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How to prove $$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx=\text{Im}\left(\operatorname{Li}_3(1+i)\right)-\frac{\pi^3}{32}-G\ln2 \ ?$$ where $\operatorname{Li}_3(x)=\sum\limits_{n=1}^\infty\frac{x^n}{n^3}$ is the trilogarithm and $G=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}$ is Catalan's constant

Trying the algebraic identity $\ 4ab=(a+b)^2-(a-b)^2\ $ where $\ a=\ln(1-x)$ and $b=\ln(1+x)\ $is not helpful here and the integral will be more complicated.

Also, applying IBP or substituting $x=\frac{1-y}{1+y}$ is not that useful either.

All approaches are appreciated.

Ali Shadhar
  • 25,498

3 Answers3

8

Applying the algebraic identity $2ab=a^2+b^2-(a-b)^2$ gives us: $$2I=2\int_0^1 \frac{\ln(1-x)\ln(1+x)}{1+x^2}dx$$ $$=\color{red}{\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx}+\color{blue}{\int_0^1 \frac{\ln^2(1+x)}{1+x^2}}-\color{purple}{\int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right)}{1+x^2}dx}$$

For the third integral, set $\frac{1-x}{1+x}= t$ to get: $$\color{purple}{\int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right)}{1+x^2}dx}=\int_0^1 \frac{\ln^2 t}{1+t^2}dt=\sum_{n=0}^\infty (-1)^n \int_0^1 t^{2n} \ln^2 tdt=2\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}=\color{purple}{\frac{\pi^3}{16}}$$ The second integral reminds us of the following result: $$\int_0^\infty\frac{\ln^2(1+x)}{1+x^2} dx=2\Im\operatorname{Li}_3(1+i)$$ $$\Rightarrow J=\color{blue}{\int_0^1 \frac{\ln^2(1+x)}{1+x^2} dx}=2\Im\operatorname{Li}_3(1+i)-\int_1^\infty \frac{\ln^2(1+x)}{1+x^2} dx $$ Furthermore, let $x=\frac{1+u}{1-u}$ to see that: $$\int_1^\infty \frac{\ln^2(1+x)}{1+x^2} dx=\int_0^1 \frac{\ln^2 2 -2\ln 2\ln(1-u)+\ln^2(1-u)}{1+u^2}du$$ $$=\frac{\pi}{4}\ln^2 2-2\ln 2 {\int_0^1 \frac{\ln(1-u)}{1+u^2}du}+\int_0^1 \frac{\ln^2(1-u)}{1+u^2}du$$ $$\int_0^1 \frac{\ln(1-u)}{1+u^2}du=\underbrace{\int_0^1 \frac{\ln\left(\frac{1}{u}-1\right)}{1+u^2}du}_{=K}+\int_0^1 \frac{\ln u}{1+u^2}du$$ Now for $K$ let $u=\frac{1-x}{1+x}$ to get: $$K=\int_0^1 \frac{\ln\left(\frac{1}{u}-1\right)}{1+u^2}du=\int_0^1 \frac{\ln 2-\ln\left(\frac{1}{x}-1\right)}{1+x^2}dx$$ $$\Rightarrow 2K=\ln 2 \int_0^1 \frac{1}{1+x^2}dx \Rightarrow K=\frac{\pi}{8}\ln 2$$ $$\int_0^1 \frac{\ln u}{1+u^2}dt=\sum_{n=0}^\infty (-1)^n \int_0^1 u^{2n} \ln udu=-\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}=-G$$ $$\Rightarrow J=\color{blue}{2\Im \operatorname{Li}_3(1+i)-2G\ln 2 -\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx}$$

Plugging the above results into the original integral yields: $$\require{cancel} 2I=\color{red}{\cancel{\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx}}+\color{blue}{2\Im \operatorname{Li}_3(1+i)-2G\ln 2 -\cancel{\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx}}-\color{purple}{\frac{\pi^3}{16}}$$ $$\Rightarrow I=\boxed{\int_0^1 \frac{\ln(1-x)\ln(1+x)}{1+x^2}dx=\Im \operatorname{Li}_3(1+i)-\frac{\pi^3}{32}-G\ln 2}$$

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Zacky
  • 27,674
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lets start with $\displaystyle\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx=2\Im\operatorname{Li}_3(1+i)\quad$ (proved here)

\begin{align} \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx&=\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ d+\underbrace{\int_1^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx}_{\small\displaystyle x\mapsto1/x}\\ 2\Im\operatorname{Li}_3(1+i)&=2\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx-2\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}\ dx+\underbrace{\int_0^1\frac{\ln^2x}{1+x^2}\ dx}_{2\beta(3)} \end{align} then

$$\int_0^1\frac{\ln^2(1+x)-\ln x\ln(1+x)}{1+x^2}\ dx=\Im\operatorname{Li}_3(1+i)-\beta(3)\tag{1}$$

now lets start with $\ I=\displaystyle\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx$ and by setting $x=\frac{1-y}{1+y}$, we get $$I=\displaystyle\int_0^1\frac{\ln^2(1+x)-\ln x\ln(1+x)}{1+x^2}-\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx+\ln2\underbrace{\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}\ dx}_{x=(1-y)/(1+y)}+I$$ then

\begin{align} \int_0^1\frac{\ln^2(1+x)-\ln x\ln(1+x)}{1+x^2}=\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx-\ln2\underbrace{\int_0^1\frac{\ln x}{1+x^2}}_{-G}\tag{2} \end{align}

from $(1)$ and $(2)$ and substituting $\displaystyle\beta(3)=\frac{\pi^3}{32}\ $, the result follows.

Ali Shadhar
  • 25,498
1

Different approach:

Start with subbing $x\mapsto \frac{1-x}{1+x}$

$$\small{\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}dx=\ln2\underbrace{\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}dx}_{-G}-\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx+\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx}\tag1$$

where

$$\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx=\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx-\underbrace{\int_1^\infty\frac{\ln^2(1+x)}{1+x^2}dx}_{x\mapsto 1/x}$$

$$=\underbrace{\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx}_{2\ \text{Im}\operatorname{Li}_3(1+i)}-\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx+2\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx-\underbrace{\int_0^1\frac{\ln^2x}{1+x^2}dx}_{\pi^3/16}$$

$$\Longrightarrow \int_0^1\frac{\ln^2(1+x)}{1+x^2}dx=\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx+\text{Im}\operatorname{Li}_3(1+i)-\frac{\pi^3}{32}\tag2$$

Plug $(2)$ in $(1)$ we obtain

$$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx=\text{Im}\left(\operatorname{Li}_3(1+i)\right)-\frac{\pi^3}{32}-G\ln2$$

Ali Shadhar
  • 25,498