How to evaluate $\int_0^1 \frac{\operatorname{Li}_2\left(\frac{x}{x-1}\right) - \operatorname{Li}_2\left(\frac{x}{x+1}\right)}{x} \, dx $

Question

$$\int_0^1 \frac{\operatorname{Li}_2\left(\frac{x}{x-1}\right) - \operatorname{Li}_2\left(\frac{x}{x+1}\right)}{x} \, dx $$

$(\operatorname{Li}_2(z))$ for $(|z| < 1)$ is: $\operatorname{Li}_2(z) = -\sum_{n=1}^\infty \frac{z^n}{n^2}$

Series representation

$$\int_0^1 \frac{-\left(\sum_{n=1}^\infty \frac{(\frac{x}{x-1})^n}{n^2}\right) + \left(\sum_{n=1}^\infty \frac{(\frac{x}{x+1})^n}{n^2}\right)}{x} dx $$

$(\operatorname{Li}_2(z))$ series converge for $|z| < 1$ \begin{align*}&-\sum_{n=1}^\infty \int_0^1 \frac{1}{n^2} \left(\frac{x^n}{(x-1)^n} - \frac{x^n}{(x+1)^n}\right) dx \\ &=-\int_0^1 \frac{x^n}{n^2} \left(\frac{1}{(x-1)^n} - \frac{1}{(x+1)^n}\right) dx\\ & =-\int_0^1 \frac{x^n}{n^2} \left(\frac{\sum_{k=0}^n \binom{n}{k}x^k - \sum_{k=0}^n \binom{n}{k}(-1)^kx^k}{(x^2 - 1)^n}\right) dx\\ & =-\int_0^1 \frac{x^n}{n^2} \left(\frac{\sum_{k=0}^n \binom{n}{k}x^k}{(x^2 - 1)^n} - \frac{\sum_{k=0}^n \binom{n}{k}(-1)^kx^k}{(x^2 - 1)^n}\right) dx \end{align*}

Answer 1

From the definition of the polylogarithm, you directly get the recursion $\int_{0}^{z}\frac{\mathrm{Li}_s(t)}{t}\mathrm{d}t = \mathrm{Li}_{s+1}(z)$. Also using that $\mathrm{Li}_2(x) +\mathrm{Li}_2\left(\frac{x}{x-1}\right) = - \frac{\ln^2(1-x)}{2}$ we get \begin{align*} \int_{0}^{1}\frac{\mathrm{Li}_2\left( \frac{x}{x-1}\right)}{x}\,\mathrm{d}x & = - \int_{0}^{1}\frac{\mathrm{Li}_2(x)}{x}\,\mathrm{d}x -\frac12 \int_{0}^{1}\frac{\ln^2(1-x)}{x}\,\mathrm{d}x = -\mathrm{Li}_3(1)-\frac12 \int_{0}^{1}\frac{\ln^2(1-x)}{x}\,\mathrm{d}x \end{align*} and since $\int_{0}^{1}\frac{\ln^2(1-x)}{x}\,\mathrm{d}x = 2\zeta(3)$, combining this with $\mathrm{Li}_s(1) = \zeta(s)$ (which is easily seen from the series definition) we get $$ \int_{0}^{1}\frac{\mathrm{Li}_2\left( \frac{x}{x-1}\right)}{x}\,\mathrm{d}x =-\zeta(3) - \frac{2\zeta(3)}{2} =-2\zeta(3) $$ Similarlly, since plugging $x \to -x$ in the dilogarithm formula gives $\mathrm{Li}_2(-x) +\mathrm{Li}_2\left(\frac{x}{x+1}\right) = - \frac{\ln^2(1+x)}{2}$, so $$ \int_{0}^{1}\frac{\mathrm{Li}_2\left( \frac{x}{x+1}\right)}{x}\,\mathrm{d}x = - \int_{0}^{1}\frac{\mathrm{Li}_2(-x)}{x}\,\mathrm{d}x -\frac12 \int_{0}^{1}\frac{\ln^2(1+x)}{x}\,\mathrm{d}x \overset{\color{blue}{x\to -x}}{=} -\mathrm{Li}_3(-1) - \frac12 \int_{0}^{1}\frac{\ln^2(1+x)}{x}\,\mathrm{d}x $$ Recalling that $\mathrm{Li}_s(-1) = \left(2^{1-s}-1\right)\zeta(s) $ and that $\int_{0}^{1}\frac{\ln^2(1+x)}{x}\,\mathrm{d}x = \frac{\zeta(3)}{4}$ we get $$ \int_{0}^{1}\frac{\mathrm{Li}_2\left( \frac{x}{x+1}\right)}{x}\,\mathrm{d}x = \frac{3\zeta(3)}{4} -\frac{\zeta(3)}{8} = \frac{5\zeta(3)}{8} $$ so subtracting the $2$ integral evaulations gives $$ \boxed{\int_{0}^{1}\frac{\mathrm{Li}_2\left( \frac{x}{x-1}\right)-\mathrm{Li}_2\left( \frac{x}{x+1}\right)}{x}\,\mathrm{d}x = -\frac{21\zeta(3)}{8}} $$

---

We know that $\mathrm{Li}_1(z) = -\ln(1-z)$ comparing with the Taylor series for the natural log.

Noticing that $\int_{0}^{x} \frac{\mathrm{Li}_s\left(\frac{t}{t-1}\right)}{t(1-t)}\mathrm{d}{t} \overset{\color{purple}{t\to\frac{t}{t-1}}}{=} \int_{0}^{\frac{x}{x-1}}\frac{\mathrm{Li}_s(t)}{t}\mathrm{d}{t} {=} \mathrm{Li}_{s+1}\left(\frac{x}{x-1}\right) $. This gives \begin{align} \mathrm{Li}_2(x) +\mathrm{Li}_2\left(\frac{x}{x-1}\right) & = \int_{0}^{x} \frac{\mathrm{Li}_1(t)}{t}\mathrm{d}{t} + \int_{0}^{x} \frac{\mathrm{Li}_1(\frac{t}{t-1})}{t(1-t)}\mathrm{d}{t} \notag \\ & = \int_{0}^{x} -\frac{\ln(1-t)}{t} + \frac{\ln(1-t)}{t(1-t)}\mathrm{d}{t}\notag \\ &= - \frac{\ln^2(1-x)}{2} \end{align} obtaining the formula given at the beginning.

Answer 2

Consider $$I=\int \frac{\text{Li}_2\left(\frac{x}{x-1}\right)}{x}\,dx$$ Let $$\frac{x}{x-1}=t \quad\implies \quad x=\frac{t}{t-1} \implies \quad dx=-\frac{dt}{(t-1)^2}$$ which makes $$I=\int \frac{\text{Li}_2(t)}{(1-t) t}\,dt=\int \frac{\text{Li}_2(t)}{t}\,dt+\int\frac{\text{Li}_2(t)}{1-t}\,dt $$ which are simple.

Do the same for the second part; for each of them, go back to $x$ before using the bounds.

You should get nice results for each piece.

Answer 3

\begin{align}J&=\int_0^1 \frac{\text{Li}_2\left(\frac{x}{x-1}\right) - \text{Li}_2\left(\frac{x}{x+1}\right)}{x} \, dx\\ &\overset{\text{IBP}}=\int_0^1 \left(\frac{\ln(1+x)\ln x}{x}-\frac{\ln(1+x)\ln x}{1+x}-\frac{\ln(1-x)\ln x}{x}-\frac{\ln(1-x)\ln x}{1-x}\right)dx\\ &\overset{\text{IBP}}=-\frac{1}{2}\underbrace{\int_0^1\frac{\ln^2 x}{1+x}dx}_{=A}+\frac{1}{2}\underbrace{\int_0^1 \frac{\ln^2(1+x)}{x}dx}_{=B}-\frac{1}{2}\underbrace{\int_0^1 \frac{\ln^2 x}{1-x}dx}_{=2\zeta(3)}-\frac{1}{2}\int_0^1 \underbrace{\frac{\ln^2(1-x)}{x}dx}_{=1-u}\\ &=-\frac{1}{2}A+\frac{1}{2}B-2\zeta(3)\\ A&=\int_0^1 \frac{\ln^2 x}{1-x}dx-\underbrace{\int_0^1 \frac{2x\ln^2 x}{1-x^2}dx}_{u=x^2}=\frac{3}{4}\int_0^1 \frac{\ln^2 x}{1-x}dx=\boxed{\frac{3}{2}\zeta(3)}\\ B&\overset{u=\frac{1}{1+x}}=\int_{\frac{1}{2}}^1\frac{\ln^2 u}{u(1-u)}du=\frac{1}{3}\ln^3 2+\underbrace{\int_{\frac{1}{2}}^1\frac{\ln^2 u}{1-u}du}_{=C}=\frac{1}{3}\ln^3 2+2\zeta(3)-\int_0^{\frac{1}{2}}\frac{\ln^2 u}{1-u}du\\ &=\frac{1}{3}\ln^3 2+2\zeta(3)-\underbrace{\int_0^{\frac{1}{2}}\frac{\ln^2\left( \frac{u}{1-u}\right)}{1-u}du}_{z=\frac{u}{1-u}}+\int_0^{\frac{1}{2}}\frac{\ln^2\left(1-u\right)}{1-u}du-2\underbrace{\int_0^{\frac{1}{2}}\frac{\ln(1-u)\ln u}{1-u}du}_{\text{IBP}}\\ &=\frac{2}{3}\ln^3 2+2\zeta(3)-A-\ln^3 2-\underbrace{\int_0^{\frac{1}{2}}\frac{\ln^2(1-u)}{u}du}_{z=1-u}\\ &=\frac{1}{2}\zeta(3)-\frac{1}{3}\ln^3 2-C\\ 2B&=\frac{1}{2}\zeta(3)\\ B&=\boxed{\frac{1}{4}\zeta(3)}\\ J&=-\frac{1}{2}\times \frac{3}{2}\zeta(3)+\frac{1}{2}\times \frac{1}{4}\zeta(3)-2\zeta(3)=\boxed{-\frac{21}{8}\zeta(3)}\\ \end{align}

NB: I assume: \begin{align}\text{Li}_2(x)&=-\int_0^x\frac{\ln(1-t)}{t}dt\\ \int_0^1 \frac{\ln^2 x}{1-x}dx&=2\zeta(3) \end{align}