Finding $\int_{1}^{\infty} \frac{1}{1+x^2} \frac{\operatorname{Li}_2\left ( \frac{1-x}{2} \right ) }{\pi^2+\ln^2\left(\frac{x-1}{2}\right)}\text{d}x$

Question

Prove the integral $$\int_{1}^{\infty} \frac{1}{1+x^2} \frac{\operatorname{Li}_2\left ( \frac{1-x}{2} \right ) }{ \pi^2+\ln^2\left ( \frac{x-1}{2} \right ) }\text{d}x =\frac{96C\ln2+7\pi^3}{12(\pi^2+4\ln^2(2))} -\frac{\pi}{24}(3+4\pi)$$ where $\operatorname{Li}_2$ is dilogarithm and $C$ is Catalan's constant.
(I checked it high precision. So I believe that it's absolutely true.)
How we prove that? I tried to use functional equations of $\operatorname{Li}_2$ but I get nothing useful. Any suggestion will be appreciated.

Answer 1

$$I = \int_{1}^{\infty} \frac{1}{1+x^2} \frac{\operatorname{Li}_2 \left(\frac{1-x}{2}\right)}{\pi^2 + \ln^2 \left(\frac{x-1}{2}\right)} \, dx = 2 \int_{0}^{\infty} \frac{\operatorname{Li}_2 (-x)}{1+(2x+1)^2} \frac{1}{\pi^2 + \ln^2(x)}\, dx$$

Recall $$\operatorname{Li}_2 (-x) = -\int_{0}^{\infty} \frac{x t}{e^t + x} \, dt$$ $$\frac{1}{\pi^2+\ln^2(x)} = \frac{1}{\pi} \int_{0}^{\infty}\cos(\ln(x)u)e^{-\pi u} \,du$$ So after writing $\cos(\ln (x) u) = \Re \left(e^{i (\ln (x) u)}\right) = \Re \left(x^{i u}\right)$ we have $$2 \int_{0}^{\infty} \frac{\operatorname{Li}_2 (-x)}{1+(2x+1)^2} \frac{1}{\pi^2 + \ln^2(x)}\, dx = \Re \frac{2}{\pi}\int_{0}^{\infty} e^{-\pi u} \int_{0}^{\infty} t \int_{0}^{\infty} \frac{-x^{i u +1}}{(e^t+x)(1+(2x+1)^2)} \, dx \, dt \, du$$ $$= \Re\frac{2}{\pi}\int_{0}^{\infty}e^{-\pi u}\int_{0}^{\infty}\frac{\pi t\,\text{csch} \,(\pi u)\left(2 i e^t e^{t i u}-2^{-i u/2}e^{-\pi u/4}\left(-1+(1+i)e^t+e^{\pi u/2}(1-(1-i)e^t)\right)\right)}{4(1-2e^t+2e^{2t})}\,dt\,du$$ We can separate this integral, take the real part and then interchange the order of integration to give $$-2\color{red}{\overbrace{\int_{0}^{\infty}\frac{te^t}{1-2e^t+2e^{2t}}\int_{0}^{\infty}\frac{\sin(tu)}{e^{2\pi u}-1}\, du \,dt}^{J}} -\color{green}{\overbrace{\int_{0}^{\infty}\frac{te^t}{1-2e^t+2e^{2t}}\int_{0}^{\infty}\frac{\sin(\frac{u}{2}\ln(2))(e^{\pi u/4}+e^{-\pi u/4})}{e^{2\pi u}-1}\,du\,dt}^{K}}+\color{blue}{\underbrace{\int_{0}^{\infty}\frac{te^t-t}{1-2e^t+2e^{2t}}\int_{0}^{\infty}\frac{\cos(\frac{u}{2}\ln(2))(e^{\pi u/4}-e^{-\pi u/4})}{e^{2\pi u}-1}\,du\,dt}_{L}}$$

We begin by evaluating $\color{red}{J}$:

Recall $$\frac{1}{e^{2\pi u}-1} = \sum_{n=1}^{\infty} e^{-2 \pi n u}$$ By the Mittag-Leffler series of $\cot$ $$ \implies \int_{0}^{\infty} \frac{\sin (t u)}{e^{2\pi u}-1} \, du = \frac{t \coth \left(\frac{t}{2}\right)-2}{4t}$$ $$\implies J = \int_{0}^{\infty}\frac{te^t}{1-2e^t+2e^{2t}}\frac{t \coth \left(\frac{t}{2}\right)-2}{4t} \,dt = \frac{\pi\ln(2)}{32}-\frac{G}{4}-\frac{\pi}{8}+\frac{11\pi^2}{192}+\frac{\ln^2(2)}{16}$$

We evaluate $\color{green}{K}$ next:

Recall $$\frac{1}{e^{2\pi u}-1} = \sum_{n=1}^{\infty} e^{-2 \pi n u}$$ $$\implies K= \int_{0}^{\infty}\frac{te^t}{1-2e^t+2e^{2t}} \sum_{n=1}^{\infty} \int_{0}^{\infty}\sin \left(\frac{u}{2}\ln(2)\right)(e^{\pi u/4}+e^{-\pi u/4}) e^{-2 \pi n u}\,du\,dt$$ It can be fairly trivially shown that $$\sum_{n=1}^{\infty} \int_{0}^{\infty}\sin \left(\frac{u}{2}\ln(2)\right)(e^{\pi u/4}+e^{-\pi u/4}) e^{-2 \pi n u}\,du = \frac{1}{2}-\frac{8\ln(2)}{\pi^2+4\ln^2(2)}$$ $$\int_{0}^{\infty}\frac{te^t}{1-2e^t+2e^{2t}} \, dt = G-\frac{\pi\ln(2)}{8}$$ $$\implies K = \frac{G}{2}-\frac{8G\ln(2)}{\pi^2+4\ln^2(2)}+\frac{\pi\ln^2(2)}{\pi^2+4\ln^2(2)}-\frac{\pi\ln(2)}{16}$$

With a similar initial idea, we evaluate $\color{blue}{L}$ now:

Recall once more $$\frac{1}{e^{2\pi u}-1} = \sum_{n=1}^{\infty} e^{-2 \pi n u}$$ $$\implies L= \int_{0}^{\infty}\frac{te^t-t}{1-2e^t+2e^{2t}} \sum_{n=1}^{\infty} \int_{0}^{\infty}\cos \left(\frac{u}{2}\ln(2)\right)(e^{\pi u/4}-e^{-\pi u/4}) e^{-2 \pi n u}\,du\,dt$$

Again, it can be fairly trivially shown that

$$\sum_{n=1}^{\infty} \int_{0}^{\infty}\cos \left(\frac{u}{2}\ln(2)\right)(e^{\pi u/4}-e^{-\pi u/4}) e^{-2 \pi n u}\,du = \frac{4\pi}{\pi^2+4\ln^2(2)}-1$$ $$\int_{0}^{\infty}\frac{te^t-t}{1-2e^t+2e^{2t}} \, dt = \frac{5\pi^2}{96}-\frac{\ln^2(2)}{8}$$

$$\implies L = \frac{\pi^3}{3\left(\pi^2+4\ln^2(2)\right)}-\frac{\pi}{8}-\frac{5\pi^2}{96}+\frac{\ln^2(2)}{8}$$

Finally, we have then

$$I = -2 \color{red}{\left(\frac{\pi\ln(2)}{32}-\frac{G}{4}-\frac{\pi}{8}+\frac{11\pi^2}{192}+\frac{\ln^2(2)}{16}\right)} - \color{green}{\left(\frac{G}{2}-\frac{8G\ln(2)}{\pi^2+4\ln^2(2)}+\frac{\pi\ln^2(2)}{\pi^2+4\ln^2(2)}-\frac{\pi\ln(2)}{16}\right)} + \color{blue}{\left(\frac{\pi^3}{3\left(\pi^2+4\ln^2(2)\right)}-\frac{\pi}{8}-\frac{5\pi^2}{96}+\frac{\ln^2(2)}{8}\right)} = \frac{\left(96G\ln(2)+7\pi^3\right)}{12(\pi^2+4\ln^2(2))}-\frac{\pi}{24}(3+4\pi)$$ $\square$

Answer 2

$$I=2\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\pi ^2+\ln ^2x}}\mathrm{d}x$$ Consider $$f\left( z \right) =\frac{\mathrm{Li}_2\left( z \right)}{1+\left( 2z-1 \right) ^2}\frac{1}{\ln z}$$ Use a key-shaped contour with $0$ and $1$ as the keyholes: contour

When $x<0$, we have $$\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\ln x+\pi \mathrm{i}}}\mathrm{d}x-\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\ln x-\pi \mathrm{i}}}\mathrm{d}x\\=-2\pi \mathrm{i}\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\ln ^2x+\pi ^2}}\mathrm{d}x$$ When $x>1$, we have $$\int_1^{\infty}{\frac{1}{1+\left( 2x-1 \right) ^2}\frac{1}{\ln x}\left( \frac{\pi ^2}{6}-\int_1^x{\frac{\ln \left( t-1 \right)}{t}}\mathrm{d}t+\pi \mathrm{i}\ln x \right)}\mathrm{d}x \\ -\int_1^{\infty}{\frac{1}{1+\left( 2x-1 \right) ^2}\frac{1}{\ln x}\left( \frac{\pi ^2}{6}-\int_1^x{\frac{\ln \left( t-1 \right)}{t}}\mathrm{d}t-\pi \mathrm{i}\ln x \right)}\mathrm{d}x \\ =2\pi \mathrm{i}\int_1^{\infty}{\frac{1}{1+\left( 2x-1 \right) ^2}}\mathrm{d}x=\frac{\pi ^2\mathrm{i}}{4}$$ This is because $$\mathrm{Li}_2\left( x \right) =-\int_0^x{\frac{\ln \left( 1-t \right)}{t}}\mathrm{d}t=\frac{\pi ^2}{6}-\int_1^x{\frac{\ln \left( 1-t \right)}{t}}\mathrm{d}t \\ =\frac{\pi ^2}{6}-\int_1^x{\frac{\ln \left( t-1 \right)}{t}}\mathrm{d}t\pm \pi \mathrm{i}\ln x\,\,\left( x>1 \right) $$ Then, use the residue theorem, we get $$-2\pi \mathrm{i}\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\ln ^2x+\pi ^2}}\mathrm{d}x+\frac{\pi ^2\mathrm{i}}{4}=2\pi \mathrm{i}\left[ \mathrm{Res}\left[ f\left( z \right) ,1 \right] +\mathrm{Res}\left[ f\left( z \right) ,\frac{1\pm \mathrm{i}}{2} \right] \right] $$ And $$\begin{align*}&\mathrm{Res}\left[ f\left( z \right) ,1 \right] =\frac{\pi ^2}{12} \\ &\mathrm{Res}\left[ f\left( z \right) ,\frac{1-\mathrm{i}}{2} \right] =\frac{\mathrm{i}}{4}\frac{\mathrm{Li}_2\left( \frac{1-\mathrm{i}}{2} \right)}{\ln \left( \frac{1-\mathrm{i}}{2} \right)}=\frac{\mathrm{i}}{4}\frac{\frac{5\pi ^2}{96}-\frac{\ln ^22}{8}+\mathrm{i}\left( \frac{\pi \ln 2}{8}-C \right)}{-\frac{\ln 2}{2}-\frac{\pi \mathrm{i}}{4}} \\ &\mathrm{Res}\left[ f\left( z \right) ,\frac{1+\mathrm{i}}{2} \right] =-\frac{\mathrm{i}}{4}\frac{\mathrm{Li}_2\left( \frac{1+\mathrm{i}}{2} \right)}{\ln \left( \frac{1+\mathrm{i}}{2} \right)}=-\frac{\mathrm{i}}{4}\frac{\frac{5\pi ^2}{96}-\frac{\ln ^22}{8}-\mathrm{i}\left( \frac{\pi \ln 2}{8}-C \right)}{-\frac{\ln 2}{2}+\frac{\pi \mathrm{i}}{4}}\end{align*}$$ So $$\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\pi ^2+\ln ^2x}}\mathrm{d}x=\frac{7\pi ^3}{24\left( \pi ^2+4\ln ^22 \right)}+\frac{4C\ln 2}{\pi ^2+4\ln ^22}-\frac{\pi ^2}{12}-\frac{\pi}{16}$$ Finally $$\boxed{I=\frac{7\pi ^3}{12\left( \pi ^2+4\ln ^22 \right)}+\frac{8C\ln 2}{\pi ^2+4\ln ^22}-\frac{\pi ^2}{6}-\frac{\pi}{8}}$$