Help needed with an integral $\int_0^1 \frac{\ln(x) \ln(1+x) }{1+x^2} dx$.

Question

I am trying to evaluate the integral $$ \int_0^1 \frac{\ln(x) \ln(1+x) }{1+x^2} dx $$ Integration by parts gives \begin{eqnarray*} \int_0^1 \frac{\ln(x) \tan^{-1}(x) }{1+x} dx + \int_0^1 \frac{\ln(x) \ln(1+x) }{1+x^2} dx + \int_0^1 \frac{\tan^{-1}(x) \ln(1+x) }{x} dx =0. \end{eqnarray*} The first integral has been done here Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$ \begin{eqnarray*} \int_0^1 \frac{\ln(x) \tan^{-1}(x) }{1+x} dx = -\frac{ \pi^3}{64} + \frac{1}{2} K \ln(2). \end{eqnarray*} Any hints or solutions on how to approach the other integrals will be gratefully recieved.

Answer 1

Denote $I= \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx $ and $J=\int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx$

\begin{align} &\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx \overset{x\to\frac{1-x}{1+x}}= J -2 I-2G\ln2+\frac{\pi^3}{16}\\ &\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}dx = J+ \int_1^\infty \frac{\ln^2(1+x)}{1+x^2}\overset{x\to \frac1x}{dx} = 2J-2I+\frac{\pi^3}{16} \end{align}

Eliminate $J$

\begin{align} I=&\ \frac12\int_0^\infty \frac{\ln^2(1+x)}{1+x^2} \overset{t=1+x}{dx}-\int_0^1 \frac{\ln^2(1-x)}{1+x^2}\overset{t=1-x}{dx}-2G\ln2+\frac{\pi^3}{32}\\ =& \ \frac12\int_0^\infty\frac{\ln^2t}{1+(t-1)^2}dt -\frac32\int_0^1 \frac{\ln^2(1-x)}{1+x^2}{dx}-2G\ln2+\frac{\pi^3}{32}\tag1 \end{align}

where $\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx =2\Im\text{Li}_3\left(\frac{1+i}2\right) $

\begin{align} &\int_0^\infty \frac{\ln^2t}{1+(t-1)^2}dt =\int_0^\infty \frac{(t^2+2t+2)\ln^2t}{t^4+4}dt =\frac{7\pi^3}{64}+\frac{3\pi}{16}\ln^22\\ \end{align}

Plug into (1) to obtain $$ I=-3\Im\text{Li}_3\left(\frac{1+i}2\right) -2G\ln2 +\frac{3\pi}{32}\ln^22+\frac{11\pi^3}{128}$$

Answer 2

Using Feynman's trick.

$$I(a)=\int_0^1 \frac{\log(x) \log(1+ax) }{1+x^2}\, dx$$ $$I'(a)=\int_0^1\frac{x \log (x)}{\left(x^2+1\right) (a x+1)}\,dx$$ $$I'(a)= \frac 1 {a^2+1} \int_0^1\Bigg(\frac{x \log (x)}{ x^2+1}+\frac{a \log (x)}{x^2+1}-\frac{a \log (x)}{a x+1}\Bigg) \,dx$$

$$J_1=\int_0^1 \frac{x \log (x)}{ x^2+1} \,dx=-\frac{\pi ^2}{48}$$ $$J_2=a\int_0^1 \frac{ \log (x)}{ x^2+1} \,dx=-a C$$ $$J_3=-a\int_0^1 \frac{ \log (x)}{ ax+1} \,dx=-a \text{Li}_2(-a)$$ Simplify and recombine $$I'(a)=\frac{J_1+J_2+J_3}{1+a^2}=-\frac{a C}{a^2+1}-\frac{\pi ^2}{48\left(a^2+1\right)}-\frac{\text{Li}_2(-a)}{a^2+1}$$ $$I(a)=\int_0^1 I'(a)\, da=-\frac{1}{2} C \log (2)-\frac{\pi ^3}{192}+\int_0^1 \frac {J_3}{1+a^2} \,da$$

The last integral is not the most pleasant (but Wolfram Alpha gives the antiderivative- see here) $$\int_0^1 \frac {J_3}{1+a^2} \,da=\frac{35 \pi ^3}{384}-\frac{3}{2} C \log (2)+$$ $$\frac i{64}\left(192 \text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)-105 \zeta (3)-4 \log ^3(2)-6 i \pi \log ^2(2)+5 \pi ^2 \log (2)\right)$$

So, as a total

$$K=\int_0^1 \frac{\log(x) \log(1+ax) }{1+x^2}\, dx=\frac{11 \pi ^3}{128}-2 C \log (2)+\frac{3}{32} \pi \log ^2(2)+$$ $$\frac i{64} \left(192 \text{Li}_3\left(\frac{1+i}{2}\right)-105 \zeta (3)-4 \log ^3(2)+5 \pi ^2 \log (2)\right)$$