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

Question

How to show that

$$\int_0^1\frac{\ln^2(1+x)+2\ln(x)\ln(1+x^2)}{1+x^2}dx=\frac{5\pi^3}{64}+\frac{\pi}{16}\ln^2(2)-4\,\text{G}\ln(2)$$

without breaking up the integrand since we already know:

$$\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx=4\,\mathfrak{J}\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^2(2)-2\,\text{G}\ln(2)$$

and

$$\int_0^1\frac{\ln(x)\ln(1+x^2)}{1+x^2}dx=-2\,\Im\operatorname{Li_3}(1+i)+\frac{3\pi^3}{32}+\frac{\pi}8\ln^2(2)-\text{G}\ln(2).$$

We can see that the imaginary parts got cancelled out leaving us only a real value and this fact pushed me to propose such a question. I tried integration by parts and subbing $x\to (1-x)/(1+x)$ but no use.

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The two integrals are given in (here) and (here) respectively.

Answer 1

Evaluate \begin{align} &\int_0^1\frac{2\ln y\tan^{-1}y}{1-y^2}dy = \int_0^1 \ln y \tan^{-1}y\ d\left(\ln \frac{1+y}{1-y}\right) \\ =& \int_0^1 \frac{\ln y\ln \frac{1-y}{1+y}}{1+y^2}dy +\int_0^1 \frac{\tan^{-1}y \ln \frac{1-y}{1+y}}{y} \overset{\frac{1-y}{1+y}\to y}{dy} = \frac12\int_0^1 \frac{\ln y\ln \frac{1-y}{1+y}}{1+y^2}dy -\frac{\pi^3}{32} \end{align} and \begin{align} &\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}dx =\int_0^1\frac{\ln x}{1+x^2}\left(\int_0^1 \frac{2x^2y}{1+x^2y^2}dy\right) dx\\ =& \ \int_0^1\int_0^1\frac{2y}{1-y^2}\bigg( \overset{t=xy}{\frac{\ln xy}{1+x^2y^2}}-\frac{\ln x}{1+x^2} - \frac{\ln y}{1+x^2y^2} \bigg)dxdy\\ =&\int_0^1 \frac2{1-y^2}\left(\int_1^y\frac{\ln t}{1+t^2}dt\right)\overset{ibp}{dy} +2\ln2\int_0^1\frac{\ln y}{1+y^2}dy -\int_0^1\frac{2\ln y\tan^{-1}y}{1-y^2}dy\\ = & \ \frac12 \int_0^1 \frac{\ln y\ln \frac{1-y}{1+y}}{1+y^2}dy - 2\ln2 G+\frac{\pi^3}{32} \\ \end{align} Then, express the original integral as \begin{align} I=&\int_0^1\frac{\ln^2(1+x)+2\ln x\ln(1+x^2)}{1+x^2}dx\\ =& \int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx + \int_0^1 \frac{\ln x\ln \frac{1-x}{1+x}}{1+x^2}dx - 4\ln2 G+\frac{\pi^3}{16} \tag1\\ \end{align}

Note that \begin{align} &\int_0^1 \frac{2\ln x\ln(1-x)}{1+x^2}dx\\ =&\ \int_0^1\frac{\ln^2(1-x)}{1+x^2}dx -\int_0^1\frac{\ln^2\frac x{1-x}}{1+x^2}\overset{\frac x{1-x} \to x}{dx }+\int_0^1\frac{\ln^2x}{1+x^2}dx\\ =&\int_0^1\frac{\ln^2(1-x)}{1+x^2}dx -\int_0^\infty \frac{\ln^2x}{1+(x+1)^2}dx+\frac{\pi^3}{16}\tag2\\ \\ &\int_0^\infty \frac{\ln^2x}{1+(x-1)^2}dx\\ = &\int_0^1\frac{\ln^2(1-x)}{1+x^2}dx + \int_0^1\frac{\ln^2(1+x)}{1+x^2} dx +\int_1^\infty \frac{\ln^2(1+x)}{1+x^2} \overset{x\to 1/x}{dx}\\ =& \int_0^1\frac{\ln^2(1-x)}{1+x^2}dx +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+\frac{\pi^3}{16}\tag3 \\ \end{align}

and take (2) - (3) to get \begin{align} & \int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx +\int_0^1 \frac{\ln x\ln \frac{1-x}{1+x}}{1+x^2}dx \\ =& \ \frac12 \int_0^\infty \frac{\ln^2x}{1+(x-1)^2}dx - \frac12\int_0^\infty \frac{\ln^2x}{1+(x+1)^2}dx =\int_0^\infty {\frac{2x \ln^2 x}{4+x^4}}\overset{x^2\to x} {dx}\tag4\\ \end{align}

Substitute (4) into (1) to obtain

\begin{align} I= \frac14\int_0^\infty {\frac{\ln^2 x}{4+x^2}} {dx} - 4\ln2 G+\frac{\pi^3}{16} = \frac{5\pi^3}{64}+\frac{\pi}{16}\ln^22-4\ln2G \end{align}