Integral $\int\limits_0^1 \frac{\ln x\ln ( 1 - zx )}{1 - x} dx$

Question

How can I evaluate following logarithmic integral:

$$\int\limits_0^1 \frac{\ln x\ln ( 1 - zx )}{1 - x} dx$$

Answer 1

Let $$ I(z)=\int_0^1 \frac{\ln x\ln ( 1 - zx )}{1 - x}\ dx\qquad;\qquad\text{for }\ z<1 $$ then \begin{align} I'(z)&=-\int_0^1 \frac{x\ln x}{(1 - x)(1 - zx)}\ dx\\ &=-\frac{1}{1-z}\int_0^1\left[\frac{x\ln x}{1-x}-\frac{zx\ln x}{1-zx}\right]\ dx\\ &=-\frac{1}{1-z}\int_0^1\left[\sum_{n=0}^\infty x^{n+1}\ln x-\sum_{n=0}^\infty (zx)^{n+1}\ln x\right]\ dx\\ &=\frac{1}{1-z}\left[\sum_{n=0}^\infty \frac{1}{(n+2)^2}-\sum_{n=0}^\infty \frac{z^{n+1}}{(n+2)^2}\right]\\ &=\frac{1}{1-z}\left[\frac{\pi^2}{6}-1-\frac{\operatorname{Li}_2(z)}{z}+1\right]\\ I(z)&=-\frac{\pi^2}{6}\ln(1-z)-\int\frac{\operatorname{Li}_2(z)}{z(1-z)}\ dz\\ &=-\frac{\pi^2}{6}\ln(1-z)-\int\frac{\operatorname{Li}_2(z)}{z}\ dz-\int\frac{\operatorname{Li}_2(z)}{1-z}\ dz\\ &=\color{blue}{-\frac{\pi^2}{6}\ln(1-z)-\operatorname{Li}_3(z)-2\operatorname{Li}_3(1-z)+2\operatorname{Li}_2(1-z)\ln(1-z)}\\&\quad\ \color{blue}{+\operatorname{Li}_2(z)\ln (1-z)+\ln z\ln^2(1-z)+2\zeta(3)}, \end{align} where $I(0)=0$ implying $C=2\operatorname{Li}_3(1)=2\zeta(3)$.

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Notes :

$$\int\frac{\operatorname{Li}_k(x)}{x}\ dx=\operatorname{Li}_{k+1}(x)+C$$ and $$\int\frac{\operatorname{Li}_2(x)}{1-x}\ dx=2\operatorname{Li}_3(1-z)-2\operatorname{Li}_2(1-z)\ln(1-z)-\operatorname{Li}_2(z)\ln (1-z)-\ln z\ln^2(1-z)+C$$

Answer 2

It appears that if $z$ is a negative integer number, $z=-m$, $$\begin{eqnarray*} I = &-&\zeta(2)\log(m+1)+\frac{1}{2}\log m\log^2(m+1)-\frac{1}{2}\log^3(m+1)\\&-&\log(m+1)\operatorname{Li}_2\left(\frac{1}{m+1}\right)-\operatorname{Li}_3\left(\frac{1}{m+1}\right)+\operatorname{Li}_3\left(\frac{m}{m+1}\right)+\zeta(3).\end{eqnarray*}$$ Proof is straightforward through Euler-Landen's identities.