Integral $\int\limits_0^1 \frac{\ln x\ln ( 1 - zx )}{1 - x} dx$ How can I evaluate following logarithmic integral:
$$\int\limits_0^1 \frac{\ln x\ln ( 1 - zx )}{1 - x} dx$$
 A: 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 $C=2\operatorname{Li}_3(1)=2\zeta(3)$.

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