A generalized "Rare" integral involving $\operatorname{Li}_3$

Question

In my previous post, it can be shown that $$\int_{0}^{1} \frac{\operatorname{Li}_2(-x)- \operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x) -\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} } =\frac{\pi^3}{48\sqrt{2} }.$$ But how we verify this? $$\int_{0}^{1} \frac{\operatorname{Li}_3(1-z)+\operatorname{Li}_3 \left ( \frac{1}{1+z} \right ) +\frac{\pi^2}{3}\ln(1+z) -\frac{\pi z}{2}\ln(1+z)^2-\frac{1}{6}\ln(1+z)^3 - \frac{\pi z}{2}\ln(z)^2+\pi z\ln(z)\ln(1+z) }{1+z^2} \frac{\text{d}z}{\sqrt{1-z^2} } =\frac{35\pi\zeta(3)}{64\sqrt{2} }+\frac{\pi^3}{32\sqrt{2} }\ln(2).$$ Where $\operatorname{Li}_3$ is trilogarithm and $\zeta(3)=\operatorname{Li}_3(1)$ in the principal branch. The same method seems not quite powerful. Any suggestion will be appreciated.

Answer 1

A year and a half later...

Firstly, some generalisations that I'm sure the OP is aware of/may appreciate:

$$\int_{0}^{1}\frac{\operatorname{Li}_4\left(\frac{1}{1+z}\right)-\operatorname{Li}_4(1-z)+\frac{1}{24} \ln(1+z)^4+\frac{\pi z}{6}\ln(1+z)^3-\frac{\pi z}{6}\ln(z)^3-\frac{\pi^2}{6}\ln(1+z)^2+\frac{\pi z}{2}\ln(z)^2\ln(1+z)-\frac{\pi z}{2}\ln(z)\ln(1+z)^2}{1+z^2}\frac{\mathrm{d}z}{\sqrt{1-z^2}}=\frac{3\pi^5}{5120\sqrt{2}}-\frac{\pi^3\ln(2)^2}{48\sqrt{2}}+\frac{5\pi\ln(2)^4}{384\sqrt{2}}+\frac{5\pi}{16\sqrt{2}}\operatorname{Li}_4\left(\frac{1}{2}\right).$$

$$\int_{0}^{1}\frac{\operatorname{Li}_5(1-z)+\operatorname{Li}_5\left(\frac{1}{1+z}\right)-\frac{1}{120}\ln(1+z)^5-\frac{\pi z}{24}\ln(1+z)^4-\frac{\pi z}{24}\ln(z)^4+\frac{\pi z}{6}\ln(z)\ln(1+z)^3+\frac{\pi^2}{18}\ln(1+z)^3+\frac{\pi z}{6}\ln(z)^3\ln(1+z)-\frac{\pi z}{4}\ln(z)^2\ln(1+z)^2+\frac{\pi^4}{45}\ln(1+z)}{1+z^2}\frac{\mathrm{d}z}{\sqrt{1-z^2}}=\frac{2139\pi\zeta (5)}{4096\sqrt{2}}-\frac{\pi\ln(2)^5}{768 \sqrt{2}}+\frac{\pi^3\ln(2)^3}{288\sqrt{2}}+\frac{97\pi^5\ln(2)}{18432 \sqrt{2}}+\frac{5 \pi}{32\sqrt{2}}\operatorname{Li}_5\left(\frac{1}{2}\right).$$

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Although I imagine the OP is after an approach involving contour integration, I shall present an elementary real analytic approach instead. We shall make use of the following two representations:

$$\tag{1}\frac{(-1)^{n-1}}{n!}\int_{0}^{1}\frac{\ln(t)^n\ln\left|1-2t+t^2(1-x^2)\right|}{t}\, dt = \Re\left(\operatorname{Li}_{n+2}\left(1+x\right)+\operatorname{Li}_{n+2}\left(1-x\right)\right)$$ for $n\in\mathbb{N}_0$, and $$\tag{2}\int_{0}^{\infty}\frac{\ln\left|1-2t+\frac{t^2}{1+u^2}\right|}{2u^2+1}\, du=\frac{\pi}{\sqrt{2}}\ln\left|\frac{1+\frac{\sqrt{2}}{2}\sqrt{1-2t}-t}{1+\frac{\sqrt{2}}{2}}\right|.$$ We leave the proof of $(1)$ as an exercise and prove $(2)$ at the end.

Now begin by recognising that we can write the given integral in the following way through trilogarithm identities: \begin{align*}I&=\int_{0}^{1}\frac{\Re\left(\operatorname{Li}_3\left(1+x\right)+\operatorname{Li}_3\left(1-x\right)\right)-\frac{\pi x}{2}\ln\left(1+\frac{1}{x}\right)^2}{(1+x^2)\sqrt{1-x^2}}\,dx \\ &=\color{red}{\underbrace{\int_{0}^{1}\frac{\Re\left(\operatorname{Li}_3\left(1+x\right)+\operatorname{Li}_3\left(1-x\right)\right)}{(1+x^2)\sqrt{1-x^2}}\,dx}_A} - \color{blue}{\underbrace{\int_{0}^{1} \frac{\frac{\pi x}{2}\ln\left(1+\frac{1}{x}\right)^2}{(1+x^2) \sqrt{1-x^2}} \, dx}_B} \end{align*}

Now let us focus our attention on $\color{red}{A}$: \begin{align*}A&:=\int_{0}^{1}\frac{\Re\left(\operatorname{Li}_3\left(1+x\right)+\operatorname{Li}_3\left(1-x\right)\right)}{(1+x^2)\sqrt{1-x^2}}\,dx \\ &\stackrel{(1)}{=}\int_{0}^{1}\int_{0}^{1}\frac{1}{(x^2+1)\sqrt{1-x^2}}\frac{\ln(t)\ln\left|1-2t+t^2(1-x^2)\right|}{t}\, dt\, dx \end{align*} Now enforce the substitution $u=\frac{x}{\sqrt{1-x^2}}$ to arrive at \begin{align*}A &= \int_{0}^{1}\int_{0}^{\infty}\frac{\ln(t)\ln\left|1-2t+\frac{t^2}{1+u^2}\right|}{t(2u^2+1)}\,du\,dt \\ &\stackrel{(2)}{=}\frac{\pi}{\sqrt{2}}\int_{0}^{1}\frac{\ln(t)\ln\left|\frac{1+\frac{\sqrt{2}}{2}\sqrt{1-2t}-t}{1+\frac{\sqrt{2}}{2}}\right|}{t}\,dt \\ &=\frac{\pi}{2\sqrt{2}}\int_{0}^{1}\frac{\ln(t)\left(2\ln\left|\frac{1+\frac{\sqrt{2}}{2}\sqrt{1-2t}-t}{1+\frac{\sqrt{2}}{2}}\right|+\ln\left|\frac{1-\frac{\sqrt{2}}{2}\sqrt{1-2t}-t}{1-\frac{\sqrt{2}}{2}}\right|-\ln\left|\frac{1-\frac{\sqrt{2}}{2}\sqrt{1-2t}-t}{1-\frac{\sqrt{2}}{2}}\right|\right)}{t}\,dt \\ &\stackrel{\text{IBP}}{=}\frac{\pi}{2\sqrt{2}}\int_{0}^{1}\frac{\ln(t)\ln\left(2t^2-2t+1\right)}{t}\,dt+\Re\left(\frac{\pi}{2\sqrt{2}}\int_{0}^{1}\frac{t \ln(t)^2}{\sqrt{\frac{1}{2}-t}\,(2t^2-2t+1)}\,dt\right) \\ &=\frac{\pi}{2\sqrt{2}}\int_{0}^{1}\frac{\ln(t)\ln\left(2t^2-2t+1\right)}{t}\, dt+\frac{\pi}{2\sqrt{2}}\int_{0}^{1/2}\frac{t\ln(t)^2}{\sqrt{\frac{1}{2}-t}\,(2t^2-2t+1)}\, dt \end{align*} Now on the second integral, enforce the substitution $\frac{1}{t}=1+\frac{1}{x}$ to get: \begin{align*}A &= \frac{\pi}{2\sqrt{2}}\int_{0}^{1}\frac{\ln(t)\ln\left(2t^2-2t+1\right)}{t}\, dt+\color{blue}{\underbrace{\int_{0}^{1} \frac{\frac{\pi x}{2}\ln\left(1+\frac{1}{x}\right)^2}{(1+x^2) \sqrt{1-x^2}} \, dx}_{B}} \end{align*} i.e. \begin{align*}I &= \frac{\pi}{2\sqrt{2}}\int_{0}^{1}\frac{\ln(t)\ln(2t^2-2t+1)}{t}\, dt \\ &\stackrel{(1)}{=}\frac{\pi}{\sqrt{2}}\Re\left(\operatorname{Li}_3\left(1+i\right)\right)\\ &= \frac{35\pi\zeta(3)}{64\sqrt{2}}+\frac{\pi^3}{32\sqrt{2}}\ln(2),\end{align*} where we use $\Re\left(\operatorname{Li}_3(1+i)\right)=\frac{\pi^2}{32}\ln(2)+\frac{35}{64}\zeta(3)$, and thus obtain the desired result. $\square$

To summarise what we have done, we have trivialised the given integral by simplifying and reducing it to a much easier logarithmic integral that I'm sure you are familiar with methods of evaluating. Your previous question that I gave an answer to can be attacked in exactly the same way by using dilogarithm identities to reduce the integral to a much simpler logarithmic integral expression, which I have now extended my original answer to include this method.

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It remains to show $(2)$, i.e. that $\forall t\in\mathbb{C}$, $$\int_{0}^{\infty}\frac{\ln\left|1-2t+\frac{t^2}{1+u^2}\right|}{2u^2+1}\, du=\frac{\pi}{\sqrt{2}}\ln\left|\frac{1+\frac{\sqrt{2}}{2}\sqrt{1-2t}-t}{1+\frac{\sqrt{2}}{2}}\right|.$$

Proof. Let \begin{align*}F(t)&:=\int_{0}^{\infty}\frac{\ln\left|1-2t+\frac{t^2}{1+u^2}\right|}{2u^2+1}\, du \\ &= -\int_{0}^{\infty}\frac{\ln(1+u^2)}{2u^2+1}\,du+\int_{0}^{\infty}\frac{\ln\left|(1-t)^2+(1-2t)u^2\right|}{2u^2+1}\, du \\ &=-\int_{0}^{\infty}\frac{\ln(1+u^2)}{2u^2+1}\,du+\frac{\pi}{\sqrt{2}}\ln|1-t|+\int_{0}^{\infty}\frac{\ln\left|1+\frac{(1-2t)u^2}{(1-t)^2}\right|}{2u^2+1}du \end{align*} since $\int_{0}^{\infty}\frac{1}{2u^2+1}\,du=\frac{\pi}{2\sqrt{2}}$. Now use the representation: $$\log\left|1+\frac{a^2}{b^2}\right|=2\Re\int_{0}^{\infty}\frac{(1-\cos(ax))e^{-bx}}{x}\, dx$$ for $a\in\mathbb{R}$ and $\Re(b) \geq 0$, $b\neq 0$. This gives:

\begin{align*} F(t) &= -\int_{0}^{\infty}\frac{\ln(1+u^2)}{2u^2+1}\,du+\frac{\pi}{\sqrt{2}}\ln|1-t|+2\Re\int_{0}^{\infty}\int_{0}^{\infty}\frac{(1-\cos(ux))\exp\left(-\frac{(1-t)x}{\sqrt{1-2t}}\right)}{x(2u^2+1)}\,dx\,du \\ &= -\int_{0}^{\infty}\frac{\ln(1+u^2)}{2u^2+1}\,du+\frac{\pi}{\sqrt{2}}\ln|1-t|+\frac{\pi}{\sqrt{2}}\Re\int_{0}^{\infty}\frac{\left(1-\exp\left(-\frac{x}{\sqrt{2}}\right)\right)\exp\left(-\frac{(1-t)x}{\sqrt{1-2t}}\right)}{x}\, dx \\ &= -\int_{0}^{\infty}\frac{\ln(1+u^2)}{2u^2+1}\,du+\frac{\pi}{\sqrt{2}}\ln|1-t|+\frac{\pi}{\sqrt{2}}\Re\ln\left(\frac{\frac{1}{\sqrt{2}}+\frac{1-t}{\sqrt{1-2t}}}{\frac{1-t}{\sqrt{1-2t}}}\right) \end{align*} by recognising the rightmost integral as a Frullani-type integral. Hence, $$F(t)=-\int_{0}^{\infty}\frac{\ln(1+u^2)}{2u^2+1}\,du+\frac{\pi}{\sqrt{2}}\ln\left|1+\sqrt{\frac{1}{2}-t}-t\right|.$$ Now using the fact that $F(0)=0$, we can conclude the value of the remaining integral and determine that $$F(t)=\int_{0}^{\infty}\frac{\ln\left|1-2t+\frac{t^2}{1+u^2}\right|}{2u^2+1}\, du=\frac{\pi}{\sqrt{2}}\ln\left|\frac{1+\frac{\sqrt{2}}{2}\sqrt{1-2t}-t}{1+\frac{\sqrt{2}}{2}}\right|,$$ as was to be shown. $\square$