Background
I am trying to solve the following integral:
\begin{align*}
\int_{1}^{\infty} \frac{\log(x)}{x^2(x^2-1)} \, dx.
\end{align*}
This is a follow-up question from this post, and the indefinite integral of said integral was answered by Claude Leibovici with the following answer:
\begin{align*}
\int\frac{\log (x)}{x^2 \left(x^2-1\right)}\,dx = f(x) = -\frac{1}{2} [\text{Li}_2(1-x)+\text{Li}_2(-x)]-\frac{1}{2} \log (x+1) \log (x)+\frac{\log (x)}{x}+\frac{1}{x}.
\end{align*}
I am new to the dilogarithm, so I studied some of its identities, which are
\begin{align*}
&\mathrm{Li}_2(x)+\mathrm{Li}_2(-x)=\frac{1}{2} \mathrm{Li}_2(x^2),\\
&\mathrm{Li}_2(1-x)+\mathrm{Li}_2\left(1-\frac{1}{x}\right)=-\frac{\log^2(x)}{2},\\
&\mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)=\frac{\pi^2}{6}-\log(x)\log(1-x),\\
&\mathrm{Li}_2(-x)-\mathrm{Li}_2(1-x)+\frac{1}{2}\mathrm{Li}_2(1-x^2)=-\frac{\pi^2}{12}-\log(x)\log(x+1),\\
&\mathrm{Li}_2(x)+\mathrm{Li}_2\left(\frac{1}{x}\right)=-\frac{\pi^2}{6}-\frac{\log^2(-x)}{2}.
\end{align*}
I also know that
\begin{align*}
& \mathrm{Li}_2(0) = 0,\\
& \mathrm{Li}_2(1) = \frac{\pi^2}{6},\\
& \mathrm{Li}_2(-1) = -\frac{\pi^2}{12}.
\end{align*}
Question
So, I am now trying to prove that
\begin{align*}
\lim_{x \to \infty} f(x) = \frac{\pi^2}{6}.
\end{align*}
To be more precise, I am trying to show that
\begin{align}
\lim_{x \to \infty} \left[ \mathrm{Li}_2(1-x) + \mathrm{Li}_2(-x) + \log(x) \log(x+1) \right] = -\frac{\pi^2}{3}. \tag{1}
\end{align}
I have proved that
\begin{align*}
\lim_{x \to 1} f(x) = \frac{\pi^2}{24}+1,
\end{align*}
but I still can't figure out how to use the dilogarithm identities to prove equation (1). Can anyone help me with this problem? Any hints, answers, suggestions, or feedbacks are much appreciated. Thank you!
Best Answer
We can use these identities: \begin{align*} &\mathrm{Li}_2(x)+\mathrm{Li}_2(-x)=\frac{1}{2} \mathrm{Li}_2(x^2) \tag{1} \\ &\mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)=\frac{\pi^2}{6}-\log(x)\log(1-x)\tag{2} \\ &\mathrm{Li}_2(x)+\mathrm{Li}_2\left(\frac{1}{x}\right)=-\frac{\pi^2}{6}-\frac{\log^2(-x)}{2} \tag{3}. \end{align*}
We can suppose that $x>1$. Additionally, take the principal value of $\log(x)=\ln|x|+i\arg(x)$ i.e. $-\pi<\arg(x)\leq \pi$.
Then equations (3) and (2) become:
$$ \mathrm{Li}_2(x) + \mathrm{Li}_2\left(\frac{1}{x}\right) = \frac{\pi^2}{3}-\frac{1}{2}\ln^2(x)-i\pi\ln(x) \tag{4}$$
$$\mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)=\frac{\pi^2}{6}-\ln(x)\ln(x-1)-i\pi\ln(x) \tag{5}$$
Adding equations (1) and (5):
$$\mathrm{Li}_2(-x)+\mathrm{Li}_2(1-x)=-2\mathrm{Li}_2(x)+\frac{1}{2} \mathrm{Li}_2(x^2)+\frac{\pi^2}{6}-\ln(x)\ln(x-1)-i\pi\ln(x) \tag{6}$$
From equation (4):
$$\mathrm{Li}_2(x) = -\mathrm{Li}_2\left(\frac{1}{x}\right)+\frac{\pi^2}{3}-\frac{1}{2}\ln^2(x)-i\pi\ln(x) \tag{7} $$
$$\mathrm{Li}_2(x^2) = -\mathrm{Li}_2\left(\frac{1}{x^2}\right)+\frac{\pi^2}{3}-2\ln^2(x)-2i\pi\ln(x) \tag{8}$$
Substituting (7) and (8) in (6):
$$ \mathrm{Li}_2(-x) + \mathrm{Li}_2(1-x) = -\frac{1}{2}\mathrm{Li}_2\left(\frac{1}{x^2}\right)+2\mathrm{Li}_{2}\left(\frac{1}{x}\right)-\ln(x)\ln(x-1)-\frac{\pi^2}{3}$$
Hence, adding $\ln(x)\ln(x+1)$:
$$ \mathrm{Li}_2(-x) + \mathrm{Li}_2(1-x) + \ln(x)\ln(x+1) = -\frac{1}{2}\mathrm{Li}_2\left(\frac{1}{x^2}\right)+2\mathrm{Li}_{2}\left(\frac{1}{x}\right)+\ln(x)\ln\left(\frac{1+x}{x-1}\right)-\frac{\pi^2}{3}$$
Given that $\mathrm{Li}_{2}$ is continuous $\forall z\in\mathbb{C}\setminus(1,\infty)$ then $\displaystyle \mathrm{Li}_{2}\left(\frac{1}{x^2}\right),\mathrm{Li}_{2}\left(\frac{1}{x^2}\right) \to 0$ as $x \to \infty$
Additionally, due to the L'Hopital rule:
$$ \lim_{x\to \infty} \ln(x)\ln\left(\frac{1+x}{x-1}\right) = \lim_{x\to \infty} \frac{\ln\left(\frac{1+x}{x-1}\right)}{\frac{1}{\ln(x)}} = 0 $$
$$ \boxed{\lim_{x \to \infty}\left[\mathrm{Li}_2(-x) + \mathrm{Li}_2(1-x) + \ln(x)\ln(x+1) \right]= -\frac{\pi^2}{3}}$$