Evaluating $\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz$

Solution 1:

Let's use the classical property of the polylogarithm $\ \displaystyle z\frac{\partial\operatorname{Li}_2(z)}{\partial z}=\frac {\operatorname{Li}_1(z)}z=-\frac{\ln(1-z)}z\,$ to rewrite this as a double integral : \begin{align} I&:=\int_0^1 \frac{\operatorname{Li}_2 \left(-\frac{1}{1-z}\right)-\operatorname{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz \\ &=\int_0^1 \frac 1z\left[-\int_0^{-1/(1-z)}\frac{\ln(1-t)}tdt +\int_0^{-1/(1+z)}\frac{\ln(1-t)}tdt\right]dz \\ &=\int_0^1 \frac 1z\int_{-1/(1-z)}^{-1/(1+z)}\frac{\ln(1-t)}tdt\,dz \\ \end{align} Substitute $\,t:=t(y)=-\dfrac 1{1-y\,z}\,$ and use $\,\dfrac {dt}t=\dfrac {z\,dy}{1-y\,z}$ to get : \begin{align} I&=\int_0^1 \frac 1z\int_{1}^{-1}\ln\left(1+\frac 1{1-y\,z}\right)\frac z{1-y\,z}dy\,dz \\ &=-\int_0^1 \int_{-1}^{1}\frac{\ln\left(2-y\,z\right)-\ln\left(1-y\,z\right)}{1-y\,z}dy\,dz \\ &=\int_0^1 \int_{-1}^{1}\frac{\ln\left(1-y\,z\right)}{1-y\,z}dy\,dz -\int_0^1 \int_{-1}^{1}\frac{\ln\left(2-y\,z\right)}{1-y\,z}dy\,dz\\ &=\int_0^1 \int_0^1\frac{\ln\left(1-y\,z\right)}{1-y\,z}dy\,dz+\int_0^1 \int_0^1\frac{\ln\left(1+y\,z\right)}{1+y\,z}dy\,dz\\ &\quad -\int_0^1 \int_0^1\frac{\ln\left(2-y\,z\right)}{1-y\,z}dy\,dz-\int_0^1 \int_0^1\frac{\ln\left(2+y\,z\right)}{1+y\,z}dy\,dz\\ \end{align}

Such integrals were studied by Beukers and more generally by Guillera and Sondow in "Double integrals and infinite products for some classical constants via analytic continuations of Lerch's transcendent" and these four last integrals return fractions of $\zeta(3)$. More exactly : $$I=-\zeta(3)+\frac 18\zeta(3)-\frac 58\zeta(3)-\frac {13}{24}\zeta(3)=-\frac{49}{24}\zeta(3)$$ with the correct result (a proof for the first and third integral is in the paper : see the examples $3.26$ and $3.25$).

UPDATE (partial)
The idea is to evaluate integrals like $\;\displaystyle\int_0^1 \int_0^1\frac{\ln\left(a-y\,z\right)}{1\pm y\,z}dy\,$ by expanding the logarithm as power series so that we will need to evaluate integrals like $\displaystyle \int_0^1\int_0^1\frac{(xy)^u}{1\pm xy} dy$.

Let's rewrite a little the Lerch transcendent (see $(9)$ from the paper) : \begin{align} \Phi(z,s,u)&:=\sum_{n=0}^\infty\frac{z^n}{(n+u)^s}\\ &=\frac 1{\Gamma(s)}\int_0^\infty\frac{e^{-ut}\,t^{s-1}}{1-e^{-t}\,z}\,dt,\quad\text{setting}\;\;t:=-\ln(X)\;\text{gives}\\ &=\frac 1{\Gamma(s)}\int_0^1\frac{X^{\,u-1}\,}{1-Xz}(-\ln(X))^{s-1}\,dX\\ \end{align} Setting $X:=x\,y\;$ we have $\,x=\frac Xy$ and $\,0\le X\le y\le 1\,$ getting (cf $(29)$) : \begin{align} \int_0^1\int_0^1\frac{x^u\,y^v}{1-xyz}(-\ln(xy))^s dy\;dx&=\int_0^1\int_X^1\frac{X^u\,y^{v-u}}{1-Xz}(-\ln(X))^s \frac 1y\,dy\;dX\\ &=\int_0^1\frac{X^u}{1-Xz}(-\ln(X))^s\int_X^1 y^{v-u-1}\,dy\;dX\\ &=\frac 1{v-u}\int_0^1\frac{X^u-X^v}{1-Xz}(-\ln(X))^s\,dX\\ &=-\Gamma(s+1)\frac {\Phi(z,1+s,1+u)-\Phi(z,1+s,1+v)}{u-v}\\ \end{align} For $z=1$ and at the limit $s\to 0$ (using the series for digamma) this becomes (cf $(40)$ and $(42)$) : \begin{align} \int_0^1\int_0^1\frac{x^u\,y^v}{1-xy} dy\;dx&=\frac {\psi(1+u)-\psi(1+v)}{u-v}\quad\text{while for}\;\;v\to u\\ \int_0^1\int_0^1\frac{(xy)^u}{1-xy} dy\;dx&=\psi'(1+u) \end{align} While for $z=-1\,$ and setting $\;\displaystyle \eta(x):=\psi(x)-\psi\left(\frac{x+1}2\right)\;$ I obtained : \begin{align} \int_0^1\int_0^1\frac{x^u\,y^v}{1+xy} dy\;dx&=\frac {\eta(1+u)-\eta(1+v)}{u-v}\quad\text{and for}\;\;v\to u\\ \int_0^1\int_0^1\frac{(xy)^u}{1+xy} dy\;dx&=\eta'(1+u) \end{align}

The theorem $3.2$ from the paper should allow to conclude using $\psi(n+1)-\psi(1)=H_n$ (and $\psi'(n+1)=\dfrac{\pi^2}6$) as well as the corresponding formulae for $z=-1$ but I'll have to stop again...

An alternative method could be to use integration by parts : \begin{align} I&:=\int_0^1 \frac{\operatorname{Li}_2 \left(-\frac{1}{1-z}\right)-\operatorname{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz \\ &=\left.\ln(z)\left(\operatorname{Li}_2 \left(-\frac{1}{1-z}\right)-\operatorname{Li}_2 \left(-\frac{1}{1+z}\right)\right)\right|_0^1-\\&\quad\int_0^1\ln(z)\left[\frac{\ln\left(1+\frac 1{1-z}\right)}{z-1}-\frac{\ln\left(1+\frac 1{1+z}\right)}{z+1}\right]dz\\ &=\int_0^1\ln(z)\left[\frac{\ln\left(1+\frac 1{1-z}\right)}{1-z}+\frac{\ln\left(1+\frac 1{1+z}\right)}{1+z}\right]dz\\ \end{align} Again this gives four integrals (more or less easy to evaluate...)

Solution 2:

Integrating by parts and using the known series results, we get that $$\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz$$ $$=\int_0^1 \frac{\log (z+2) \log (z)}{z+1}dz+\underbrace{\int_0^1\frac{\log (2-z) \log (z)}{1-z} dz}_{\large \sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^2}=-5/8 \zeta (3)}-\underbrace{\int_0^1\frac{\log (1-z) \log (z)}{1-z}dz}_{\large\sum _{n=1}^{\infty } \frac{H_n}{(n+1)^2}=\zeta (3)}-\underbrace{\int_0^1\frac{\log (z+1) \log (z)}{z+1} \, dz}_{\large \sum _{n=1}^{\infty } \frac{(-1)^n H_n}{(n+1)^2}=-1/8\zeta (3)}$$

For evaluation of the first integral, check my answer here Proving that $\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3)$.