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

Question

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.

Answer 1

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

Answer 2

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:

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

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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:

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

Answer 3

\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}

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Proof of the last sum: Using $\displaystyle\sum_{m=1}^\infty \sum_{n=1}^m a_mb_n=\sum_{n=1}^\infty \sum_{m=n}^\infty a_mb_n$, we have

\begin{align} \sum_{m=1}^\infty\frac{H_m^{(r)}}{m^s}&=\sum_{m=1}^\infty\sum_{n=1}^m \frac{1}{n^r m^s}\\ &=\sum_{n=1}^\infty\left(\sum_{m=n}^\infty\frac{1}{m^s}\right)\frac1{n^r}\\ &=\sum_{n=1}^\infty\left(\sum_{m=1}^\infty\frac{1}{m^s}-\sum_{m=1}^n\frac1{m^s}+\frac1{n^s}\right)\frac1{n^r}\\ &=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{n^rm^s}-\sum_{n=1}^\infty\frac{H_n^{(s)}}{n^r}+\sum_{n=1}^\infty\frac{1}{n^{r+s}}\\ &=\zeta(r)\zeta(s)-\sum_{n=1}^\infty\frac{H_n^{(s)}}{n^r}+\zeta(r+s), \end{align}

or

$$\sum_{n=1}^\infty\frac{H_n^{(s)}}{n^r}+\sum_{n=1}^\infty\frac{H_n^{(r)}}{n^s}=\zeta(r)\zeta(s)+\zeta(r+s).$$

Setting $r=s=2$ gives $\displaystyle \sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}=\frac12\zeta^2(2)+\frac12\zeta(4)=\frac74\zeta(4).$

Answer 4

\begin{align}\text{J}&=\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx\\ &\overset{\text{IBP}}=\underbrace{\Big[-\ln(1-x)\ln x\mathrm{Li}_2(x)\Big]_0^1}_{=0}+\underbrace{\int_0^1\frac{\ln(1-x)\text{Li}_2(x)}{x}dx}_{=-\frac{1}{2}\text{Li}^2_2(1)}-\underbrace{\int_0^1 \frac{\ln^2(1-x)\ln x}{x}dx}_{\text{IBP}}\\ &=-\frac{\pi^4}{72}-\int_0^1 \frac{\ln(1-x)\ln^2 x}{1-x}dx\\ &\overset{\text{IBP}}=-\frac{\pi^4}{72}-\left[\left(\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)-\int_0^1\frac{\ln^2 t}{1-t}dt\right)\ln(1-x)\right]_0^1-\\&\int_0^1 \frac{1}{1-x}\left(\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)-\int_0^1 \frac{\ln^2 t}{1-t}dt\right)dx\\ &=-\frac{\pi^4}{72}-\int_0^1 \frac{1}{1-x}\left(\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)-\int_0^1 \frac{\ln^2 t}{1-t}dt\right)dx\\ &=-\frac{\pi^4}{72}-\int_0^1 \int_0^1 \left(\frac{x\ln^2(tx)}{(1-tx)(1-x)}-\frac{\ln^2 t}{(1-x)(1-t)}\right)dtdx\\ &=-\frac{\pi^4}{72}+\int_0^1\int_0^1\left(\frac{\ln^2(tx)}{(1-t)(1-tx)}+\frac{\ln^2 t}{(1-t)(1-x)}-\frac{\ln^2(tx)}{(1-t)(1-x)}\right)dtdx\\ &=-\frac{\pi^4}{72}+\int_0^1\int_0^1\left(\frac{\ln^2(tx)}{(1-t)(1-tx)}-\frac{\ln^2 x}{(1-t)(1-x)}\right)dtdx-\\&2\underbrace{\left(\int_0^1 \frac{\ln t}{1-t}dt\right)}_{=-\frac{\pi^2}{6}}\left(\int_0^1 \frac{\ln x}{1-x}dx\right)\\ &=-\frac{5\pi^4}{72}+\int_0^1 \left(\frac{1}{t(1-t)}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)-\frac{1}{1-t}\left(\int_0^1\frac{\ln^2 x}{1-x}dx\right)\right)dt\\ &=-\frac{5\pi^4}{72}+\int_0^1 \frac{1}{t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt-\int_0^1 \frac{1}{1-t}\left(\int_t^1 \frac{\ln^2 u}{1-u}du\right)dt\\ &\overset{\text{IBP}}=-\frac{5\pi^4}{72}-\underbrace{\int_0^1 \frac{\ln^3 t}{1-t}dt}_{=-6\zeta(4)=-\frac{\pi^4}{15}}+\underbrace{\int_0^1 \frac{\ln(1-t)\ln^2 t}{1-t}dt}_{-\frac{\pi^4}{72}-\text{J}}\\ &=-\frac{\pi^4}{60}-\text{J}\\ &=\boxed{-\dfrac{\pi^4}{120}} \end{align}

Answer 5

Evaluate the integral $I$ below in two ways

\begin{align} I=&\int_0^1 \frac{\ln^2x \ln (1-x)}{1-x}dx\\ = &\int_0^1 \frac{\ln^2x}{1-x}\left(\int_0^1 \frac {-x}{1-x y}dy\right) dx =\int_0^1 \frac{1}{1-y}\left(\int_0^1 \frac {\ln^2x}{1-yx}dx - 2Li_3(1)\right) dy \\ =& \>2 \int_0^1\frac{Li_3(y)}{y}dy + 2 \int_0^1\frac{Li_3(y)-Li_3(1)}{1-y}\>\overset{ibp}{dy} = 2Li_4(1) - Li_2^2(1)\\\\ I =&\int_0^1 \frac{\ln x \ln^2(1-x)}{x}dx \overset{ibp}=- \int_0^1 \frac{\ln x Li_2(x)}{1-x}dx - \frac12 Li_2^2(1) \end{align}

which leads to $$\int_0^1 \frac{\ln x Li_2(x)}{1-x}dx=-2Li_4(1)+\frac12Li_2^2(1)=-\frac{\pi^4}{120} $$