\ln^2(2)}{2}-\frac{\pi^2}{12} \right) \int_0^1 \int_0^t \frac{1}{t(1+a)} \, \mathrm{d}a \, \mathrm{d}t $$

The second part of the integral above is trivial, what's giving me trouble is the first part. Any help whatsoever is much appreciated!

Solution 1:

The integral is immediately derived by combining the integral result at the point $i)$, Sect. $1.27$, page $17$, from the book (Almost) Impossible Integrals, Sums, and Series and Landen's identity, and we get $$\int_0^1 \frac{\ln(1-x)}{1+x} \operatorname{Li}_2(x)\textrm{d}x=3\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{1}{4}\log^2(2)\zeta(2)-\frac{29}{16}\zeta(4)+\frac{1}{8}\log^4(2).$$ The other resulting integral is trivial, that is $\displaystyle \int_0^1 \frac{\log^3(1-x)}{1+x}\textrm{d}x=-6\operatorname{Li}_4\left(\frac{1}{2}\right).$

End of story (also subtler ways are possible)

Additional information: If also interested in the following very similar integral, $\displaystyle \int_0^1 \frac{\log(1-x)}{1+x} \operatorname{Li}_2(-x)\textrm{d}x$, one may find it calculated here.

Solution 2:

\begin{align*}J&=\int_0^1 \frac{\ln(1-t)\text{Li}_2(t)}{1+t}dt\\ &=-\int_0^1 \frac{\ln(1-t)}{1+t}\left(\int_0^t\frac{\ln(1-u)}{u}du\right)dt\\ &=-\int_0^1\int_0^1 \frac{\ln(1-t)\ln(1-tu)}{u(1+t)}dtdu\\ &\overset{x=1-tu,y=\frac{1-t}{1-tu}}=-\int_0^1\int_0^1 \frac{x\ln x\ln(xy)}{(1-x)(2-xy)}dxdy\\ &=2\int_0^1\int_0^1\frac{\ln x\ln(xy)}{(2-y)(2-xy)}dxdy-\int_0^1\int_0^1 \frac{\ln x\ln(xy)}{(1-x)(2-y)}dxdy\\ &=\underbrace{\int_0^1\int_0^1\frac{\ln^2(xy)+\ln^2 x-\ln^2 y}{(2-y)(2-xy)}dxdy}_{=\text{A}}-\underbrace{\left(\int_0^1\frac{\ln x}{1-x}dx\right)\left(\int_0^1\frac{\ln y}{2-y}dy\right)}_{=\frac{\pi^4}{72}-\frac{\pi^2\ln^2 2}{12}}-\\&\underbrace{\int_0^1\frac{\ln^2 x}{(1-x)(2-y)}dxdy}_{=2\zeta(3)\ln2 }\\ \end{align*} \begin{align*} \text{A}&=\underbrace{\int_0^1\int_0^1\frac{\ln^2(xy)}{(2-y)(2-xy)}dxdy}_{u(x)=xy}+\frac{1}{2}\underbrace{\int_0^1 \frac{\ln(2-x)\ln^2 x}{1-x}dx}_{u=1-x}-\underbrace{\int_0^1 \frac{\ln\left(\frac{2}{2-y}\right)\ln^2 y}{y(2-y)}dy}_{u=1-y}\\ &=\underbrace{\int_0^1\frac{1}{y(2-y)}\left(\int_0^y \frac{\ln^2 u}{2-u}du\right)dy}_{\text{IBP}}+\frac{1}{2}\underbrace{\int_0^1 \frac{\ln(1+u)\ln^2(1-u)}{u}du}_{=\text{B}}-\\&\underbrace{\int_0^1 \frac{\ln\left(\frac{2}{1+u}\right)\ln^2(1-u)}{1-u^2}du}_{z=\frac{1-u}{1+u}}\\ &=\left(\frac{1}{2}\left[\ln\left(\frac{y}{2-y}\right)\left(\int_0^y \frac{\ln^2 u}{2-u}du\right)\right]_0^1-\frac{1}{2}\underbrace{\int_0^1\frac{\ln^3 y}{2-y}dy}_{=-6\text{Li}_4\left(\frac{1}{2}\right)}+\frac{1}{2}\underbrace{\int_0^1\frac{\ln(2-y)\ln^2 y}{2-y}dy}_{z=\frac{y}{2-y}}\right)+\\&\frac{1}{2} \text{B}-\frac{1}{2}\underbrace{\int_0^1\frac{\ln(1+z)\ln^2\left(\frac{2z}{1+z}\right)}{z}dz}_{=\text{C}}\\ &=3\text{Li}_4\left(\frac{1}{2}\right)+\frac{1}{2}\underbrace{\int_0^1\frac{\ln\left(\frac{2}{1+z}\right)\ln^2\left(\frac{2z}{1+z}\right)}{1+z}dz}_{=\text{D}}+\frac{1}{2}\text{B}-\frac{1}{2}\text{C}\\ \end{align*} \begin{align*} \text{B}&=\frac{1}{6}\left(\underbrace{\int_0^1 \frac{\ln^3(1-u^2)}{u}du}_{z=1-u^2}-\underbrace{\int_0^1 \frac{\ln^3\left(\frac{1-u}{1+u}\right)}{u}du}_{z=\frac{1-u}{1+u}}-2\underbrace{\int_0^1 \frac{\ln^3(1+u)}{u}du}_{z=\frac{1}{1+u}}\right)\\ &=\frac{1}{6}\left(\frac{1}{2}\int_0^1 \frac{\ln^3 z}{1-z}dz-2\int_0^1 \frac{\ln^3 z}{1-z^2}dz+2\int_{\frac{1}{2}}^1 \frac{\ln^3 z}{z(1-z)}dz\right)\\ &=\frac{1}{6}\left(\frac{1}{2}\int_0^1 \frac{\ln^3 z}{1-z}dz-\int_0^1 \frac{\ln^3 z}{1-z}dz-\int_0^1 \frac{\ln^3 z}{1+z}dz+2\int_{\frac{1}{2}}^1 \frac{\ln^3 z}{1-z}dz-\frac{1}{2}\ln^4 2\right)\\ &=\frac{1}{6}\left(\frac{3}{2}\underbrace{\int_0^1 \frac{\ln^3 z}{1-z}dz}_{=-\frac{\pi^4}{15}}-\underbrace{\int_0^1 \frac{\ln^3 z}{1+z}dz}_{=-\frac{7\pi^4}{120}}-2\underbrace{\int_0^{\frac{1}{2}} \frac{\ln^3 z}{1-z}dz}_{=\frac{\pi^2\ln^2}{4}-6\text{Li}_4\left(\frac{ 1}{2}\right)-\frac{21}{4}\zeta(3)\ln 2-\frac{1}{2}\ln^4 2}-\frac{1}{2}\ln^4 2\right)\\ &=-\frac{1}{144}\pi^4-\frac{1}{12}\pi^2\ln^2 2+\frac{1}{12}\ln^4 2+\frac{7}{4}\zeta(3)\ln 2+2\text{Li}_4\left(\frac{ 1}{2}\right)\\ \text{D}&=\underbrace{\int_0^1\frac{\ln^3\left(\frac{2}{1+z}\right)}{1+z}dz}_{u=\frac{1-z}{1+z}}+2\underbrace{\int_0^1\frac{\ln z\ln^2\left(\frac{2}{1+z}\right)}{1+z}dz}_{u=\frac{1-z}{1+z}}+\int_0^1\frac{\ln^2 z\ln\left(\frac{2}{1+z}\right)}{1+z}dz\\ &=\frac{1}{4}\ln^4 2+2\int_0^1\frac{\ln\left(\frac{1-u}{1+u}\right)\ln^2\left(1+u\right)}{1+u}du-\underbrace{\int_0^1 \frac{\ln(1+u)\ln^2 u}{1+u}du}_{=-\frac{\pi^4}{24}-\frac{\pi^2\ln^2 2}{6}+4\text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{2}\zeta(3)\ln 2+\frac{1}{6}\ln^4 2}+\\&\ln 2\underbrace{\int_0^1 \frac{\ln^2 u}{1+u}du}_{=\frac{3}{2}\zeta(3)}\\ &=2\underbrace{\int_0^1 \frac{\ln(1-u)\ln^2(1+u)}{1+u}du}_{=\text{E}}+\frac{1}{24}\pi^4+\frac{1}{6}\pi^2\ln^2 2-\frac{5}{12}\ln^4 2-2\zeta(3)\ln 2-4\text{Li}_4\left(\frac{1}{2}\right)\\ \text{E}&=\frac{1}{3}\left(\underbrace{\int_0^1 \frac{\ln^3\left(\frac{1-u}{1+u}\right)}{1+u}du}_{z=\frac{1-u}{1+u}}-\underbrace{\int_0^1 \frac{\ln^3\left(1-u\right)}{1+u}du}_{z=1-u}+\int_0^1 \frac{\ln^3\left(1+u\right)}{1+u}du+3\underbrace{\int_0^1 \frac{\ln^2\left(1-u\right)\ln(1+u)}{1+u}du}_{z=\frac{1-u}{1+u}}\right)\\ &=-\frac{7}{360}\pi^4+2\text{Li}_4\left(\frac{1}{2}\right)+\frac{1}{12}\ln^4 2+\text{D}\\ \text{D}&=\frac{1}{360}\pi^4+\frac{1}{6}\pi^2\ln^2 2-\frac{1}{4}\ln^4 2-2\zeta(3)\ln 2+2\text{D}\\ \text{D}&=-\frac{1}{360}\pi^4-\frac{1}{6}\pi^2\ln^2 2+\frac{1}{4}\ln^4 2+2\zeta(3)\ln 2 \end{align*} \begin{align*} C&=\underbrace{\int_0^1 \frac{\ln^3(1+z)}{z}dz}_{u=\frac{1}{1+z}}-2\underbrace{\int_0^1 \frac{\ln^2(1+z)\ln z}{z}dz}_{\text{IBP}}+\underbrace{\int_0^1 \frac{\ln(1+z)\ln^2 z}{z}dz}_{\text{IBP}}-\\&2\ln 2\underbrace{\int_0^1 \frac{\ln^2(1+z)}{z}dz}_{u=\frac{1}{1+z}}+2\ln 2\underbrace{\int_0^1 \frac{\ln(1+z)\ln z}{z}dz}_{\text{IBP}}+\ln^2 2\underbrace{\int_0^1 \frac{\ln(1+z)}{z}dz}_{=\frac{\pi^2}{12}}\\ &=\left(\frac{1}{4}\ln^4 2-\underbrace{\int_{\frac{1}{2}}^{1}\frac{\ln^3 u}{1-u}du}_{=-\frac{\pi^4}{15}-\frac{\pi^2\ln^2 2}{4}+\frac{21\zeta(3)\ln 2}{4}+\frac{\ln^4 2}{2}+6\text{Li}_4\left(\frac{1}{2}\right)} \right)+2\underbrace{\int_0^1\frac{\ln(1+z)\ln^2 z}{1+z}dz}_{=-\frac{\pi^4}{24}-\frac{\pi^2\ln^2 2}{6}+\frac{7\zeta(3)\ln 2}{2}+\frac{\ln^4 2}{6}+4\text{Li}_4\left(\frac{1}{2}\right)}-\\&\frac{1}{3}\underbrace{\int_0^1\frac{\ln^3 z}{1+z}dz}_{=-\frac{7\pi^4}{120}}-2\ln 2\left(\frac{1}{3}\ln^3 2+\underbrace{\int_{\frac{1}{2}}^{1}\frac{\ln^2 u}{1-u}du}_{=\frac{1}{4}\zeta(3)-\frac{1}{3}\ln^3 2} \right)-\ln 2\underbrace{\int_0^1 \frac{\ln^2 z}{1+z}dz}_{=\frac{3}{2}\zeta(3)}+\frac{1}{12}\pi^2\ln^2 2\\ &=\frac{1}{360}\pi^4+\frac{1}{12}\ln^4 2-\frac{1}{4}\zeta(3)\ln 2+2\text{Li}_4\left(\frac{1}{2}\right)\\ A&=-\frac{1}{160}\pi^4-\frac{1}{8}\pi^2 \ln^2 2+\frac{1}{8}\ln^4 2+2\zeta(3)\ln 2+3\text{Li}_4\left(\frac{1}{2}\right)\\ J&=\boxed{-\dfrac{29}{1440}\pi^4-\dfrac{1}{24}\pi^2\ln^2+\dfrac{1}{8}\ln^4 2+3\text{Li}_4\left(\dfrac{1}{2}\right)} \end{align*}

Solution 3:

\begin{align} I &=\int_0^1 \frac{\ln(1-x)Li_2(x)}{1+x}dx\\ &= \int_0^1 \frac{\ln x Li_2(1-x)}{2-x}dx \overset{ibp} =-\int_0^1 \frac{\ln x}{1-x}\left(\int_0^x\frac{\ln t}{2-t} \overset{t=x y}{dt}\right)dx\\ &=\int_0^1 \int_0^1 \left( -\frac{x \ln^2x }{(1-x)(2-xy)}+\frac{2\ln x\ln y}{(2-y)(2-xy)}-\frac{\ln x \ln y} {(2-y)(1-x)}\right) dy dx \\ &=J+2K-Li_2(1)Li_2(\frac12)\tag1 \end{align} where \begin{align} J&=- \int_0^1 \int_0^1 \frac{x \ln^2x }{(1-x)(2-xy)}dy dx\\ & =\int_0^1 \frac{1}{2-y}\left(\int_0^1 \frac {2\ln^2x}{2-yx}dx - \int_0^1 \frac{\ln^2x}{1-x}dx\right) dy \\ &= 2 \int_0^1\frac{Li_3(\frac y2)}{y}dy + 2 \int_0^1\frac{Li_3(\frac y2)-Li_3(1)}{2-y}\>\overset{ibp}{dy}\\ &= 2Li_4(\frac12) - Li_2^2(\frac12)+2\ln2(Li_3(\frac12)-Li_3(1)) \end{align} and \begin{align} K &=\int_0^1 \int_0^1 \frac{\ln x \ln y}{(2-y)(2-xy)}\overset{x=t/y}{dx }dy= \int_0^1 \frac{ \ln y}{y(2-y)} \int_0^y\frac{\ln t-\ln y}{2-t}dt\>dy\\ &= \frac12 \int_0^1 \left( \frac{\ln y}y+ \frac{\ln y}{2-y}\right)\left(\int_0^y \frac{\ln t}{2-t}dt + \ln y\ln \frac{2-y}2\right)dy\\ &\overset{ibp}=-\frac1{12}\int_0^1\frac{\ln^3 y}{2-y}dy+\frac14\left(\int_0^1 \frac{\ln y}{2-y}dy\right)^2 +\frac12\int_0^1 \frac{\ln^2y\ln\frac{2-y}2}{2-y} \overset{y\to 2y}{dy}\\ &=\frac12Li_4(\frac12)+\frac14Li_2^2(\frac12)+\frac12\int_0^{1/2} \frac{\ln^2(2y)\ln(1-y)}{1-y}dy \end{align} Note

\begin{align} &\int_0^{1/2} \frac{\ln^2(2y)\ln(1-y)}{1-y}dy\\ =& \>\ln2 \int_0^{1/2} \frac{\ln(2y^2)\ln(1-y)}{1-y}\>\overset{ibp}{dy} + \int_0^{1/2} \frac{\ln^2y\ln(1-y)}{1-y}\overset{y\to1-y}{dy}\\ =& 2\ln2(Li_3(1)-Li_3(\frac12))-2\ln^22Li_2(\frac12)-\frac34\ln^42 +\frac12\int_0^{1} \frac{\ln^2y\ln(1-y)}{1-y}dy \end{align} with \begin{align} &\int_0^1 \frac{\ln^2x \ln (1-x)}{1-x}dx = \int_0^1 \frac{\ln^2x}{1-x}\left(-\int_0^1 \frac x{1-x y}dy\right) dx\\ =&\int_0^1 \frac{1}{1-y}\left(\int_0^1 \frac {\ln^2x}{1-yx}dx - \int_0^1 \frac{\ln^2x}{1-x}dx\right) dy \\ =& 2 \int_0^1\frac{Li_3(y)}{y}dy + 2 \int_0^1\frac{Li_3(y)-Li_3(1)}{1-y}\>\overset{ibp}{dy} = 2Li_4(1) - Li_2^2(1) \end{align}

Substitute the results above for $J$ and $K$ into (1) to obtain $$I= 3Li_4(\frac12) + Li_4(1) - \frac12\left(Li_2(\frac12) + Li_2(1)\right)^2-2\ln^22 Li_2(\frac12)-\frac34\ln^42 $$