Evaluate $\int\limits_0^\infty \frac{\ln^2(1+x)}{1+x^2}\ dx$

Question

This problem was already solved here (in different closed form).

But how can you prove $\ \displaystyle\int\limits_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx=2\Im\left(\operatorname{Li}_3(1+i)\right)\ $

Where $\displaystyle \operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}\ $ is the the trilogarithm.

Answer 1

The following solution is by Cornel Valean:

\begin{gather*} \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\textrm{d}x\overset{x= \frac1y}{=}\int_0^\infty\frac{\ln^2\left(\frac{y}{1+y}\right)}{1+y^2}\textrm{d}y\\ \overset{\frac{y}{1+y}=x}{=}\int_0^1\frac{\ln^2(x)}{x^2+(1-x)^2}\textrm{d}x\\ \left\{\text{write $\frac{1}{x^2+(1-x)^2}=\Im \frac{1+i}{1-(1+i)x}$}\right\}\\ =\Im \int_0^1\frac{(1+i)\ln^2(x)}{1-(1+i)x}\textrm{d}x\\ =2\ \Im\{\operatorname{Li}_{3}(1+i)\}. \end{gather*}

Answer 2

\begin{align}J=\int\limits_0^\infty \frac{\ln^2(1+x)}{1+x^2}\ dx\end{align} Perform the change of variable $y=\dfrac{1}{1+x}$,

\begin{align}J&=\int\limits_0^1 \frac{\ln^2 x}{2x^2-2x+1}\ dx\\ &=\int_0^1 \frac{\ln^2 x}{\Big(1-(1+i)x\Big)\Big(1-(1-i)x\Big)}\,dx\\ &=\frac{1}{2i}\left(\int_0^1 \frac{(1+i)\ln^2 x}{1-(1+i)x}\,dx-\int_0^1 \frac{(1-i)\ln^2 x}{1-(1-i)x}\,dx\right)\\ &=2\times \frac{1}{2i}\left(\text{Li}_3(1+i)-\text{Li}_3(1-i)\right)\\ &=2\times \frac{1}{2i}\left(\text{Li}_3(1+i)-\overline{\text{Li}_3(1+i)}\right)\\ &=2\Im\Big(\text{Li}_3(1+i)\Big) \end{align}

Since, for $\Im(a)\neq 0$,

\begin{align}\int_0^1 \frac{a\ln^2 x}{1-ax}\,dx=2\text{Li}_{3}(a)\end{align}

NB:

It can be proved easily that the identity used is true for $|a|<1$ using Taylor's expansion and usual definition of $\text{Li}_{3}(a)$ for $|a|<1$. The two functions are analytic not only for $|a|<1$ and thus, the identity can be extended.

Answer 3

letting $\ \small{\displaystyle x=\frac{1-y}{y}}\ $ gives $\ \displaystyle I=\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx=\int_0^1\frac{\ln^2x}{2x^2-2x+1}\ dx=\frac12\int_0^1\frac{\ln^2x}{(a-x)(b-x)}\ dx$

where $\ a=\frac12(1+i)\ $ and $\ b=\frac12(1-i)$

then $\ \displaystyle I=\frac1{2(a-b)}\int_0^1\ln^2x\left(\frac1{b-x}-\frac1{a-x}\right)\ dx=\frac1{2(a-b)}\left(2\operatorname{Li}_3\left(\frac1b\right)-2\operatorname{Li}_3\left(\frac1a\right)\right)$

plugging $a$ and $b$, we get $\qquad\boxed{I=-i\left(\operatorname{Li}_3(1+i)-\operatorname{Li}_3(1-i)\right)=2\text{Im}\operatorname{Li}_3(1+i)}$