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Possibly evaluate the integral? $$ \int_{0}^{1} \operatorname{Li}_3\left [ \left ( \frac{x(1-x)}{1+x} \right ) ^2 \right ] \text{d}x. $$ I came across this when playing with Legendre polynomials, and I am curious about the existence of this closed-form of the simple-looking integral. The relevant functional equations are much easier to arrive at, but leaving a massive calculation to reform the integral, out of my reach. I am appreciated for your help.
I somehow believe its existence, as I have derived the following expression:
Denoting $\chi_2(x)=\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)^2}$ and defining $$ f(x)=\operatorname{Li}_3(x^2) +4\ln\left ( 1-x^2 \right ) +8\left ( \frac{\text{artanh}(x)+\chi_2(x)}{x} -1 \right ), $$ one have $$ \int_{0}^{1} f\left ( \frac{x(1-x)}{1+x} \right ) \text{d}x =\frac{21}{2}\zeta(3)-4\ln(2)^2-16+\pi^2-\frac{2\pi^2}3\ln(2). $$ The idea is to find a function, whose Legendre-Fourier coefficients consisting of $$ \int_{0}^{1} \left ( \frac{x\left ( 1-x \right ) }{1+x} \right )^n \text{d} x, $$ and therefore it's an obvious task.

Setness Ramesory
  • 7,127

1 Answers1

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Not really an answer only solution from Mathematica (with MZIntegrate-package)

Using 2 times integration-by-part:

$$\int_0^1 \text{Li}_3\left(\left(\frac{x (1-x)}{1+x}\right)^2\right) \, dx=\\4 \int_0^1 \frac{(-1+x (2+x)) \left(x+\log \left(-1+x^2\right)\right) \left(2 \log (1+x)-\log \left(1+2 x+2 x^3-x^4\right)\right)}{x \left(-1+x^2\right)} \, dx=\\8-2 (1+2 C) \pi -8 \sqrt{2} \sinh ^{-1}(1)+4 \coth ^{-1}\left(\sqrt{2}\right) \sinh ^{-1}(1)-7 \sinh ^{-1}(1)^2-\frac{32}{3} \sinh ^{-1}(1)^3+\frac{64}{3} \sqrt{2} L_{8,2}(3)+\pi ^2 \left(\frac{1}{4}-\frac{2 \log (2)}{3}\right)+8 \sinh ^{-1}(1)^2 \log (2)-4 \log ^2(2)+\log (4096)-2 \text{Li}_2\left(1-\sqrt{2}\right)+\left(2-32 \sinh ^{-1}(1)\right) \text{Li}_2\left(-1+\sqrt{2}\right)-32 \text{Li}_3\left(-1+\sqrt{2}\right)+\frac{29 \zeta (3)}{2}$$

where: $ L_{8,2}(3)$ is the Dirichlet L-function.

Mathematica code:

 {NIntegrate[PolyLog[3, (x (1 - x)/(1 + x))^2], {x, 0, 1}], 
 8 - 2 (1 + 2 Catalan) \[Pi] - 8 Sqrt[2] ArcSinh[1] + 
 4 ArcCoth[Sqrt[2]] ArcSinh[1] - 7 ArcSinh[1]^2 - (32 ArcSinh[1]^3)/
 3 + 64/3 Sqrt[2]
 DirichletL[8, 2, 3] + \[Pi]^2 (1/4 - (2 Log[2])/3) + 
 8 ArcSinh[1]^2 Log[2] - 4 Log[2]^2 + Log[4096] - 
 2 PolyLog[2, 1 - Sqrt[2]] + (2 - 32 ArcSinh[1]) PolyLog[
 2, -1 + Sqrt[2]] - 32 PolyLog[3, -1 + Sqrt[2]] + (29 Zeta[3])/2 // N}


{0.015611, 0.015611}
Mariusz Iwaniuk
  • 6,863
  • Thanks for answering. Could you help me with $\int_{0}^{1} \frac{2}{1+2x} \left ( \ln(x)\ln(1+x)-\frac{1}{2}\ln(x)^2+ \operatorname{Li}_2\left (-x \right ) -\frac{\pi^2}{12} \right )\text{d}x?$ Just a result. I don't have MMA on hand now. – Setness Ramesory Nov 24 '24 at 02:55
  • @SetnessRamesory Gives: $-\frac{7 \zeta (3)}{6}-\frac{1}{6} \pi ^2 \ln (3)$. – Mariusz Iwaniuk Nov 24 '24 at 10:50
  • Apologize for disturbing you again. Mind integrating the couple? $$\int_{0}^{1} \left ( \frac{4}{1+2y}+\frac{2}{1-y} \right ) \ln(y)\left ( \operatorname{Li}2\left (
    -y\right ) +\ln (y)\ln(1+y)-\frac{1}{2} \ln(y)^2 +\frac{\pi^2}{12} \right ) \text{d}y,$$$$\int    {0}^{1} \left ( \frac{12}{1+2y}-\frac{6}{1+y} \right ) \ln(y)\left ( \operatorname{Li}_2\left (
    -y\right ) +\ln (y)\ln(1+y)-\frac{1}{2} \ln(y)^2 +\frac{\pi^2}{12} \right ) \text{d}y.$$     – Setness Ramesory Nov 30 '24 at 10:55
  • @SetnessRamesory For first: $-\frac{32 \text{Li}_4\left(\frac{1}{2}\right)}{3}+\frac{26 \pi ^4}{135}-\frac{4 \log ^4(2)}{9}+\frac{4}{9} \pi ^2 \log ^2(2)$ and second: $-32 \text{Li}_4\left(\frac{1}{2}\right)+\frac{26 \pi ^4}{45}-\frac{4 \log ^4(2)}{3}+\frac{4}{3} \pi ^2 \log ^2(2)$ – Mariusz Iwaniuk Nov 11 '25 at 10:45