# Integral $\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx$

I would like to know how to evaluate the integral $$\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx$$ I tried expanding the integrand as a series but made little progress as I do not know how to evaluate the resulting sum. \begin{align} \int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx &=\int^1_0\sum_{n \ge 1}\frac{x^n}{n^2}\sum_{k \ge 0}x^k\ln{x}dx\\ &=\sum_{n \ge 1}\frac{1}{n^2}\sum_{k \ge 0}\int^1_0x^{n+k}\ln{x}dx\\ &=-\sum_{n \ge 1}\frac{1}{n^2}\sum_{k \ge 0}\frac{1}{(n+k+1)^2} \end{align} I am aware that a similar question has been answered here, however, I find that the answers are not detailed enough for someone who has a shallow understanding on Euler sums, such as myself, to fully comprehend. Hence, I would like to seek your help on the techniques that can be used to evaluate this integral. Thank you.

• Note that, if you bring $n^{-2}$ inside the second summation, then it looks like you can do a partial fractions expansion. That might lead to a tractable double sum... – Semiclassical Aug 11 '14 at 3:25
• There is a standard "Fubini-style"-trick when dealing with $$\sum_{n\geq 1}\frac{H_n^{(k)}}{n^k},$$ as explained below. – Jack D'Aurizio Aug 11 '14 at 3:50

$$2\sum_{n\geq 1}\frac{1}{n^2}\sum_{m>n}\frac{1}{m^2}=\left(\sum_{n\geq 1}\frac{1}{n^2}\right)^2 -\sum_{n\geq 1}\frac{1}{n^4}=\zeta(2)^2-\zeta(4)=\frac{\pi^4}{60},$$ hence the value of your integral is just $-\frac{\pi^4}{120}$. Pretty nice.

• +1 This answer is clean and concise. Forgive my foolishness but I have trouble seeing how you got $$2\sum_{n\geq 1}\frac{1}{n^2}\sum_{m>n}\frac{1}{m^2}=\left(\sum_{n\geq 1}\frac{1}{n^2}\right)^2 -\sum_{n\geq 1}\frac{1}{n^4}$$. Do you mind explaning this slightly more? Thanks a lot. – SuperAbound Aug 11 '14 at 4:01
• Just write down $\left(\sum\frac{1}{n^2}\right)^2=\sum_{n,m}\frac{1}{m^2 n^2}$ and divide the sum into three cases, $n<m,n>m$ (the contribute is symmetric) then $n=m$. – Jack D'Aurizio Aug 11 '14 at 4:03

I present an alternative evaluation of the integral that does not make use of Euler sums at all. Instead, I apply Felix Martin's wonderful procedure of evaluation via derivatives of beta functions and polygamma functions, a technique which he has mastered. For example, see here.

Applying the reflection substitution about the interval $[0,1]$, $x\mapsto1-x$, and using Euler's dilogarithm identity,

$$\operatorname{Li}_2{(1-x)}=\zeta{(2)}-\operatorname{Li}_2{(x)}-\ln{(1-x)}\ln{(x)},~~~\text{(Euler)},$$

we can split up the integral into a sum of three integrals, the first two of which have simple anti-derivatives in terms of the dilogarithm function (see appendix below):

\begin{align} \int_{0}^{1}\frac{\ln{(x)}\operatorname{Li}_2{(x)}}{1-x}\mathrm{d}x &=\int_{0}^{1}\frac{\ln{(1-x)}\operatorname{Li}_2{(1-x)}}{x}\mathrm{d}x\\ &=\zeta{(2)}\int_{0}^{1}\frac{\ln{(1-x)}}{x}\mathrm{d}x-\int_{0}^{1}\frac{\ln{(1-x)}\operatorname{Li}_2{(x)}}{x}\mathrm{d}x-\int_{0}^{1}\frac{\ln^2{(1-x)}\ln{(x)}}{x}\mathrm{d}x\\ &=\zeta{(2)}\left[-\operatorname{Li}_2{(1)}\right]-\left[-\frac12\operatorname{Li}_2{(1)}^2\right]-\int_{0}^{1}\frac{\ln^2{(1-x)}\ln{(x)}}{x}\mathrm{d}x\\ &=-\frac12\zeta{(2)}^2-\int_{0}^{1}\frac{\ln^2{(1-x)}\ln{(x)}}{x}\mathrm{d}x\\ &=-\frac{\pi^4}{72}-\int_{0}^{1}\frac{\ln^2{(1-x)}\ln{(x)}}{x}\mathrm{d}x. \end{align}

The last integral can evaluated as derivatives of a beta function.

\begin{align} \int_{0}^{1}\frac{\ln^2{(1-x)}\ln{(x)}}{x}\mathrm{d}x &=\lim_{\mu\to 0}\lim_{\nu\to 1}\int_{0}^{1}x^{\mu-1}(1-x)^{\nu-1}\ln^2{(1-x)}\ln{(x)}\mathrm{d}x\\ &=\lim_{\mu\to 0}\lim_{\nu\to 1}\frac{\partial}{\partial\mu}\frac{\partial^2}{\partial\nu^2}\int_{0}^{1}x^{\mu-1}(1-x)^{\nu-1}\mathrm{d}x\\ &=\lim_{\mu\to 0}\lim_{\nu\to 1}\frac{\partial}{\partial\mu}\frac{\partial^2}{\partial\nu^2}\operatorname{B}{(\mu,\nu)}\\ &=\lim_{\mu\to 0}\lim_{\nu\to 1}\frac{\partial}{\partial\mu}\operatorname{B}{(\mu,\nu)}\left[(\psi{(\nu)}-\psi{(\mu+\nu)})^2+\psi^{(1)}{(\nu)}-\psi^{(1)}{(\mu+\nu)}\right]\\ &=\lim_{\mu\to 0}\lim_{\nu\to 1}\operatorname{B}{(\mu,\nu)}\left[\left(\psi{(\mu)}-\psi{(\mu+\nu)}\right) \left(\left(\psi{(\nu)}-\psi{(\mu+\nu)}\right)^2 - \psi^{(1)}{(\mu+\nu)}+\psi^{(1)}{(\nu)}\right)-2\left(\psi{(\nu)}-\psi{(\mu+\nu)}\right)\psi^{(1)}{(\mu+\nu)}-\psi^{(2)}{(\mu+\nu)}\right]\\ &=\lim_{\mu\to 0}\operatorname{B}{(\mu,1)}\left[\left(\psi{(\mu)}-\psi{(\mu+1)}\right)\left(\left(\psi{(1)}-\psi{(\mu+1)}\right)^2-\psi^{(1)}{(\mu+1)}+\psi^{(1)}{(1)}\right)-2\left(\psi{(1)}-\psi{(\mu+1)}\right)\psi^{(1)}{(\mu+1)}-\psi^{(2)}{(\mu+1)}\right]\\ &=\lim_{\mu\to 0}\frac{1}{\mu}\left[-\frac{1}{\mu}\left(H_{\mu}^2-\psi^{(1)}{(\mu+1)}+\psi^{(1)}{(1)}\right)+2H_{\mu}\psi^{(1)}{(\mu+1)}-\psi^{(2)}{(\mu+1)}\right]\\ &=\lim_{\mu\to 0}\left[\frac{2H_{\mu}\psi^{(1)}{(\mu+1)}}{\mu}-\frac{H_{\mu}^2-\psi^{(1)}{(\mu+1)}+\zeta{(2)}}{\mu^2}-\frac{\psi^{(2)}{(\mu+1)}}{\mu}\right]\\ &=\lim_{\mu\to 0}\left[\frac{2H_{\mu}\psi^{(1)}{(\mu+1)}}{\mu}-\frac{H_{\mu}^2-\psi^{(1)}{(\mu+1)}+\zeta{(2)}+\mu\,\psi^{(2)}{(\mu+1)}}{\mu^2}\right]\\ &=2\zeta^2{(2)}-\frac{11\pi^4}{180}\\ &=-\frac{\pi^4}{180}. \end{align}

Hence, the integral come to a value of:

$$\int_{0}^{1}\frac{\ln{(x)}\operatorname{Li}_2{(x)}}{1-x}\mathrm{d}x=\frac{\pi^4}{180}-\frac{\pi^4}{72}=-\frac{\pi^4}{120}.$$

Appendix

The most common integral representation for the dilogarithm function is,

$$\operatorname{Li}_2{(z)}=-\int_{0}^{z}\frac{\ln{(1-t)}}{t}\mathrm{d}t.$$

Hence, the value of the first integral is:

$$\int_{0}^{1}\frac{\ln{(1-x)}}{x}\mathrm{d}x=-\operatorname{Li}_2{(1)}.$$

Note that the integral representation implies that $\operatorname{Li}_2{(0)}=0$. The value of the dilogarithm function at $z=1$ is given by the zeta function: $\operatorname{Li}_2{(1)}=\zeta{(2)}=\frac{\pi^2}{6}$.

The second integral may be found readily via integration by parts:

$$\int_{0}^{1}\frac{\ln{(1-x)}\operatorname{Li}_2{(x)}}{x}\mathrm{d}x=-\operatorname{Li}_2{(x)}^2\bigg{|}_{0}^{1}-\int_{0}^{1}\frac{\ln{(1-x)}\operatorname{Li}_2{(x)}}{x}\mathrm{d}x\\ \implies 2\int_{0}^{1}\frac{\ln{(1-x)}\operatorname{Li}_2{(x)}}{x}\mathrm{d}x=-\operatorname{Li}_2{(1)}^2.$$

• +1 This is a very detailed answer. I obtained this integral when I was attempting to evaluate the last integral. I considered solving it also by differentiating the beta function thrice, but that ended up being too painful for me. I see that it was absolutely no problem for you though! :) – SuperAbound Aug 11 '14 at 23:28
• @Superabound Oh, make no mistake, it was painful for me too! It just so happens I gained ALOT of experience doing these kinds of integrals when I was working on my answer to this question. I added a little more detail of the steps to evaluating that third derivative of the beta function, if that helps. – David H Aug 11 '14 at 23:34
• You can arrive faster to the main result without the Euler Reflection Formula as follows: $$\int_{0}^{1}{\ln\left(x\right){\rm Li}_{2}\left(x\right) \over 1 - x}\,{\rm d}x =\int_{0}^{1}\ln\left(1 - x\right)\left[{{\rm Li}_{2}\left(x\right) \over x} + \ln\left(x\right){\rm Li}_{2}'\left(x\right)\right]\,{\rm d}x =-\int_{0}^{1}{\rm Li}_{2}'\left(x\right){\rm Li}_{2}\left(x\right)\,{\rm d}x - \int_{0}^{1}{\ln\left(x\right)\ln^{2}\left(1 - x\right) \over x}\,{\rm d}x$$ – Felix Marin Aug 26 '14 at 5:18

\begin{align} I&=\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-x}\ dx=\sum_{n=1}^\infty\left(H_n^{(2)}-\frac1{n^2}\right)\int_0^1x^{n-1}\ln x\ dx=\sum_{n=1}^\infty\frac1{n^4}-\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^2}\\ &=\zeta(4)-\frac74\zeta(4)=-\frac34\zeta(4) \end{align}