# 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!

• I wonder how you wrote your MathJax code. You repeatedly used \displaystyle when an actual display would have been even better, and less onerous to you while typing, and you repeatedly used the \tag command in a way that does not result in any actual tags. (You also repeatedly put not space between f(x) and dx.) Aug 13, 2021 at 21:55
• @MichaelHardy Maybe I'm misinterpreting it's usage, but can't the command 'tag*{}' be used to center align text, I know that it's primary use is to mark equations, but can't it be also used to center align text? Aug 13, 2021 at 22:09
• It does have that effect in MathJax (I've never tried it in genuine LaTeX) but it's a lot simpler to just do the whole thing in display. Aug 13, 2021 at 22:17
• As an application of the techniques in arxiv.org/abs/hep-ph/9905237, Mathematica with the code HPL = SpecialFunctionsHarmonicPolyLog; SpecialFunctionsShuffleProductExpand[HPL[{1}, x] HPL[{2}, x]] /. HPL[{m__}, x] -> HPL[{-1, m}, 1] // Simplify gives $-3\mathrm{Li}_4(1/2) + 29 \pi^4 / 1440 - \log^4(2) / 8 + \pi^2 \log^2(2) / 24$, which is about 0.577998. Aug 14, 2021 at 9:34
• the mathematica code:$\text{MZIntegrate} \left [ \mathrm{Log}\left [1- x \right ] /(1+x) \text{PolyLog}[2,x],\{x,0,1\} \right ]$ also gives the value. Aug 14, 2021 at 10:19

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.

• The result is: $\displaystyle \int_0^1 \frac{\ln(1-x)}{1+x} \operatorname{Li}_2(x)\textrm{d}x= -\frac{29}{1440}\pi^4-\frac{1}{24}\pi^2\ln^2 2+3\text{Li}_4\left(\frac{1}{2}\right)+\frac{1}{8}\ln^4 2$
– FDP
Aug 14, 2021 at 21:08
• @FDP The same result, except that I (often) find it neater to use zeta values. Aug 14, 2021 at 22:23

\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*}

\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 $$I= 3Li_4(\frac12) + Li_4(1) - \frac12\left(Li_2(\frac12) + Li_2(1)\right)^2-2\ln^22 Li_2(\frac12)-\frac34\ln^42$$

• Nicely done! A minor suggestion: You should probably add \DeclareMathOperator{\Li}{Li} for better readability. Aug 23, 2021 at 8:45