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How to integrate $$\int_{0}^{1} \frac{x \operatorname{Li}_2(1 - x)}{1 + x^2} \, dx$$

My try to integrate $$\text{I}=\int_{0}^{1} \frac{x \operatorname{Li}_2(1 - x)}{1 + x^2} \, dx$$

\begin{aligned} &= -\frac{1}{2} \int_{0}^{1} \frac{\ln(x) \ln(1 + x^2)}{1 - x} \, dx \\ &= -\frac{1}{2} \int_{0}^{1} \frac{\ln(x) \ln(1 + x^2)}{1 - x^2} \, dx \\ &\quad -\frac{1}{2} \int_{0}^{1} \frac{x \ln(x) \ln(1 + x^2)}{1 - x^2} \, dx \end{aligned}

Can someone help me to integrate last two integrals?

Sbsty
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2 Answers2

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Note that \begin{aligned} \int_{0}^{1} \frac{x \operatorname{Li}_2(1 - x)}{1 + x^2} \, dx \overset{ibp}=-\frac{1}{2} \int_{0}^{1} \frac{\ln x\ln(1 + x^2)}{1 - x} \, dx \\ \end{aligned} where the integral on the right hand side is given by $$ \int_{0}^{1} \frac{\ln x\ln(1 + x^2)}{1 - x} \, dx= -\frac{3\pi^2}{16}\ln2-\frac\pi2 G+2\zeta(3) $$

Quanto
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\begin{align}J&=\int_0^1\frac{x\text{Li}_2(1-x)}{1+x^2}dx\\ &=\int_0^1\frac{x}{1+x^2}\left(\int_0^x\frac{\ln t}{1-t}dt+\zeta(2)\right)dx\\ &=\int_0^1\int_0^1\frac{x^2\ln(tx)}{(1+x^2)(1-tx)}dtdx+\frac{1}{2}\zeta(2)\ln 2\\ &\int_0^1\left(\frac{\ln(tx)}{(1-tx)(1+t^2)}-\frac{\ln(tx)+tx\ln(tx)}{(1+t^2)(1+x^2)}\right)dtdx+\frac{1}{2}\zeta(2)\ln 2\\ &\overset{\text{Fubini}}=\int_0^1\frac{1}{t(1+t^2)}\left(\int_0^t\frac{\ln z}{1-z}dz\right)dt+\frac{\pi\text{G}}{2}-\ln 2\underbrace{\int_0^1\frac{x\ln x}{1+x^2}dx}_{u=x^2}+\frac{1}{2}\zeta(2)\ln 2\\ &\overset{\text{IBP}}=\frac{\zeta(2)\ln 2}{2}-2\zeta(3)+\frac{1}{2}\int_0^1\frac{\ln(1+t^2)\ln t}{1-t}dt +\frac{\pi\text{G}}{2}+\frac{5}{8}\zeta(2)\ln 2\\ &=\frac{9\zeta(2)\ln 2}{8}-2\zeta(3) +\frac{\pi\text{G}}{2}+\frac{1}{2}\int_0^1\frac{\ln(1+t^2)\ln t}{1-t}dt \end{align}

On the other hand, \begin{align}J&\overset{\text{IBP}}=\frac{1}{2}\Big[\ln(1+x^2)\text{Li}_2(1-x)\Big]_0^1-\frac{1}{2}\int_0^1\frac{\ln(1+x^2)\ln x}{1-x}dx\\ &=-\frac{1}{2}\int_0^1\frac{\ln(1+x^2)\ln x}{1-x}dx\\ \end{align} Therefore, \begin{align}J&=\frac{9\zeta(2)\ln 2}{8}-2\zeta(3) +\frac{\pi\text{G}}{2}-J\\ &=\frac{9\zeta(2)\ln 2}{16}-\zeta(3) +\frac{\pi\text{G}}{4}=\boxed{\frac{3\pi^2\ln 2}{32}-\zeta(3) +\frac{\pi\text{G}}{4}}\\ \end{align}

FDP
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