TITLE: Integral $\int_0^1\frac{\ln x}{x^2+1}\cdot\ln\left(\frac{3\,x^2+1}{x^2+3}\right)dx$
QUESTION [27 upvotes]: 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.

REPLY [9 votes]: $\newcommand{\+}{^{\dagger}}
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$\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}