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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.

Ali Shadhar
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    A magic way (incorrect though): $$\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx\overset{1+x=\frac{1}{t}}=\int_0^1 \frac{\ln^2 t}{1+2t(t-1)}dt=\Im\left(-\frac{2}{1+i}\int_0^1 \frac{\ln^2 t}{1-\frac{2t}{1+i}}dt \right)$$ $$=-\Im\left(\sum_{n\ge 0}\left(\frac{2}{1+i}\right)^{n+1}\int_0^1 t^n \ln^2 t dt\right)=-2\Im\left(\sum_{n\ge 1}\frac{\left(1-i\right)^n}{n^3}\right)=-2\Im\operatorname{Li}_3(1-i)$$ – Zacky Jun 14 '19 at 14:12
  • @Zacky: A modification of your computation gives the result. – FDP Jun 14 '19 at 15:53
  • @FDP It is not correct to expand into series there, since $\sum_{n\ge 1}\limits \frac{(1-i)^n}{n^3} $ diverges. Magically the results happens to be correct, but that is only luck. Do you think it can be salvaged somehow? – Zacky Jun 14 '19 at 16:16
  • I think it's a good idea to clean up the interior of $ln^2$. – FDP Jun 14 '19 at 16:41
  • I want to use, for $|a|<1,r\geq 1$, an integer : \begin{align}\int_0^1 \frac{a\ln^r x}{1-ax},dx=(-1)^r!\text{Li}_{r+1}(a)\end{align}This identity is easy to prove when $|a|<1$ – FDP Jun 14 '19 at 16:46
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    @Zacky nice . It's interesting how the series diverges and still gives the right result. We better investigate into it. – Ali Shadhar Jun 14 '19 at 18:06
  • \begin{align}\Im\Big(\text{Li}{3}(1+i)\Big)=\frac{7}{128}\pi^3+\frac{3}{32}\pi^2\ln 2-\Im\left(\text{Li}{3}\left(\frac{1+i}{2}\right)\right)\end{align} – FDP Jun 14 '19 at 18:26
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    $\left|\frac{1+i}{2}\right|<1$, the usual definition of $Li_3$ can be used. – FDP Jun 14 '19 at 18:31
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    I'm voting to close this question as off-topic because it's a challenge problem, not a true question. – Adrian Keister Jul 01 '19 at 13:15
  • @Adrian Keister :) – Ali Shadhar Jul 01 '19 at 16:46

3 Answers3

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\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.

FDP
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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 $\qquad\boxed{I=-i\left(\operatorname{Li}_3(1+i)-\operatorname{Li}_3(1-i)\right)=2\text{Im}\operatorname{Li}_3(1+i)}$

Ali Shadhar
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  • Behind your computation you use too an identity obtained by analytic continuation. – FDP Jun 14 '19 at 18:39
  • yes as our solutions are very similar. – Ali Shadhar Jun 14 '19 at 18:42
  • One can obtain probably an elementary computation using $\Im\left(\text{Li}_{3}\left(\frac{1+i}{2}\right)\right)$ – FDP Jun 14 '19 at 18:48
  • the link i provided in the body shows that term in the calculations. – Ali Shadhar Jun 14 '19 at 18:57
  • I have obtained the equality above using GP PARI. lindep rulez ! – FDP Jun 14 '19 at 19:02
  • using $\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)+\operatorname{Li}_3\left(1-\frac1x\right)=\zeta(3)+\frac16\ln^3x+\zeta(2)\ln x-\frac12\ln^2x\ln(1-x)$ with taking $x=1+i\ $ gives the result you mentioned. – Ali Shadhar Jun 14 '19 at 20:00
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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*}

Ali Shadhar
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