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

Question

$$I=\int_0^1 \frac{\tan^{-1}(x)\ln^2(x)}{1+x}\,dx$$

I have not been able to find any similar integrals to refer and solve this one.

$$\int_0^1 \frac{\tan^{-1}(x)\ln^2(x)}{1+x}\,dx\underset{ibp}=-\color{green}{\int_0^1\frac{\ln(x)\ln(1+x)}{1+x^2}\,dx}-2\color{red}{\int_0^1\frac{\tan^{-1}(x)\ln(x)\ln(1+x)}{x}\,dx}$$

Both the integrals seem to be hard to evaluate and going into some polylogarithmic forms.

The second integral with another round of IBP,

$$\int_0^1\frac{\tan^{-1}(x)\ln(x)\ln(1+x)}{x}\,dx=-\frac{3\pi\zeta(3)}{16}-\int_{0}^{1}\frac{\text{Li}_3(-x)}{1+x^2}\,dx+\color{blue}{\int_{0}^{1}\frac{\ln(x)\text{Li}_2(-x)}{1+x^2}\,dx}$$

Luckily, the first integral can be evaluated from here but the second integral was the dead-end.

Is there a better solution that avoids this path?

Also, I would appreciate any hints on solving those coloured integrals

Answer 1

\begin{align} I=&\int_0^1 \frac{\tan^{-1}x\ln^2 x}{1+x}\,dx\\ =& \int_0^1 \tan^{-1}x \ d\bigg(\int_0^x \frac{\ln^2 t}{1+t}dt\bigg) \overset{ibp}= \frac\pi4 \int_0^1 \frac{\ln^2 t}{1+t}dt-K\tag1\\ \end{align} where $\int_0^1 \frac{\ln^2 t}{1+t}dt=\frac32\zeta(3) $ \begin{align} K=& \int_0^1 \frac1{1+x^2}\int_0^x \frac{\ln^2 t}{1+t}\overset{t=xy}{dt} \ dx \\ =& \int_0^1 \int_0^1\frac{x\ln^2 xy}{(1+x^2)(1+xy)} dy\ dx\\ =& \int_0^1 \int_0^1 \frac{(x+y)\ln^2xy}{(1+y^2 )(1+x^2)}\overset{x\leftrightarrows y}{dx dy}-\int_0^1 \int_0^1\frac{y\ln^2 xy}{(1+y^2)(1+xy)}\overset{t=xy}{dx}dy\\ =& \int_0^1 \int_0^1 \frac{(x+x)\ln^2xy}{(1+y^2 )(1+x^2)}dx \ dy-K\\ =& \int_0^1 \int_0^1 \frac{x(\ln^2x+2\ln x\ln y+\ln^2y)}{(1+y^2 )(1+x^2)}dx \ dy\\ =&\int_0^1 \frac{x\ln^2x}{1+x^2}dx \int_0^1\frac1{1+y^2}dy +2 \int_0^1 \frac{x\ln x}{1+x^2}dx \int_0^1 \frac{\ln y}{1+y^2}dy\\ &\>\>\>\>\> +\int_0^1 \frac{x}{1+x^2}dx\int_0^1 \frac{\ln^2y}{1+y^2}dy \\ =&\ \frac\pi4\cdot \frac3{16}\zeta(3)+2\left(-\frac{\pi^2}{48}\right)(-G)+ \frac12\ln2 \cdot\frac{\pi^3}{16} \end{align} Substitute $K$ into (1) to obtain $$I= \frac{21\pi}{64}\zeta(3)-\frac{\pi^2}{24}G-\frac{\pi^3}{32}\ln2$$

Answer 2

A similar path, denote $$ P(x) = \int_{0}^{x} \frac{\ln^2\!t}{1+t}\,\mathrm{d}t = \int_{0}^{1} \frac{x\ln^2(xt)}{1+xt}\,\mathrm{d}t, \quad P(1) = \int_{0}^{1} \frac{\ln^2\!t}{1+t}\,\mathrm{d}t = \frac{3\zeta(3)}{2} $$ hence $$ \begin{aligned} I &= P(x)\arctan x\bigr|_{x=0}^{1} - \int_{0}^{1} \frac{P(x)}{1+x^2}\,\mathrm{d}x \\ &=\frac{3\pi}{8}\zeta(3) - \int_{0}^{1} \int_{0}^{1} \frac{x\ln^2(xt)}{(1+xt)(1+x^2)}\,\mathrm{d}t\mathrm{d}x \\ &=\frac{3\pi}{8}\zeta(3) - \int_{0}^{1} \frac{\ln^2\!x}{1+x^2} \left(\int_{0}^{1} \frac{x}{1+xt}\,\mathrm{d}t\right)\mathrm{d}x - 2\int_{0}^{1} \int_{0}^{1} \frac{x\ln x\ln t}{(1+xt)(1+x^2)}\,\mathrm{d}t\mathrm{d}x\\ &\quad\quad -\int_{0}^{1} \ln^2\!t \left(\int_{0}^{1} \frac{x}{(1+xt)(1+x^2)}\,\mathrm{d}x\right)\mathrm{d}t \end{aligned} $$ where the first one $$ \int_{0}^{1} \frac{\ln^2\!x}{1+x^2} \left(\int_{0}^{1} \frac{x}{1+xt}\,\mathrm{d}t\right)\mathrm{d}x = \int_{0}^{1} \frac{\ln^2\!x\ln(1+x)}{1+x^2} \,\mathrm{d}x $$ the second one is what your question focus $$ \begin{aligned} \int_{0}^{1} \int_{0}^{1} \frac{x\ln x\ln t}{(1+xt)(1+x^2)}\,\mathrm{d}t\mathrm{d}x &= \int_{0}^{1} \frac{\ln x}{1+x^2}\left(\int_{0}^{1} \frac{x\ln t}{1+xt}\,\mathrm{d}t\right)\mathrm{d}x \\ &= \int_{0}^{1} \frac{\ln x}{1+x^2}\left(-\int_{0}^{1} \frac{\ln(1+xt)}{t}\,\mathrm{d}t\right)\mathrm{d}x \\ &= \color{blue}{\int_{0}^{1} \frac{\ln x\operatorname{Li}_2(-x)}{1+x^2}\,\mathrm{d}x} \end{aligned} $$ on the other hand $$ \begin{aligned} &\int_{0}^{1} \int_{0}^{1} \frac{x\ln x\ln t}{(1+xt)(1+x^2)}\,\mathrm{d}t\mathrm{d}x \\ =& \int_{0}^{1} {\ln t}\left(\int_{0}^{1} \frac{x\ln x}{(1+xt)(1+x^2)}\,\mathrm{d}x\right)\mathrm{d}t \\ =& \int_{0}^{1} \frac{\ln t}{1+t^2} \left(\int_{0}^{1} \left(\frac{t+x}{1+x^2}-\frac{t}{1+xt}\right)\ln x\,\mathrm{d}x\right)\mathrm{d}t \\ =& \int_{0}^{1} \frac{t\ln t}{1+t^2} \,\mathrm{d}t\int_{0}^{1} \frac{\ln x}{1+x^2}\,\mathrm{d}x + \int_{0}^{1} \frac{\ln t}{1+t^2} \,\mathrm{d}t\int_{0}^{1} \frac{x\ln x}{1+x^2}\,\mathrm{d}x - \color{blue}{\int_{0}^{1} \frac{\ln t\operatorname{Li}_2(-t)}{1+t^2}\,\mathrm{d}t} \end{aligned} $$ which shows $$ \int_{0}^{1} \int_{0}^{1} \frac{x\ln x\ln t}{(1+xt)(1+x^2)}\,\mathrm{d}t\mathrm{d}x = \color{blue}{\int_{0}^{1} \frac{\ln x\operatorname{Li}_2(-x)}{1+x^2}\,\mathrm{d}x} = \int_{0}^{1} \frac{t\ln t}{1+t^2} \,\mathrm{d}t\int_{0}^{1} \frac{\ln x}{1+x^2}\,\mathrm{d}x $$ at last, the third one $$ \begin{aligned} &\int_{0}^{1} \ln^2\!t \left(\int_{0}^{1} \frac{x}{(1+xt)(1+x^2)}\,\mathrm{d}x\right)\mathrm{d}t \\ = &\int_{0}^{1} \frac{\ln^2\!t}{1+t^2} \left(\int_{0}^{1} \left(\frac{t+x}{1+x^2}-\frac{t}{1+xt}\right)\mathrm{d}x\right)\mathrm{d}t \\ = &\frac{\pi}{4}\int_{0}^{1} \frac{t\ln^2\!t}{1+t^2} \,\mathrm{d}t + \frac{\ln2}{2}\int_{0}^{1} \frac{\ln^2\!t}{1+t^2} \,\mathrm{d}t - \int_{0}^{1} \frac{\ln^2\!t\ln(1+t)}{1+t^2}\,\mathrm{d}t \end{aligned} $$ notice $\int_{0}^{1} \frac{\ln^2\!t\ln(1+t)}{1+t^2}\,\mathrm{d}t$ should be canceled out, thus $$ \begin{aligned} I &= \frac{3\pi}{8}\zeta(3) - \frac{\pi}{4}\int_{0}^{1} \frac{t\ln^2\!t}{1+t^2} \,\mathrm{d}t - \frac{\ln2}{2}\int_{0}^{1} \frac{\ln^2\!t}{1+t^2} \,\mathrm{d}t - 2\int_{0}^{1} \frac{t\ln t}{1+t^2} \,\mathrm{d}t\int_{0}^{1} \frac{\ln x}{1+x^2}\,\mathrm{d}x\\ &= \frac{21\pi}{64}\zeta(3) - \frac{\pi^2}{24}G - \frac{\pi^3}{32}\ln2 \end{aligned} $$

Answer 3

In page 18 of this preprint , we have for positive integer $q$

$$\int_0^1\frac{\arctan(x)\ln^{q-1}(x)}{1+x}dx=\frac{(-1)^{q-1}(q-1)!}{4}\left(\pi\,\eta(q)-2\sum_{k=1}^{q}2^{k-q}\eta(q-k+1)\beta(k)\right),$$ where $\eta$ is the Dirichlet eta function and $\beta$ is the Dirichlet beta function.

Examples:

$$\int_0^1\frac{\arctan(x)\ln(x)}{1+x}dx=-\frac{3\pi}{32}\zeta(2)+\frac{1}{2}\ln(2)\beta(2)$$

$$\int_0^1\frac{\arctan(x)\ln^{2}(x)}{1+x}dx=\frac{21\pi}{64}\zeta(3)-\frac14\zeta(2)\beta(2)-\ln(2)\beta(3)$$

$$\int_0^1\frac{\arctan(x)\ln^{3}(x)}{1+x}dx=-\frac{315\pi}{250}\zeta(4)+\frac{9}{16}\zeta(3)\beta(2)+\frac34\zeta(2)\beta(3)+3\ln(2)\beta(4)$$