Integral $\int_0^1\frac{\ln x}{x^2+1}\cdot\ln\left(\frac{3\,x^2+1}{x^2+3}\right)dx$

Question

I need to evaluate the following integral: $$\int_0^1\frac{\ln x}{x^2+1}\cdot\ln\left(\frac{3\,x^2+1}{x^2+3}\right)dx.$$ Could you suggest how to find a closed form for it? I am not sure if there is one, but the integrand seems simple enough, so I hope it might exist.

Answer 1

My calculation shows that

$$ \int_{0}^{1} \frac{\log x}{x^{2} + 1} \log \left( \frac{t^{2}x^{2} + 1}{x^{2} + t^{2}} \right) \, dx = - \pi \chi_{2} \left( \frac{1 - t}{1 + t} \right), \tag{1} $$

where $\chi_{2}$ stands for the Legendre chi function. I will post a detailed solution later, but I should note that the idea is very simple: denote this integral as $I(t)$ and differentiate it to obtain

$$ I'(t) = -\frac{\pi \log t}{1 - t^{2}}. $$

Restricting $|t| < 1$ temporarily, this gives

$$ I(t) = I(0) + \int_{0}^{t} I'(s) \, ds = -\pi \left\{ \frac{\pi^{2}}{8} + \log t \operatorname{artanh} t - \chi_{2}(t) \right\}. $$

Then $\text{(1)}$ restricted to $|t| < 1$ follows from the following identity:

$$ \chi_{2}\left( \frac{1-t}{1+t} \right) + \chi_{2}(t) = \frac{\pi^{2}}{8} + \log t \operatorname{artanh} t. $$

Then the equality for the general $t$ follows by analytic continuation. Here is a Mathematica code for testing this:

LegendreChi[n_, z_] := z/2^n LerchPhi[z^2, n, 1/2];

t = Sqrt[3];
NIntegrate[(Log[x] Log[(t^2 x^2 + 1)/(x^2 + t^2)])/(1 + x^2), {x, 0, 1},
    WorkingPrecision -> 100]
N[-Pi LegendreChi[2, (1 - t)/(1 + t)], 100]
Clear[t];

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\color{#c00000}{\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\, \ln\pars{3x^{2} + 1 \over x^{2} + 3}\,\dd x}:\ {\large ?}}$.

Note that \begin{align}&\color{#c00000}{\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\, \ln\pars{3x^{2} + 1 \over x^{2} + 3}\,\dd x} =\int_{\infty}^{1}{\ln\pars{1/x} \over \pars{1/x}^{2} + 1}\, \ln\pars{3/x^{2} + 1 \over 1/x^{2} + 3} \,\pars{-\,{\dd x \over x^{2}}} \\[3mm]&=\int_{1}^{\infty}{\ln\pars{x} \over x^{2} + 1}\, \ln\pars{3x^{2} + 1 \over x^{2} + 3}\,\dd x \\[3mm]&\mbox{such that}\ \begin{array}{|c|}\hline\\ \quad\color{#c00000}{\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\, \ln\pars{3x^{2} + 1 \over x^{2} + 3}\,\dd x} =\half\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\, \ln\pars{3x^{2} + 1 \over x^{2} + 3}\,\dd x\quad \\ \\ \hline \end{array} \\[3mm]&\mbox{Hereafter, we'll use repeatedly the well known result}\ \int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x = 0. \\&\mbox{It's trivially shown by splitting the integral in two integrals over}\ \pars{0,1}\ \mbox{and}\ \pars{0,\infty}. \end{align}

\begin{align}&\color{#c00000}{\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\, \ln\pars{3x^{2} + 1 \over x^{2} + 3}\,\dd x} =\half\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\, \int_{3}^{^{1}/_{3}}{\dd t \over t + x^{2}} \\[3mm]&=\half\int_{3}^{^{1}/_{3}}\int_{0}^{\infty}\ln\pars{x} \pars{{1 \over x^{2} + 1} - {1 \over x^{2} + t}}\,{1 \over t - 1}\,\dd x\,\dd t \\[3mm]&=-\,\half\int_{3}^{^{1}/_{3}}{1 \over t - 1}\int_{0}^{\infty} {\ln\pars{x} \over x^{2} + t}\,\dd x\,\dd t =-\,\half\int_{3}^{^{1}/_{3}}{1 \over t - 1}\,{1 \over \root{t}} \int_{0}^{\infty}{\ln\pars{\root{t}x} \over x^{2} + 1}\,\dd x\,\dd t \\[3mm]&=-\,{\pi \over 4}\int_{3}^{^{1}/_{3}} {\ln\pars{\root{t}} \over \pars{t - 1}\root{t}}\,\dd t =-\,{\pi \over 2}\int_{\root{3}}^{1/\root{3}}{\ln\pars{t} \over t^{2} - 1}\,\dd t \\[3mm]&={\pi \over 2}\bracks{\int_{1/\root{3}}^{1} {\ln\pars{t} \over t^{2} - 1}\,\dd t -\int_{\root{3}}^{1}{\ln\pars{t} \over t^{2} - 1}\,\dd t} \\[3mm]&={\pi \over 2}\bracks{\int_{1/\root{3}}^{1} {\ln\pars{t} \over t^{2} - 1}\,\dd t -\int_{1/\root{3}}^{1}{\ln\pars{1/t} \over 1/t^{2} - 1}\, \pars{-\,{\dd t \over t^{2}}}} \end{align}

The original integral is reduced to: $$ \color{#c00000}{\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\, \ln\pars{3x^{2} + 1 \over x^{2} + 3}\,\dd x} =-\pi\int_{1/\root{3}}^{1}{\ln\pars{t} \over 1 - t^{2}}\,\dd t $$

Then, \begin{align}&\color{#c00000}{\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\, \ln\pars{3x^{2} + 1 \over x^{2} + 3}\,\dd x} =-\,{\pi \over 2}\int_{1/\root{3}}^{1}{\ln\pars{t} \over 1 - t}\,\dd t -{\pi \over 2}\int_{1/\root{3}}^{1}{\ln\pars{t} \over 1 + t}\,\dd t \\[3mm]&=-\,{\pi \over 2}\int_{0}^{1 - 1/\root{3}}{\ln\pars{1 - t} \over t}\,\dd t +{\pi \over 2}\int_{-1/\root{3}}^{-1}{\ln\pars{-t} \over 1 - t}\,\dd t \\[3mm]&=-\,{\pi \over 2}\int_{0}^{1 - 1/\root{3}}{\ln\pars{1 - t} \over t}\,\dd t -{\pi \over 2}\bracks{\vphantom{\LARGE A}% \ln\pars{1 - t}\ln\pars{-t}}_{-1/\root{3}}^{-1} +{\pi \over 2}\int_{-1/\root{3}}^{-1}{\ln\pars{1 - t} \over t}\,\dd t \end{align}

\begin{align}&\color{#c00000}{\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\, \ln\pars{3x^{2} + 1 \over x^{2} + 3}\,\dd x} \\[3mm]&=-\,{\pi \over 2}\bracks{% \half\,\ln\pars{3}\ln\pars{1 + {1 \over \root{3}}} +\int_{0}^{1 - 1/\root{3}}\!\!\!{\ln\pars{1 - t} \over t}\,\dd t +\int^{-1/\root{3}}_{-1}{\ln\pars{1 - t} \over t}\,\dd t} \end{align}

However, with $\ds{a < b < 1}$: \begin{align} \int_{a}^{b}{\ln\pars{1 - t} \over t}\,\dd t& =-\int_{a}^{b}{{\rm Li}_{1}\pars{t} \over t}\,\dd t =-\int_{a}^{b}{\rm Li}_{2}'\pars{t}\,\dd t ={\rm Li}_{2}\pars{a} - {\rm Li}_{2}\pars{b} \end{align} where $\ds{{\rm Li}_{\rm s}\pars{z}}$ is a PolyLogarithm Function. Hereafter, we'll use well known properties of them as reported in the cited link.

\begin{align}&\color{#66f}{\large\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\, \ln\pars{3x^{2} + 1 \over x^{2} + 3}\,\dd x} \\[3mm]&=\color{#66f}{\large-\,{\pi \over 2}\left\lbrack% \half\,\ln\pars{3}\ln\pars{1 + {1 \over \root{3}}} -{\rm Li}_{2}\pars{1 - {1 \over \root{3}}}\right.} \\[3mm]&\phantom{=-\,{\pi \over 2}\bracks{}}\color{#66f}{\large% \left.\mbox{}+\ \underbrace{{\rm Li}_{2}\pars{-1}}_{\ds{\color{#c00000}{-\,{\pi^{2} \over 12}}}}\ -\ {\rm Li}_{2}\pars{-\,{1 \over \root{3}}} \right\rbrack} \approx 0.8487 \end{align}