The trilogarithm function

The strong connection between the $\operatorname{Li}_{s}$ function and $\phi$ vol. III

Today we show the proof of this integral posted by @integralsbot $$\int_{0}^{\ln \phi} \frac{x^2(e^{2x}+1)}{(e^{2x}-1)} dx = \frac{\zeta(3)}{10}$$ The proof relies on some identities of the trilogarithm and dilogarithm function. We evaluated similar integrals and series here, here.

Proof:

$$\begin{align*} \int_{0}^{\ln \phi} \frac{x^2(e^{2x}+1)}{(e^{2x}-1)} dx =& \frac{1}{8}\int_{1}^{\phi^2} \frac{\ln^2(w)(w+1)}{w(w-1)}dw \quad (w\mapsto e^{2x})\\ =& \frac{1}{8} \int_{1}^{\phi^2} \frac{\ln^2(w)}{w-1+}dw + \frac{1}{8}\int_{1}^{\phi^2} \frac{\ln^2(w)}{w(w-1)}dw \\ =& \frac{1}{8} \int_{1}^{\phi^2} \frac{\ln^2(w)}{w-1}dw + \frac{1}{8} \int_{1}^{\phi^2} \frac{\ln^2(w)}{w-1}dw - \frac{1}{8}\int_{1}^{\phi^2} \frac{\ln^2(w)}{w}dw \\ =& \frac{1}{4} \underbrace{\int_{1}^{\phi^2} \frac{\ln^2(w)}{w-1}dw}_{J} - \frac{1}{8}\underbrace{\int_{1}^{\phi^2} \frac{\ln^2(w)}{w}dw}_{K} \end{align*}$$ $$\begin{align*} K = \int_{1}^{\phi^2} \frac{\ln^2(w)}{w}dw =& \int_{0}^{\ln \phi^2 } s^2 ds \quad (s \mapsto \ln(w))\\ =& \left[\frac{s^3}{3}\right]_{0}^{\ln \phi^2}\\ =& \frac{\ln^3(\phi^2)}{3} \end{align*}$$ $$\begin{align*} J = \int_{1}^{\phi^2} \frac{\ln^2(w)}{w-1}dw = & \int_{0}^{\phi} \frac{\ln^2(1+s)}{s} ds \quad (s \mapsto w-1, \phi^2-1 = \phi) \\ \stackrel{IBP}{=}& \ln(s)\ln^2(1+s)\Big|_{0}^{\phi} - 2\int_{0}^{\phi} \frac{\ln(s)\ln(1+s)}{1+s} ds\\ =& \ln(\phi)\ln^2(1+\phi) - 2\int_{1}^{\phi^2} \frac{\ln(t-1)\ln(t)}{t}dt \quad (t \mapsto s+1 )\\ =& \ln(\phi)\ln^2(1+\phi) - 2\int_{1}^{\phi^2} \frac{\ln\left(t\left(1-\frac{1}{t}\right)\right)\ln(t)}{t}dt\\ =& \ln(\phi)\ln^2(1+\phi) - 2\int_{1}^{\phi^2} \frac{\ln^2(t)}{t}dt - 2\int_{1}^{\phi^2} \frac{\ln\left(1-\frac{1}{t}\right)\ln(t)}{t}dt\\ =& \ln(\phi)\ln^2(1+\phi) - \frac{2}{3}\ln^3(\phi^2) -2\int_{1}^{\phi^{-2}} \frac{\ln\left(1-y\right)\ln(y)}{y}dy\\ \stackrel{IBP}{=}& \ln(\phi)\ln^2(1+\phi^2) - \frac{2}{3}\ln^3(\phi^2) +2\operatorname{Li}_{2}(y)\ln(y)\Big|_{1}^{\phi^{-2}}-2\int_{1}^{\phi^{-2}} \frac{\operatorname{Li}_{2}(y)}{y}dy \\ =& \ln(\phi)\ln^2(1+\phi) - \frac{2}{3}\ln^3(\phi^2) +2\operatorname{Li}_{2}(\phi^{-2})\ln(\phi^{-2})-2\int_{1}^{0} \frac{\operatorname{Li}_{2}(y)}{y}dy -2\int_{0}^{\phi^{-2}} \frac{\operatorname{Li}_{2}(y)}{y}dy \\ =& \ln(\phi)\ln^2(1+\phi) - \frac{2}{3}\ln^3(\phi^2) +2\operatorname{Li}_{2}(\phi^{-2})\ln(\phi^{-2})+2\int_{0}^{1} \frac{\operatorname{Li}_{2}(y)}{y}dy -2\int_{0}^{\phi^{-2}} \frac{\operatorname{Li}_{2}(y)}{y}dy \\ =& \ln(\phi)\ln^2(1+\phi) - \frac{2}{3}\ln^3(\phi^2) +2\operatorname{Li}_{2}(\phi^{-2})\ln(\phi^{-2})+2\operatorname{Li}_{3}(1) -2\operatorname{Li}_{3}(\phi^{-2}) \\ \end{align*}$$ Note that the polylogarithm function is defined by a power series in $z$: $$\operatorname{Li}_{s} (z) = \sum_{k=1}^{\infty} \frac{z^k}{k^s}$$ If we put $s=3$ and $z=1$ : $$\operatorname{Li}_{3} (1) = \sum_{k=1}^{\infty} \frac{1}{k^3} = \zeta(3)$$ Hence $$J = \ln(\phi)\ln^2(1+\phi) - \frac{2}{3}\ln^3(\phi^2) +2\operatorname{Li}_{2}(\phi^{-2})\ln(\phi^{-2})+2\zeta(3) -2\operatorname{Li}_{3}(\phi^{-2})$$ Landen proved the following identity: $$\operatorname{Li}_{3}(\phi^{-2}) = \frac{4}{5}\zeta(3) + \frac{\pi^2}{15} \ln(\phi^{-2})- \frac{1}{12} \ln^3(\phi^{-2})$$ while Euler proved $$\operatorname{Li}_{2} (\phi^{-2}) = \frac{\pi^2}{15} -\ln^2(\phi)$$ Therefore $$J = \ln(\phi)\ln^2(1+\phi) - \frac{2}{3}\ln^3(\phi^2) - 2\ln^2(\phi)\ln(\phi^{-2}) + \frac{2}{5} \zeta(3) + \frac{\ln^3(\phi^{-2})}{6}$$ Additionally, from the identities: $$\ln(\phi)\ln^2(\phi+1) = 4\ln^3(\phi)$$ $$\ln^2(\phi)\ln(\phi^{-2}) = -2\ln^3(\phi)$$ $$\ln^3(\phi^{-2}) = -8\ln^3(\phi)$$ we have $$J = \frac{4}{3}\ln^3(\phi)+ \frac{2}{5} \zeta(3)$$ Hence $$I = \frac{1}{4}J -\frac{1}{8}K = \frac{1}{3}\ln^3(\phi)+ \frac{\zeta(3)}{10} - \frac{1}{3}\ln^3(\phi)$$ Therefore, we can conclude $$\boxed{ \int_{0}^{\ln \phi} \frac{x^2(e^{2x}+1)}{(e^{2x}-1)} dx = \frac{\zeta(3)}{10}}$$