Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I evaluate following logarithmic integral:

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

share|improve this question
Any condition for $z$ ? –  Claude Leibovici Sep 4 '14 at 14:38
Useful techniques. –  Mhenni Benghorbal Sep 5 '14 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 $$\int\frac{\operatorname{Li}_2(x)}{1-x}\ dx=2\operatorname{Li}_3(1-z)-2\operatorname{Li}_2(1-z)\ln(1-z)-\operatorname{Li}_2(z)\ln (1-z)-\ln z\ln^2(1-z)+C$$

share|improve this answer
(+1) great proof, as always ;) –  Jack D'Aurizio Sep 4 '14 at 15:24
(+1) for the same reason ! Beautiful answer ! –  Claude Leibovici Sep 4 '14 at 15:29
+1. A lot of work as always. –  Felix Marin Sep 9 '14 at 18:17
You should give a link to find the last integral. –  Felix Marin Jan 6 at 9:12

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.

share|improve this answer
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 '14 at 15:06
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 '14 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 '14 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 '14 at 15:22
Indeed, your formula is correct. +1 –  Tunk-Fey Sep 4 '14 at 16:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.