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This is related to this question and my partial answer to it. I've found that the proof of the 2nd identity reduces to showing that $$i\int_0^{\pi}\left[\mathrm{Li}_2\left(-1-e^{ix}\right)-\mathrm{Li}_2\left(-1-e^{-ix}\right)\right]dx=\frac{7}{3}\zeta(3) \tag{1} $$ Here $\mathrm{Li}_2(z)$ is the dilogarithm function defined by analytic continuation of the series $\sum\limits_{n=1}^{\infty}\frac{z^n}{n^2}$ to the cut plane $\mathbb{C}\backslash[1,\infty)$ and $\zeta(3)=\sum\limits_{n=1}^{\infty}\frac{1}{n^3}$ is the Apéry's constant.

I am looking for a proof of $(1)$ or an alternative proof of the 2nd identity here. Thank you in advance.

Sangchul Lee
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1 Answers1

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We give hints to show that

$$i\int_0^\pi\operatorname{Li}_2\left(-1-e^{ix}\right)-\operatorname{Li}_2\left(-1-e^{-ix}\right)dx=\frac{7\zeta(3)}{3}$$

in three steps.

First step. We show that $$ I = i\int_0^\pi \operatorname{Li}_2\left(-1-e^{ix}\right)-\operatorname{Li}_2\left(-1-e^{-ix}\right)dx = 4\int_0^1 \frac{\chi_2\left(\tfrac{t}{t+1}\right)}{t}\,dt, $$ where $\operatorname{Li}_2$ is the dilogarithm function and $\chi_2$ is the second order Legendre chi function.

We need some previous knowledge. We use an integral representation of the dilogarithm: $$ \operatorname{Li}_2(z)=-\int_0^1\frac{\ln(1-t\,z)}t\,dt.\tag{$\spadesuit$} $$

After that we use that for all $a,b>0$: $$ \int \ln\left(a+be^{\pm ix}\right)dx = x\ln(a) \pm i \operatorname{Li}_2\left(-\frac{be^{\pm ix}}{a}\right) + C. \tag{$\diamondsuit$} $$

Now the solution of your integral problem:

\begin{align} I &= i\int_0^\pi \operatorname{Li}_2\left(-1-e^{ix}\right)-\operatorname{Li}_2\left(-1-e^{-ix}\right)dx \\ &\stackrel{\spadesuit}{=} -i \int_0^\pi \int_0^1 \frac{1}{t} \left( \ln\left(1+t+te^{ix}\right) - \ln\left(1+t+te^{-ix}\right) \right)dt\,dx \\ &= -i \int_0^1 \frac{1}{t} \int_0^\pi \ln\left(1+t+te^{ix}\right) - \ln\left(1+t+te^{-ix}\right) dx\,dt. \end{align}

By substitute $a=t+1$ and $b=t$ into $(\diamondsuit)$ we get:

\begin{align} I &= -i \int_0^1 \frac{1}{t} \left[x\ln(t+1) + i\operatorname{Li}_2\left(-\frac{te^{ix}}{t+1}\right)\right]_0^\pi - \frac{1}{t}\left[x\ln(t+1) - i\operatorname{Li}_2\left(-\frac{te^{-ix}}{t+1}\right)\right]_0^\pi \,dt \\ &= -i \int_0^1 \frac{1}{t} \left( 2i\operatorname{Li}_2\left(\frac{t}{t+1}\right) - 2i\operatorname{Li}_2\left(-\frac{t}{t+1}\right) \right) \, dt. \end{align}

By using the fact, that $\chi_\nu(z) = \frac{1}{2}\left(\operatorname{Li}_\nu(z) - \operatorname{Li}_\nu(-z)\right)$, we get: $$ I = 4\int_0^1 \frac{\chi_2\left(\tfrac{t}{t+1}\right)}{t}\,dt. $$

Second step. We show that \begin{align} I &= -2 \operatorname{Li}_3\left(-\frac{1}{2}\right)-2 \operatorname{Li}_3\left(-\frac{1}{3}\right)+2 \operatorname{Li}_3\left(\frac{1}{3}\right)-2 \operatorname{Li}_3\left(\frac{2}{3}\right)-2 \operatorname{Li}_2\left(-\frac{1}{3}\right) \ln (3) \\ &+2 \operatorname{Li}_2\left(\frac{1}{3}\right) \ln (3)-2 \operatorname{Li}_2\left(\frac{2}{3}\right) \ln (3)+\frac{\ln ^3(2)}{3}-\frac{2 \ln^3(3)}{3}+\ln^2(3) \ln (2)+\frac{\pi^2}{3} \ln (2). \end{align}

To show this we have to calculate the following two integrals: \begin{align} \int_0^1 \frac{\operatorname{Li}_2\left(\frac{t}{t+1}\right)}{t}\,dt &= \frac{5 \zeta (3)}{8},\\ \int_0^1 \frac{\operatorname{Li}_2\left(-\frac{t}{t+1}\right)}{t}\,dt &= \operatorname{Li}_3\left(-\frac{1}{2}\right)+\operatorname{Li}_3\left(-\frac{1}{3}\right)- \operatorname{Li}_3\left(\frac{1}{3}\right)+\operatorname{Li}_3\left(\frac{2}{3}\right) \\ &+\operatorname{Li}_2\left(-\frac{1}{3}\right) \ln (3)-\operatorname{Li}_2\left(\frac{1}{3}\right) \ln (3)+\operatorname{Li}_2\left(\frac{2}{3}\right) \ln (3)\\ &-\frac{\ln^3(2)}{6}+\frac{\ln^3(3)}{3}-\frac{1}{2} \ln (2)\ln^2(3)-\frac{1}{6} \pi^2 \ln (2) +\frac{5 \zeta (3)}{8} \end{align}

By using the substitution $x=t/(t+1)$, and after that, by using $(\spadesuit)$ we get: $$ \int_0^1 \frac{\operatorname{Li}_2\left(\pm \frac{t}{t+1}\right)}{t}\,dt = - \int_0^{1/2}\frac{\operatorname{Li}_2(\pm x)}{x(x-1)}\,dx = \int_0^{1/2}\int_0^1 \frac{\ln(1 \mp tx)}{tx(x-1)}\,dt\,dx. $$ I let the rest to the reader. Mathematica is able to evaluate both integrals.

Third step. We show that $$ I = \frac{7\zeta(3)}{3}. $$

To prove that \begin{align} & -2 \operatorname{Li}_3\left(-\frac{1}{2}\right)-2 \operatorname{Li}_3\left(-\frac{1}{3}\right)+2 \operatorname{Li}_3\left(\frac{1}{3}\right)-2 \operatorname{Li}_3\left(\frac{2}{3}\right)-2 \operatorname{Li}_2\left(-\frac{1}{3}\right) \ln (3) \\ &+2 \operatorname{Li}_2\left(\frac{1}{3}\right) \ln (3)-2 \operatorname{Li}_2\left(\frac{2}{3}\right) \ln (3)+\frac{\ln ^3(2)}{3}-\frac{2 \ln^3(3)}{3}+\ln^2(3) \ln (2)+\frac{\pi^2}{3} \ln (2) \end{align} equals to $$ \frac{7\zeta(3)}{3}, $$ is quivalent to prove that $$ 2\operatorname{Li}_3\left(-\frac{1}{2}\right)+\operatorname{Li}_3\left(-\frac{1}{3}\right)+2\operatorname{Li}_3\left(\frac{2}{3}\right)+\operatorname{Li}_2\left(-\frac{1}{3}\right) \ln(3)+2\operatorname{Li}_2\left(\frac{2}{3}\right) \ln(3)$$ equals to $$ \frac{\pi^2}{3}\ln(2)+\frac{1}{3}\ln^3(2)-\frac{1}{3} \ln^2(3) \ln\left(\frac{27}{8}\right)-\frac{\zeta(3)}{6}. $$ This work is done by you, so eventually you've answered your own question.

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