How to show $\int_0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx=\ln(2)G$

Question

I am trying to prove that $$\int_0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx=\ln(2)G,$$ where $G$ is the Catalan constant and $\operatorname{Li}_2(x)$ is the dilogarithm function, which is proposed by my friend Sujeethan Balendran and solved in two methods by my friend Cornel Valean in his second book "More (Almost) Impossible Integrals, Sums, and Series" page 345. I want to try it differently using as less results and relations as possible and here is my attempt:

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$$\int_0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx=\int_0^1\frac{1}{1+x^2}\left(\int_0^1\frac{-\frac{1+x^2}{2}\ln(y)}{1-\frac{1+x^2}{2}y}dy\right)dx$$

$$=\int_0^1 \ln(y)\left(\int_0^1\frac{dx}{(1+x^2)y-2}\right)dy$$

$$=\int_0^1\ln(y)\left(\frac{1}{y}\sqrt{\frac{y}{y-2}}\arctan \sqrt{\frac{y}{y-2}}\right)dy\quad \text{let} \quad\sqrt{\frac{y}{y-2}}=ix$$

$$=\int_0^1\frac{-2\operatorname{arctanh}(x)\ln\left(\frac{2x^2}{1+x^2}\right)}{1+x^2}dx$$

$$=\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2x^2}{1+x^2}\right)}{1+x^2}dx\quad \text{let} \quad\frac{1-x}{1+x}\to x$$

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

$$=2\int_0^1\frac{\ln(x)\ln(1-x)}{1+x^2}dx-\int_0^1\frac{\ln(x)\ln(1+x^2)}{1+x^2}dx$$

These two integrals are given here and here respectively and combining them gives $\ln(2)G.$

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Question: Is there an elegant way to prove $\displaystyle\int_0^1\frac{\ln(x)\ln\left(\frac{(1-x)^2}{1+x^2}\right)}{1+x^2}dx=\ln(2)G$ without breaking the integrand or is there a different way to evaluate $\displaystyle\int_0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx$?

Thank you,

Answer 1

Substitute $x=\tan t$ to rewrite the integral as

$$\int_0^1\frac{\ln x\ln\frac{(1-x)^2}{1+x^2}}{1+x^2}\ dx = 2\int_0^{\frac\pi4}\ln(\tan t)\ln(\cos t-\sin t)\ dt $$ where the transformed integral is evaluated elementarily here.

Answer 2

Solution by Sujeethan Balendran:

By the change of variable $x=\tan\left(\frac{t}{2}\right)$, we have $$\int_0^1\frac{\ln(x)}{1+x^2}\ln\left(\frac{1+x}{1-x}\right)dx=\frac12\int_0^{\pi/2}\ln\left(\tan\frac{x}{2}\right)\ln\left(\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)dx$$

use the Fourier series of $\ln(\tan x)$

$$=\frac12\int_0^{\pi/2}\ln\left(\tan\frac{x}{2}\right)\left(2\sum_{n=1}^\infty\frac{(-1)^{n-1}\sin((2n-1)x)}{2n-1}\right)dx$$

$$=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2n-1}\left(\int_0^{\pi/2}\sin((2n-1)x)\ln\left(\tan\frac{x}{2}\right)dx\right)$$

$$=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2n-1}\left(\int_0^{\pi/2} d\left(\frac{1-\cos((2n-1)x)}{2n-1}\right)\ln\left(\tan\frac{x}{2}\right)dx\right)$$

$$=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2n-1}\left( \frac{1-\cos((2n-1)x)}{2n-1}\ln\left(\tan\frac{x}{2}\right)\bigg|_0^{\pi/2}-2\int_0^{\pi/2}\frac{\sin^2\left(\frac{(2n-1)x}{2}\right)}{(2n-1)\sin x}dx\right)$$

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

shift the index then separate the first term

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

$$=-\ln(2)+2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n+1)^2} \left(\frac{H_n+\overline{H}_n+\ln(2)}{2}\right)$$

$$=-\ln(2)+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n+1)^2} \left(H_n+\overline{H}_n\right)+\ln(2)\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n+1)^2}$$

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

$$=\ln(2)\sum_{n=0}^\infty\frac{(-1)^{n-1}}{(2n+1)^2}-\int_0^1 \ln(x)\left(\sum_{n=1}^\infty(-1)^{n-1}H_n x^{2n}+\sum_{n=1}^\infty(-1)^{n-1}\overline{H}_n x^{2n}\right)dx$$

$$=-\ln(2)G-\int_0^1 \ln(x)\left(\frac{\ln(1+x^2)}{1+x^2}-\frac{\ln(1-x^2)}{1+x^2}\right)dx$$

$$=-\ln(2)G+\int_0^1\frac{\ln(x)}{1+x^2}\ln\left(\frac{1-x^2}{1+x^2}\right)dx$$

So we have

$$\int_0^1\frac{\ln(x)}{1+x^2}\ln\left(\frac{1+x}{1-x}\right)dx=-\ln(2)G+\int_0^1\frac{\ln(x)}{1+x^2}\ln\left(\frac{1-x^2}{1+x^2}\right)dx$$

or

$$\int_0^1\frac{\ln(x)}{1+x^2}\ln\left(\frac{(1-x)^2}{1+x^2}\right)dx=\ln(2)G$$

Note: $\overline{H}_n$ is defined as $\sum_{k=1}^n\frac{(-1)^{k-1}}k$