Evaluating $\int _0^1\frac{\ln ^2\left(x^2+1\right)}{x+1}\:dx$ using real methods

Question

I've stumbled upon that interesting integral here, the OP managed to transform the integral into something more approachable using contour integration and proved that $$\int _0^1\frac{\ln ^2\left(x^2+1\right)}{x+1}\:dx\:=\:-\pi G+\frac{5}{2}\zeta \left(3\right)+\frac{2}{3}\ln ^3\left(2\right)-\frac{\pi ^2}{24}\ln \left(2\right)$$ I can't come up with a way to attack this one with only real methods, i'd appreciate any help possible.

Answer 1

Integrate by parts

$$\int _0^1\frac{\ln ^2(x^2+1)}{x+1}~dx =\ln^32 -4I \tag1$$ where $I$ is given below, along with its twin $J$ $$I=\int_0^1\frac{x\ln(1+x)\ln(x^2+1)}{1+x^2}~dx,\>\>\>\>\> J=\int_0^1\frac{x\ln(1-x)\ln(x^2+1)}{1+x^2}~dx $$

Then, evaluate their sum and difference as follows \begin{align} J+I=& \int_0^1\frac{x\ln(1-x^2)\ln(x^2+1)}{1+x^2}~{dx} \>\>\>\>\>\>\> \left( x^2\to\frac{1-x}{1+x} \right) \\ =& \frac12\int_0^1 \frac{\ln^2\frac{1+x}2}{1+x}dx +\frac{\ln2}2 \int_0^1 \frac{\ln x}{1+x}dx+\frac14\int_0^1 \frac{\ln^2(1+x)}{x}dx\\ =&\frac16\ln^32-\frac{\pi^2}{24}\ln2+\frac1{16}\zeta(3)\\ \\ J-I=& \int_0^1\frac{x\ln\frac{1-x}{1+x}\ln(x^2+1)}{1+x^2}~{dx} \>\>\>\>\>\>\> \left( \frac{1-x}{1+x} \to x\right) \\ =& \>\ln2\int_0^1 \frac{\ln x}{1+x}dx +\frac{\ln2}2 \int_0^1 \frac{\ln(1+x^2)}{x} \overset{x^2\to x}{dx} +\frac14 \int_0^1 \frac{\ln^2(1+x^2)}{x} \overset{x^2\to x}{dx}\\ & -2\int_0^1 \frac{\ln x\ln(1+x)}{1+x}\overset{ibp}{dx} - \int_0^1 \frac{\ln (1+x)\ln(1+x^2)}{x}dx\\ = & \>\frac{3\ln2}4\int_0^1 \frac{\ln x}{1+x}dx +\frac98 \int_0^1 \frac{\ln^2(1+x)}{x} dx - P =\frac9{32}\zeta(3) - \frac{\pi^2}{16}\ln2-P\\ \end{align} where $P$ and its twin $Q$ are $$P= \int_0^1 \frac{\ln (1+x)\ln(1+x^2)}{x}dx,\>\>\>\>\> Q= \int_0^1 \frac{\ln (1-x)\ln(1+x^2)}{x}dx $$ Similarly, their sum and difference are (see here for evaluation details) \begin{align} &Q+P=\int_0^1 \frac{\ln (1-x^2)\ln(1+x^2)}{x}dx=-\frac5{16}\zeta(3)\\ &Q-P=\int_0^1 \frac{\ln \frac{1-x}{1+x} \ln(1+x^2)}{x}dx=-\pi G- \frac7{4}\zeta(3)\\ \end{align}

which results in $P= \frac{\pi}{2}G-\frac{33}{32}\zeta(3) $. Substitute into $I-J$ and $I+J$ above to get $$ I= \frac{\pi}4 G-\frac5{8}\zeta(3) + \frac{\pi^2}{96}\ln2+ \frac1{12}\ln^32 $$ and then into (1) to obtain

$$\int _0^1\frac{\ln ^2(1+x^2)}{1+x}\>dx=-\pi G+\frac{5}{2}\zeta (3)-\frac{\pi ^2}{24}\ln2 +\frac{2}{3}\ln^32 $$

Answer 2

Lemma 1 :$$\displaystyle{L1:\frac{{L{i_n}\left( {a \cdot z} \right)}}{z} = \frac{d}{{dz}}\left( {L{i_{n + 1}}\left( {a \cdot z} \right)} \right)}$$ and $$\displaystyle{\frac{{Log\left( {1 - a \cdot z} \right)}}{z} = - \frac{d}{{dz}}\left( {L{i_2}\left( {a \cdot z} \right)} \right)} $$http://mathworld.wolfram.com/Polylogarithm.html

Lemma 2 :$$\displaystyle{L2:{\log ^2}\left( {1 + i} \right) \cdot \log \left( {1 - i} \right) = \left( {\frac{1}{8} {{\log }^3}2 + \frac{{{\pi ^2}}}{{32}}\log 2} \right) + i\left( {\frac{\pi }{{16} }{{\log }^2}2 + \frac{{{\pi ^3}}}{{64}}} \right)}$$(actions).

Lemma 3 :$$\displaystyle{L3:L{i_2}\left( i \right) = - \frac{{{\pi ^2}}}{{48}} + i \cdot G}$$ and $$\displaystyle{L{i_3}\left( i \right) = - \frac{3}{{32}}\zeta \left( 3 \right) + i\frac{{{\pi ^3}}}{ {32}}}$$ They arise from the more general relationship $$\displaystyle{L{i_s}\left( { \pm i} \right) = - {2^{ - s}}\eta \left( s \right) \pm i \cdot \beta \left( s \right )}$$

https://en.wikipedia.org/wiki/Polylogarithm

Lemma 4 :$$ \begin{aligned} &L_4: 2\operatorname{Re}\left(L_3\left(\frac{1 + i}{2}\right)\right) \\ &= \frac{{\log^3 2}}{{24}} - \frac{{5\pi^2\log 2}}{{96}} + \frac{{35}}{{32}}\zeta(3) \end{aligned} $$

Because from the more general equation

$$ \begin{aligned} &L_3(z) + L_3(1 - z) + L_3\left(\frac{-z}{1 - z}\right) \\ &= \zeta(3) + \frac{{\pi^2\log(1 - z)}}{6} - \frac{{\log z \cdot \log^2(1 - z)}}{2} + \frac{{\log^3 z}}{6} \end{aligned} $$

proved here (relation 9), for $z = \frac{1 + i}{2}$ follows

$$ \begin{aligned} &L_3\left(\frac{1 + i}{2}\right) + L_3\left(\frac{1 - i}{2}\right) + L_3(-i) \\ &= \frac{1}{6}\log^3\left(\frac{1 - i}{2}\right) - \frac{1}{2}\log\left(\frac{1 + i}{2}\right)\log^2\left(\frac{1 - i}{2}\right) \\ &\quad + \frac{\pi^2}{6}\log\left(\frac{1 - i}{2}\right) + \zeta(3) \\ &\Rightarrow L_3\left(\frac{1 + i}{2}\right) + L_3\left(\frac{1 - i}{2}\right) + L_3(-i) \\ &= \left(\frac{\log^3 2}{24} - \frac{5\pi^2\log 2}{96} + \zeta(3)\right) - i\frac{\pi^3}{32} \end{aligned} $$

But $L_3(i) = -\frac{3}{32}\zeta(3) + i\frac{\pi^3}{32} \Rightarrow L_3(-i) = -\frac{3}{32}\zeta(3) - i\frac{\pi^3}{32}$, so

$$ \begin{aligned} &L_3\left(\frac{1 + i}{2}\right) + L_3\left(\frac{1 - i}{2}\right) \\ &= \frac{\log^3 2}{24} - \frac{5\pi^2\log 2}{96} + \frac{35}{32}\zeta(3) \\ &\Rightarrow 2\operatorname{Re}\left(L_3\left(\frac{1 + i}{2}\right)\right) \\ &= \frac{\log^3 2}{24} - \frac{5\pi^2\log 2}{96} + \frac{35}{32}\zeta(3) \end{aligned} $$

Lemma 5 : $$\displaystyle{L5:\log \left( {2\sin x} \right) = \sum\limits_{n = 1}^\infty {\frac{{\cos 2nx}}{n}} }$$

On our topic ..

$$\displaystyle{ = \frac{1}{4}\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + {x^2}} \right)}}{{ 1 + x}}dx} - \int\limits_0^1 {\frac{{Arc{{\tan }^2}x}}{{1 + x}}dx} + i \cdot A \Rightarrow } $$ $$\displaystyle{\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + {x^2}} \right)}}{{1 + x}}dx} = 4 \left( {\int\limits_0^1 {\frac{{Arc{{\tan }^2}x}}{{1 + x}}dx} + {\mathop{\rm Re}\nolimits} \left ( {\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + ix} \right)}}{{1 + x}}dx} } \right)} \right )}$$

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Starting $$ \begin{aligned} &\int\limits_0^1 \frac{\log^2(1 + ix)}{1 + x}\,dx \\ &= \mathop = \limits ^{1 + ix = y} -\frac{i}{{1 + i}}\int\limits_1^{1 + i} \frac{\log^2y}{1 - \frac{i}{{1 + i}}y}\,dy \\ &= \int\limits_1^{1 + i} \log^2y \left(\log\left(1 - \frac{i}{{1 + i}}y\right)\right)'\,dy \\ \end{aligned} $$

$$\displaystyle{ = \left[ {{{\log }^2}y \cdot \log \left( {1 - \frac{i}{{1 + i}}y} \right)} \right]_1^ {1 + i} - 2\int\limits_1^{1 + i} {\frac{{\log y}}{y}\log \left( {1 - \frac{i}{{1 + i}} y} \right)dy} = } $$ $$\displaystyle{\mathop = \limits^{L1} = {\log ^2}\left( {1 + i} \right) \cdot \log \left( {1 - i} \right) + 2\int\ limits_1^{1 + i} {\log y{{\left( {L{i_2}\left( {\frac{i}{{1 + i}}y} \right)} \right)}{'} }dy} = }$$

$$\displaystyle{ = \mathop = \limits^{L1} = {\log ^2}\left( {1 + i} \right) \cdot \log \left( {1 - i} \right) + 2\log \left( {1 + i} \right)L{i_2}\left( i \right) - 2\int\limits_1^{1 + i} {{{\left( {L{i_3}\left( {\ frac{i}{{1 + i}}y} \right)} \right)}{'}}dy} = } $$ $$\displaystyle{{\log ^2}\left( {1 + i} \right) \cdot \log \left( {1 - i} \right) + 2\log \left( {1 + i} \right) L{i_2}\left( i \right) - 2L{i_3}\left( i \right) + 2L{i_3}\left( {\frac{{1 + i}}{2}} \right) = }$$

$$\displaystyle{ = \mathop = \limits^{L2,L3} = \left( {\frac{1}{8}{{\log }^3}2 + \frac{{{\pi ^2}}} {{96}}\log 2 - \frac{\pi }{2} \cdot G + \frac{3}{{16}}\zeta \left( 3 \right)} \right) + i\left( {\frac{\pi }{{16}}{{\log }^2}2 + G \cdot \log 2 - \frac{{11{\pi ^3}}}{{192}}} \right ) + 2L{i_3}\left( {\frac{{1 + i}}{2}} \right)}$$

Therefore $$\displaystyle{{\mathop{\rm Re}\nolimits} \left( {\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + ix} \right)}} {{1 + x}}dx} } \right) = \left( {\frac{1}{8}{{\log }^3}2 + \frac{{{\pi ^2}}}{{ 96}}\log 2 - \frac{\pi }{2} \cdot G + \frac{3}{{16}}\zeta \left( 3 \right)} \right) + 2{\mathop{\ rm Re}\nolimits} \left( {L{i_3}\left( {\frac{{1 + i}}{2}} \right)} \right) = } $$ $$\displaystyle{\mathop = \limits^{L4} = \frac{1}{6}{\log ^3}2 - \frac{{{\pi ^2}}}{{24}}\log 2 - \frac{\pi }{2} \cdot G + \frac{{41}}{{32}}\zeta \left( 3 \right)}$$

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Also $$\displaystyle{\int\limits_0^1 {\frac{{Arc{{\tan }^2}x}}{{1 + x}}dx} = \int\limits_0^1 {Arc{{\tan }^ 2}x\left( {\log \left( {1 + x} \right)} \right)'dx} = \frac{{{\pi ^2}\log 2}}{{16}} - 2 \int\limits_0^1 {\frac{{Arc\tan x \cdot \log \left( {1 + x} \right)}}{{1 + {x^2}}}dx} = }$$

$$\displaystyle{ = \mathop = \limits^{x \to \tan x} = \frac{{{\pi ^2}\log 2}}{{16}} - 2\int\limits_0^{\pi / 4} {x \cdot \log \left( {1 + tanx} \right)dx} = \frac{{{\pi ^2}\log 2}}{{16}} - 2\int\limits_0^{ \pi /4} {x \cdot \log \left( {\frac{{\sin x + \cos x}}{{\cos x}}} \right)dx} = }$$

$$\displaystyle{ = \frac{{{\pi ^2}\log 2}}{{16}} - \int\limits_0^{\pi /4} {x \cdot \log \left( {\frac{{ {{\left( {\sin x + \cos x} \right)}^2}}}{{{{\cos }^2}x}}} \right)dx} = \frac{{{\pi ^2}\log 2}}{{16}} - \int\limits_0^{\pi /4} {x \cdot \log \left( {\frac{{2\left( {1 + \sin 2x} \right)}}{{1 + \cos 2x}}} \right)dx} = }$$

$$\displaystyle{ = \frac{{{\pi ^2}\log 2}}{{16}} - \frac{1}{4}\int\limits_0^{\pi /2} {x \cdot \log \left( {\frac{{2\left( {1 + \sin x} \right)}}{{1 + \cos x}}} \right)dx} = \frac{{{\pi ^2} \log 2}}{{32}} - \frac{1}{4}\int\limits_0^{\pi /2} {x \cdot \log \left( {1 + \sin x} \right)dx } }$$ $$\displaystyle{ + \frac{1}{4}\int\limits_0^{\pi /2} {x \cdot \log \left( {1 + \cos x} \right)dx} = }$$

$$\displaystyle{ = \frac{{{\pi ^2}\log 2}}{{32}} - \frac{\pi }{8}\int\limits_0^{\pi /2} {\log \left ( {1 + \cos x} \right)dx} + \frac{1}{2}\int\limits_0^{\pi /2} {x \cdot \log \left( {1 + \cos x} \ right)dx} = }$$ $$\displaystyle{\frac{{{\pi ^2}\log 2}}{{32}} - \frac{\pi }{8}\int\limits_0^{\pi /2} {\log \left( {2{{\cos }^2}\frac{x}{2}} \right)dx} + \frac{1}{2}\int\limits_0^{\pi /2} {x \cdot \log \left( {2{{\cos }^2}\frac{x}{2}} \right)dx} = }$$

$$\displaystyle{ = \frac{{{\pi ^2}\log 2}}{{32}} - \frac{\pi }{8}\int\limits_0^{\pi /2} {\left( { \log \left( {4{{\cos }^2}\frac{x}{2}} \right) - \log 2} \right)dx} + \frac{1}{2}\int\limits_0 ^{\pi /2} {x \cdot \left( {\log \left( {4{{\cos }^2}\frac{x}{2}} \right) - \log 2} \right) dx} = }$$

$$\displaystyle{ = \frac{{{\pi ^2}\log 2}}{{32}} - \frac{\pi }{2}\int\limits_0^{\pi /4} {\log \left ( {2\cos x} \right)dx} + 4\int\limits_0^{\pi /4} {x \cdot \log \left( {2\cos x} \right)dx} = }$$ $$\displaystyle{\frac{{{\pi ^2}\log 2}}{{32}} + \frac{{3\pi }}{2}\int\limits_{\pi /4}^{\pi /2} {\log \left( {2\sin x} \right)dx} - 4\int\limits_{\pi /4}^{\pi /2} {x \cdot \log \left( {2 \sin x} \right)dx} }$$

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But $$\displaystyle{\int\limits_{\pi /4}^{\pi /2} {\log \left( {2\sin x} \right)dx} = - \sum\limits_{n = 1}^\ infty {\frac{1}{n}\int\limits_{\pi /4}^{\pi /2} {\cos 2nx\;dx} } = \frac{1}{2}\sum\limits_{ n = 1}^\infty {\frac{1}{{{n^2}}}\sin \frac{{n\pi }}{2}} = \frac{1}{{4i}}\sum \limits_{n = 1}^\infty {\frac{1}{{{n^2}}}\left( {{e^{in\pi /2}} - {e^{ - in\pi / 2}}} \right)} = }$$

$$\displaystyle{ = \frac{1}{{4i}}\sum\limits_{n = 1}^\infty {\frac{{{i^n} - {{\left( { - i} \right)} ^n}}}{{{n^2}}} = } \frac{1}{{4i}}\left( {L{i_2}\left( i \right) - L{i_2}\left( { - i} \right)} \right) = \frac{1}{{4i}}2iG = \frac{1}{2}G}$$

and $$ \begin{aligned} &\int\limits_{\pi /4}^{\pi /2} {x\log \left( {2\sin x} \right)dx} \\ &= - \sum\limits_{n = 1}^ \infty {\frac{1}{n}\int\limits_{\pi /4}^{\pi /2} {x\cos 2nx\;dx} } \\ &= - \sum\limits_{n = 1}^ \infty {\frac{1}{{8{n^3}}}\left( { - 2\cos \frac{{n\pi }}{2} + 2{{\left( { - 1} \right)}^n} - n\pi \sin \frac{{n\pi }}{2}} \right)} \\ \end{aligned} $$

$$ \begin{aligned} &= \frac{1}{4}\sum\limits_{n = 1}^\infty {\frac{1}{{n^3}}\cos \frac{{n\pi }}{2}} - \frac{1}{4}\sum\limits_{n = 1}^\infty {\frac{1}{{n^3}}{{\left( { - 1} \right)}^n}} + \frac{\pi }{8}\sum\limits_{n = 1}^\infty {\frac{1}{{n^2}}\sin \frac{{n\pi }}{2}} \\ &= \frac{1}{8}\left( {L{i_3}\left( i \right) + L{i_3}\left( { - i} \right)} \right) - \frac{1}{4}L{i_3}\left( { - 1} \right) + \end{aligned} $$

$$ \begin{aligned} &+ \frac{\pi }{{16i}}\left( {L{i_2}\left( i \right) - L{i_2}\left( { - i} \right)} \right) \\ &= - \frac{7}{{32}}L{i_3}\left( { - 1} \right) + \frac{\pi }{{16i}}\left( {L{i_2}\left( i \right) - L{i_2}\left( { - i} \right)} \right) \\ &= \frac{{21}}{{128}}\zeta \left( 3 \right) + \frac{{\pi \cdot G}}{8} \end{aligned} $$

So $$\displaystyle{\int\limits_0^1 {\frac{{Arc{{\tan }^2}x}}{{1 + x}}dx} = \frac{{{\pi ^2}\log 2} }{{32}} + \frac{{3\pi }}{2}\int\limits_{\pi /4}^{\pi /2} {\log \left( {2\sin x} \right )dx} - 4\int\limits_{\pi /4}^{\pi /2} {x \cdot \log \left( {2\sin x} \right)dx} = }$$

$$\displaystyle{ = \frac{{{\pi ^2}\log 2}}{{32}} + \frac{{3\pi }}{2}\left( {\frac{1}{2}G } \right) - 4\left( {\frac{{21}}{{128}}\zeta \left( 3 \right) + \frac{{\pi \cdot G}}{8}} \right) \Rightarrow \int\limits_0^1 {\frac{{Arc{{\tan }^2}x}}{{1 + x}}dx} = \frac{{{\pi ^2}\log 2}} {{32}} + \frac{{\pi G}}{4} - \frac{{21}}{{32}}\zeta \left( 3 \right)}$$

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Finally $$\displaystyle{\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + ix} \right)}}{{1 + x}}dx} = 4\left( { \int\limits_0^1 {\frac{{Arc{{\tan }^2}x}}{{1 + x}}dx} + {\mathop{\rm Re}\nolimits} \left( {\int \limits_0^1 {\frac{{{{\log }^2}\left( {1 + ix} \right)}}{{1 + x}}dx} } \right)} \right) = }$$

$$ \begin{aligned} &= 4\left( \left( \frac{{{\pi ^2}\log 2}}{{32}} + \frac{{\pi G}}{4} - \frac{{21}}{{32}}\zeta \left( 3 \right) \right) + \left( \frac{1}{6}{{\log }^3}2 - \frac{{\pi ^2}}{{24}}\log 2 - \frac{\pi }{2} \cdot G + \frac{{41}}{{32}}\zeta \left( 3 \right) \right) \right) \\ &= \frac{5}{2}\zeta \left( 3 \right) + \frac{2}{3}{\log ^3}2 - \pi \cdot G - \frac{{{\pi ^2}\log 2}}{{24}}. \end{aligned} $$ :)

, relationships

$$ \begin{aligned} 1) &\quad \int\limits_0^1 {\frac{{{{\log }^2}\left( {1 + ix} \right)}}{{1 + x}}dx} \\ &= \left( \frac{1}{8}{{\log }^3}2 + \frac{{{\pi ^2}}}{{96}}\log 2 - \frac{\pi }{2} \cdot G + \frac{3}{{16}}\zeta \left( 3 \right) \right) \\ &\quad + i\left( \frac{\pi }{{16}}{{\log }^2}2 + G \cdot \log 2 - \frac{{11{\pi ^3}}}{192} \right) + 2L{i_3}\left( {\frac{{1 + i}}{2}} \right). \end{aligned} $$

$$\displaystyle{\left. 2 \right)\quad \int\limits_0^1 {\frac{{Arc{{\tan }^2}x}}{{1 + x}}dx} = \frac{{{\pi ^2}\ log 2}}{{32}} + \frac{{\pi G}}{4} - \frac{{21}}{{32}}\zeta \left( 3 \right)}$$ and

$$\displaystyle{\left. 3 \right)\quad 2{\mathop{\rm Re}\nolimits} \left( {L{i_3}\left( {\frac{{1 + i}}{2}} \right)} \right) = \frac{{{{\log }^3}2}}{{24}} - \frac{{5{\pi ^2}\log 2}}{{96}} + \frac{{35} }{{32}}\zeta \left( 3 \right)}$$