calculusharmonic-numbersintegrationpolylogarithmreal-analysis
How to prove in a simpe way that
$$\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx=\frac{13}{8}\ln2\zeta(2)-\frac{33}{32}\zeta(3)\ ?$$
where $\operatorname{Li}_2$ is the dilogarithm function.
If we integrate by parts, the integral boils down to
$$\int_0^1\left(\frac{1}{x}-\frac{2x}{1+x^2}\right)\left[2\ln(1-x)\ln(1+x)-\ln(1+x)\ln(1+x^2)\right]dx$$
and that seems long and complicated, any other ideas?
Considering the algebraic identity \begin{align*} &(a-b)^3b = a^3b - 3a^2b^2 + 3ab^3 - b^4 = -2a^3b +3(a^3b+ab^3) -3a^2b^2 -b^4\\ &\Longrightarrow \ \ \ 2a^3b = -{b^4 \over 2} -{b^4 + 6a^2b^2\over 2} + 3(a^3b+ab^3) - (a-b)^3b \end{align*} with $a = \ln(1-x)$ and $b= \ln (1+x)$ it follows that \begin{align*} 2\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =& - \frac 1 2\int_0^1 {\ln^4(1+x)\over x}d x \\ &-\frac 12 \int_0^1 \frac{\ln^4(1+x) + 6\ln^2(1-x)\ln^2(1+x)}{x}dx\\ &+3\int_0^1 \frac{\ln^3(1-x)\ln(1+x) + \ln(1-x)\ln^3(1+x)}{x}dx\\ &- \int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}dx\\ =:& -I_1 - I_2 + I_3 -I_4. \end{align*}
For $I_1$, make substitution $y = \frac x {1+x}$ to get: \begin{align*} I_1 =& \frac 1 2 \int_0^{\frac 12} \frac{\ln^4(1-y)}{y(1-y)} dy \\ =& \frac 1 2\underbrace{ \int_0^{\frac 12} \frac{\ln^4(1-y)}{y} dy}_{z=1-y}+ \frac 1 2 \int_0^{\frac 12} \frac{\ln^4(1-y)}{1-y} dy\\ =& \frac 1 2 \int_{\frac 1 2 }^1 \frac{\ln^4 z} {1-z} dz + \frac {\ln^5 2}{10}\\ =& \frac 12 \sum_{n=1}^\infty \int_{\frac 1 2}^1 z^{n-1}\ln^4 z\ dz + \frac {\ln^5 2}{10}\\ =& \frac 12 \sum_{n=1}^\infty \frac{\partial^4}{\partial n^4}\left[\frac 1 n - \frac 1 {n2^n}\right] + \frac {\ln^5 2}{10}\\ =& \frac 12 \sum_{n=1}^\infty \left[\frac{24}{n^5} - \frac {24}{n^52^n} - \frac{24 \ln 2}{n^42^n}-\frac{12\ln^2 2}{n^3 2^n}-\frac{4\ln^3 2}{n^2 2^n} - \frac{\ln^4 2}{n2^n}\right] + \frac {\ln^5 2}{10}\\ =&12\zeta(5) - 12\text{Li}_5(1/2) - 12\ln 2 \text{Li}_4(1/2) -6\ln^2 2 \text{Li}_3(1/2) -2\ln^3 2\text{Li}_2(1/2)-\frac {2}{5}\ln^5 2\\ =&\boxed{-12\Big(\text{Li}_5(1/2) + \ln 2\text{Li}_4(1/2)-\zeta(5)\Big)-{21 \over 4}\zeta(3)\ln^2 2 +{1\over 3} \pi^2 \ln^3 2-{2 \over 5} \ln^5 2} \end{align*} where the well-known values \begin{align*}\text{Li}_2(1/2) = {\pi^2 \over 12}-{\ln^2 2\over 2} , \qquad \text{Li}_3(1/2) ={7\zeta(3) \over 8} -{\pi^2 \ln 2\over 12} + {\ln^3 2 \over 6} \end{align*} are used.
Actually, $I_2$ was already evaluated by the OP here using the algebraic identity $$b^4 + 6a^2b^2 = \frac {(a-b)^4} 2+\frac{(a+b)^4}{2} -a^4.$$ It holds that $$ \boxed{I_2 = \frac {21}{8} \zeta(5).} $$
In fact, the value of $I_3$ can also be found in the previous answer of @Przemo's. For $I_3$, one can use the algebraic relation $3(a^3b + ab^3) =\frac 3 8 \left[ (a+b)^4 - (a-b)^4\right]$. This gives \begin{align*} I_3=& \underbrace{\frac 3 8 \int_0^1 \frac{\ln^4(1-x^2)}{x} dx}_{x^2 = y} - \underbrace{\frac 3 8 \int_0^1 \frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x} dx}_{\frac{1-x}{1+x} = y}\\ =&\frac 3 {16}\underbrace{\int_0^1 \frac{\ln^4(1-y)}{y} dy }_{1-y\mapsto y}- \frac 3 4 \int_0^1 \frac{\ln^4 y}{1-y^2} dy\\ =&\frac 3 {16}\int_0^1 \frac{\ln^4 y}{1-y} dy - \frac 3 4 \sum_{n=0}^\infty \int_0^1 y^{2n} \ln^4 y \ dy\\ =&\frac 3 {16}\sum_{n=1}^\infty \int_0^1 y^{n-1}\ln^4 y \ dy - \frac 3 4 \sum_{n=0}^\infty \frac {24}{(2n+1)^5}\\ =&\frac 3 {16}\sum_{n=1}^\infty \frac{24}{n^5} - 18 \sum_{n=0}^\infty \frac {1}{(2n+1)^5}\\ =&\frac {9}{2} \zeta(5)- 18\cdot \frac {31}{32}\zeta(5)\\ =&\boxed{-\frac{207}{16}\zeta(5)} \end{align*} as can be found in @Przemo's answer.
For $I_4$, make substitution $ \frac{1-x}{1+x}\mapsto x$ to get \begin{align*} I_4 = &2\int_0^1 \frac{\ln^3 x \ln\left(\frac 2 {1+x}\right)}{1-x^2} dx \\ =&2\ln 2 \int_0^1 \frac{\ln^3 x}{1-x^2} dx - \underbrace{2\int_0^1\frac{\ln^3 x \ln(1+x)}{1-x^2} dx }_{=:J}\\ =& 2\ln 2\sum_{n=0}^\infty \int_0^1 x^{2n} \ln^3 x\ dx - J\\ =& - 12\ln 2 \underbrace{\sum_{n=0}^\infty \frac 1 {(2n+1)^4}}_{\frac{15}{16}\zeta(4) = \frac{\pi^4}{96}} - J \\ =& -\frac{\pi^4 \ln 2}{8} - J. \end{align*} \begin{align*} J = &\int_0^1\frac{2\ln^3 x \ln(1+x)}{1-x^2} dx \\ =& \underbrace{\int_0^1 \frac{\ln^3 x \ln(1+x)}{1+x}dx}_{=:A} + \int_0^1 \frac{\ln^3 x \ln(1+x)}{1-x}dx\\ =& A + \int_0^1 \frac{\ln^3 x \ln(1-x^2)}{1-x}dx -\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\ =&A + \int_0^1 \frac{(1+x)\ln^3 x \ln(1-x^2)}{1-x^2}dx -\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\ =&A + \underbrace{\int_0^1 \frac{\ln^3 x \ln(1-x^2)}{1-x^2}dx }_{=:B}+\underbrace{\int_0^1 \frac{x\ln^3 x \ln(1-x^2)}{1-x^2}dx}_{x^2 \mapsto x}-\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\ =&A + B - \underbrace{\frac {15}{16} \int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx}_{=:C}\\ =&A + B - C. \end{align*}
For $A$, we can use the McLaurin series of $$ \frac{\ln (1+x)}{1+x} = \sum_{n=0}^\infty (-1)^{n-1}H_n x^n $$ ($H_0= 0$) to get \begin{align*} A = & \sum_{n=0}^\infty (-1)^{n-1}H_n \int_0^1 x^n\ln^3 x \ dx \\ =&6 \sum_{n=0}^\infty \frac{(-1)^{n}H_n}{(n+1)^4}\\ =&6 \sum_{n=0}^\infty \frac{(-1)^{n}H_{n+1}}{(n+1)^4} - 6\sum_{n=0}^\infty \frac{(-1)^{n}}{(n+1)^5}\\ =&6 \sum_{n=1}^\infty \frac{(-1)^{n-1}H_{n}}{n^4} - 6\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^5}\\ =& 6\left(\frac{59}{32}\zeta(5) - \frac{\pi^2\zeta(3)}{12}\right)-6\cdot \frac{15}{16}\zeta(5)\\ =& \frac{87}{16}\zeta(5) - \frac{\pi^2 \zeta(3)}{2}. \end{align*} Here, the known value of $ \sum_{n=1}^\infty (-1)^{n-1}{H_n \over n^4}$ is used.
For $B$, make substitution $u = x^2$ to get \begin{align*} B =& \frac 1 {16} \int_0^1 \frac{\ln^3 u \ln(1-u)}{\sqrt u (1-u)} du \\ =& \frac 1 {16} \left[\frac{\partial^4}{\partial x^3\partial y} \text{B}(x,y)\right]_{x=\frac 1 2, y = 0^+} \end{align*} where $\text{B}(\cdot,\cdot)$ is Euler's Beta function. We can use the fact that \begin{align*} \lim_{y\to 0^+}\frac{\partial^2}{\partial x\partial y} \text{B}(x,y) = -\frac 1 2 \psi''(x) + \psi'(x) \big[\psi(x) + \gamma\big] \end{align*} to get \begin{align*} B =& \frac 1 {16}\frac{d^2}{dx^2}\left[-\frac 1 2 \psi''(x) + \psi'(x) \big[\psi(x) + \gamma\big]\right]_{x=\frac 1 2}\\ =&\frac 1 {16} \left[-\frac 1 2 \psi''''(1/2) + \psi'''(1/2)\big[\psi(1/2) + \gamma\big] + 3\psi'(1/2)\psi''(1/2)\right]\\ =& \frac 1 {16}\left[-21\pi^2 \zeta(3) + 372\zeta(5) - 2\pi^4 \ln 2\right] \end{align*} which can be evaluated using the series representations of polygamma functions $$\psi(x) +\gamma = - \frac 1 x +\sum_{n=1}^\infty \frac 1 n - \frac 1 { n+x},\\ \psi'(x) = \sum_{n=0}^\infty \frac 1 {(n+x)^2}$$ and the derived fact that $\psi(\tfrac 1 2 )+\gamma = -2\ln 2$ and $\psi^{(k)}(\tfrac 1 2)=(-1)^{k+1}k!(2^{k+1}-1)\zeta(k+1)$ for $k\ge 1$.
For $C$, we can use the same method as used in the evaluation of $B$. It holds that \begin{align*} C =& \frac {15}{16} \left[\frac{\partial^4}{\partial x^3\partial y} \text{B}(x,y)\right]_{x=1, y = 0^+}\\ =&\frac {15} {16}\left[-\frac 1 2 \psi''''(1) + \psi'''(1)\big[\psi(1) + \gamma\big] + 3\psi'(1)\psi''(1)\right]\\ =&\frac{15}{16}\left[12\zeta(5) -6\zeta(2)\zeta(3)\right]\\ =&\frac {45}{4}\zeta(5) -\frac {15\pi^2 \zeta(3)}{16} \end{align*} where $\psi(1) +\gamma = 0$, $\psi'(1) = \zeta(2)$, $\psi''(1) = -2\zeta(3)$ and $\psi''''(1) = -24\zeta(5)$ are used.
Combining $A,B,C$, we have that $$J =A+B-C= \frac{279}{16}\zeta(5) -\frac{7\pi^2\zeta(3)}{8} - \frac{\pi^4 \ln 2}{8}$$ and $$ \boxed{I_4 = -\frac{\pi^4 \ln 2}{8} - J = -\frac{279}{16}\zeta(5)+\frac{7\pi^2\zeta(3)}{8}} $$
Finally, these evaluate $\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =\frac 1 2\big[-I_1-I_2+I_3-I_4\big]$ as follows.
\begin{align*} \int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =&\ 6\text{Li}_5(1/2) + 6\ln 2\ \text{Li}_4(1/2)-\frac{81}{16}\zeta(5)-{7\pi^2 \over 16}\zeta(3)\\ &+\frac{21\ln^2 2}{8}\zeta(3)- \frac{1}{6}\pi^2\ln^3 2+\frac{1}{5}\ln^5 2. \end{align*}
Using the identity given in the OP, we get the desired integral $I$
\begin{align*} \int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx = &-2\text{Li}_5(1/2) -2\ln 2\ \text{Li}_4(1/2)+\frac{27}{32}\zeta(5) +\frac{7\pi^2}{48}\zeta(3)-\frac{7\ln^2 2}{8}\zeta(3) \\ &-\frac{\pi^4\ln 2}{144} +\frac{\pi^2\ln^3 2}{12} - \frac{7\ln^5 2}{60}. \end{align*}
Here is the Cornel's reduction of the $\operatorname{arctanh}$ integral to manageable integrals by using clever integrations by parts and exploiting Dilogarithm reflection formula in $(5)$ here http://mathworld.wolfram.com/Dilogarithm.html.
$$I=\frac{1}{16} \underbrace{\int_0^1 \frac{\text{Li}_2(x) \log ^2(x)}{x} \textrm{d}x}_{\text{known}}-\frac{1}{16} \underbrace{\int_0^1 \frac{\text{Li}_2(x) \log ^2(2-x)}{x} \textrm{d}x}_{\text{may use series with powers of 2 in denominator}}\\-\frac{1}{4} \underbrace{\int_0^1 \frac{\text{Li}_2(x) \log ^2(x)}{2-x} \textrm{d}x}_{\text{may use series with powers of 2 in denominator}}-\frac{3}{8} \underbrace{\int_0^1 \frac{\text{Li}_2(x) \log (x) \log (2-x)}{2-x} \textrm{d}x}_{\text{may use series with powers of 2 in denominator}}\\+\frac{5}{64} \underbrace{\int_0^1 \frac{\log ^4(1-x)}{x} \textrm{d}x}_{\text{known}}-\frac{1}{12} \underbrace{\int_0^1 \frac{\log (1-x) \log ^3(1+x)}{x} \textrm{d}x}_{\text{known}}\\-\frac{7}{16} \underbrace{\int_0^1 \frac{\log^2(1-x) \log (x) \log (1+x)}{1+x} \textrm{d}x}_{\text{Calculated in (Almost) Impossible Integrals, Sums, and Series}}-\frac{1}{4}\underbrace{\int_0^1\frac{\log^3(x)\log(1-x)}{2-x}\textrm{d}x}_{\text{may use series with powers of 2 in denominator}}\\-\frac{7}{32} \underbrace{\int_0^1 \frac{\log ^2(1-x) \log ^2(1+x)}{x} \textrm{d}x}_{\text{known}}.$$
Some of these integrals are already known. The harder ones can be easily reduced to advanced alternating harmonic series of weight $5$ or to advanced harmonic series of weight $5$ with powers of $2$ in denominator (and these ones are all already known). Also, playing with these integrals it's possible to get even more shortcuts. With this form of the main integral to calculate you're done.
A NOTE: The integral $\displaystyle \int_0^1 \frac{\log^2(1-x) \log (x) \log (1+x)}{1+x} \textrm{d}x$ may be found calculated in the book, (Almost) Impossible Integrals, Sums, and Series, pages $527$-$528$. In general, the integrals above where it is suggested a reduction to the series with powers of $2$ in denominator may also be rearranged and reduced to alternating harmonic series of weight $5$. End of story.
Best Answer
$$I:=\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx\overset{\large x\to\frac{1-x}{1+x}}=\int_0^1 \frac{1}{1+x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx$$ $$J:=\int_0^1 \frac{1}{1-x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx$$ $$I=\frac12\left((I+J)+(I-J)\right)=\boxed{\int_0^1 \frac{1}{1+x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx=\frac{13}{8}\zeta(2)\ln 2-\frac{33}{32}\zeta(3)}$$ $$J=\frac12\left((I+J)-(I-J)\right)=\boxed{\int_0^1 \frac{1}{1-x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx=\frac58\zeta(2)\ln 2+\frac{19}{32}\zeta(3)}$$
$$I-J=-2\int_0^1\frac{x}{1-x^2}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx\overset{\large \frac{1-x^2}{1+x^2}\to x}=\color{blue}{\int_0^1 \frac{\operatorname{Li}_2(x)}{1+x}dx}-\color{red}{\int_0^1 \frac{\operatorname{Li}_2(x)}{x}dx}$$ $$\overset{\color{blue}{IBP}}=\color{blue}{\zeta(2)\ln 2+\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx}-\color{red}{\sum_{n=1}^\infty \frac{1}{n^2}\int_0^1 x^{n-1}dx}=\boxed{\zeta(2)\ln 2-\frac{13}{8}\zeta(3)}$$ See here for the first integral.
$$I+J=2\int_0^1 \frac{1}{1-x^2}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx\overset{\large \frac{1-x^2}{1+x^2}\to x}=\int_0^1 \frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}dx$$ Now we're going to use the following result: $$\int_0^1 \frac{\ln t}{t-\frac{1}{x}}dt=- \sum_{n=1}^\infty x^n\int_0^1 t^{n-1}\ln t\,dt=-\sum_{n=1}^\infty x^n \left(\frac{1}{n}\right)'=\sum_{n=1}^\infty \frac{x^n}{n^2}=\operatorname{Li}_2(x)$$ $$\Rightarrow I+J=- \int_0^1\int_0^1 \frac{\ln t}{(1-tx)\sqrt{1-x^2}}dx dt\overset{x\to \sin x}=- \int_0^1\ln t\int_0^\frac{\pi}{2} \frac{1}{1-t\sin x}dxdt$$ $$=-\int_0^1\ln t \ \frac{\frac{\pi}{2}+\arcsin t}{\sqrt{1-t^2}} dt\overset{t=\sin x}=-\int_0^\frac{\pi}{2}\left(\frac{\pi}{2}+x\right)\ln (\sin x) dx=\boxed{\frac94\zeta(2)\ln 2 -\frac7{16}\zeta(3)}$$ See here and here for the above integrals.
Also to prove a leftover integral from above we will consider: $$f(a)=\int_0^\pi \frac{dx}{1+\sin a\sin x}\overset{\tan\frac{x}{2}=y}=\int_0^\infty \frac{dy}{1+y^2+2\sin a y}$$ $$=\int_0^\infty \frac{dy}{(\sin a+y)^2+\cos^2 a}=\frac{1}{\cos a}\arctan\left(\frac{\sin a+y}{\cos a}\right)\bigg|_0^\infty=\frac{\frac{\pi}{2}-a}{\cos a}$$ $$\Rightarrow \int_0^\frac{\pi}{2}\frac{1}{1-t\sin x}dx=\frac12 f(-\arcsin t)=\frac{\frac{\pi}{2}+\arcsin t}{\sqrt{1-t^2}}$$