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How can I evaluate following logarithmic integral:

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

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1  
Any condition for $z$ ? –  Claude Leibovici Sep 4 at 14:38
    
Useful techniques. –  Mhenni Benghorbal Sep 5 at 1:20

2 Answers 2

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

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(+1) great proof, as always ;) –  Jack D'Aurizio Sep 4 at 15:24
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(+1) for the same reason ! Beautiful answer ! –  Claude Leibovici Sep 4 at 15:29
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+1. A lot of work as always. –  Felix Marin Sep 9 at 18:17

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.

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I suppose that there is no solution for $z>1$. It seems that if $m=1$ the result is $\zeta(3)$. Do you agree ? –  Claude Leibovici Sep 4 at 15:06
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For $z=-1$ I get $$I=\zeta(3)-\frac{\pi^2}{4}\log 2.$$ This follows by writing $\log(1+x)$ as $\log(1-x^2)-\log(1-x)$, many posts on MSE are devoted to such logarithmic integrals ;) –  Jack D'Aurizio Sep 4 at 15:08
    
Sorry ! It was for $z=1$ that I think that the result is $\zeta(3)$. By the way, how did you get this general result ? Cheers :-) –  Claude Leibovici Sep 4 at 15:18
    
I just computed the integral on WA for different values of $m$, then interpolated the answer. By differentiating and using Euler-Landen identities it is easy to check that the "interpolated" formula is right. –  Jack D'Aurizio Sep 4 at 15:22
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Indeed, your formula is correct. +1 –  Tunk-Fey Sep 4 at 16:37

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