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I'm interested in this integral: $$I=\int_0^1\ln^3\!\left(1+x+x^2\right)dx.\tag1$$ Can we prove that $$\begin{align}I&\stackrel{\color{gray}?}=\frac32\ln^33-9\ln^23+36\ln3+2\pi^2\ln3-\frac{4\pi^2}3+\left(8-\ln^23-4\ln3\right)\cdot\frac{\pi\sqrt3}2\\&-48-\frac{7\pi^3}{6\sqrt3}+(2-3\ln3)\cdot\psi^{(1)}\!\left(\tfrac13\right)+36\,{_4F_3}\!\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,\tfrac34\right)\end{align}\tag2$$ of find a simpler closed form?

Also, can we prove that $${_4F_3}\!\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,\tfrac34\right)\stackrel{\color{gray}?}=\frac{71\pi^3}{1296\sqrt3}+\frac{5\pi}{48\sqrt3}\ln^23-\frac1{\sqrt3}\,\Im\operatorname{Li}_3\!\left[\frac{(-1)^{\small1/6}}{\sqrt3}\right]\tag3$$ or find a simpler expression for it?

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  • $\begingroup$ I should mention that I found these conjectured forms using PSLQ algorithm and a set of candidate terms taken from known closed forms of similar integrals and simple variations of those terms, and then verified numerically with a high precision. $\endgroup$ – Vladimir Reshetnikov Sep 10 '15 at 0:52
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    $\begingroup$ Have you tried the obvious ? Writing $x^2+x+1$ as $\dfrac{1-x^3}{1-x}$, and then using $\ln\dfrac ab=\ln a-\ln b,$ or writing $x^2+x+1=(x-z)(x-\bar z)$, where $z$ and $\bar z$ are the cube roots of unity. $\endgroup$ – Lucian Sep 10 '15 at 2:54
  • $\begingroup$ How did you find this conjectured closed-form for the hypergeometric series? $\endgroup$ – user153012 Sep 14 '15 at 9:55
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    $\begingroup$ It was a pure guess inspired by this answer. Coefficient were found using PSLQ algorithm. $\endgroup$ – Vladimir Reshetnikov Sep 14 '15 at 16:50
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A Recurrence Relation

I will use the notation $$\mathcal{A}_n=\int^1_0\ln^n(1+x+x^2)\ {\rm d}x\ \ \ , \ \ \ \mathcal{B}_n=\int^\frac{\pi}{3}_\frac{\pi}{6}\ln^n\left(\frac{3}{4\cos^2{x}}\right)\ {\rm d}x$$ Integrating by parts and applying the substitution $\displaystyle x+\frac{1}{2}\mapsto \frac{\sqrt{3}}{2}\tan{x}$, it is evident that $$\mathcal{A}_n=n\sqrt{3}\mathcal{B}_{n-1}-2n\mathcal{A}_{n-1}+\frac{3}{2}(\ln{3})^n$$ We may use this recurrence to compute $\mathcal{A}_n$ for small positive integer values of $n$.


Evaluation of $\mathcal{A}_1$

We immediately have $$\mathcal{A}_1=1\times\sqrt{3}\times\frac{\pi}{6}-2\times 1\times 1+\frac{3}{2}\ln{3}=\frac{\pi}{2\sqrt{3}}+\frac{3}{2}\ln{3}-2$$


Evaluation of $\mathcal{A}_2$

We first compute $\mathcal{B}_1$ by exploiting a Fourier series. \begin{align} \mathcal{B}_1 &=\frac{\pi}{6}\ln{3}-2\int^\frac{\pi}{3}_\frac{\pi}{6}\ln(2\cos{x})\ {\rm d}x\\ &=\frac{\pi}{6}\ln{3}+2\sum^\infty_{n=1}\frac{(-1)^n}{n}\int^\frac{\pi}{3}_\frac{\pi}{6}\cos(2nx)\ {\rm d}x\\ &=\frac{\pi}{6}\ln{3}+\sum^\infty_{n=1}\frac{(-1)^n}{n^2}\left(\sin\left(\frac{2n\pi}{3}\right)-\sin\left(\frac{n\pi}{3}\right)\right)\\ &=\frac{\pi}{6}\ln{3}-\frac{1}{12\sqrt{3}}\sum^\infty_{n=0}\left[\frac{1}{\left(n+\frac{1}{3}\right)^2}-\frac{1}{\left(n+\frac{2}{3}\right)^2}\right]\\ &=-\frac{1}{6\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{\pi^2}{9\sqrt{3}}+\frac{\pi}{6}\ln{3} \end{align} Therefore, \begin{align} \mathcal{A}_2 &=2\sqrt{3}\left(-\frac{1}{6\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{\pi^2}{9\sqrt{3}}+\frac{\pi}{6}\ln{3}\right)-4\left(\frac{\pi}{2\sqrt{3}}+\frac{3}{2}\ln{3}-2\right)+\frac{3}{2}\ln^2{3}\\ &=-\frac{1}{3}\psi_1\left(\frac{1}{3}\right)+\frac{2\pi^2}{9}+\frac{\pi}{\sqrt{3}}\ln{3}+\frac{3}{2}\ln^2{3}-\frac{2\pi}{\sqrt{3}}-6\ln{3}+8 \end{align}


Simplification of Some ${\rm Li}_2,\ {\rm Li}_3$ Terms

I will simplify the terms $$\color{red}{{\rm Li}_2(e^{-\pi i/3})},\ \color{blue}{{\rm Li}_2(1-e^{2\pi i/3})},\ \color{green}{\Im{\rm Li}_3(e^{\pi i/3})},\ \color{purple}{\Im{\rm Li}_3(e^{-\pi i/3})},\ \color{brown}{\Im{\rm Li}_3(e^{2\pi i/3})}$$ The identities (for $0<\theta<2\pi$), \begin{align} \sum^\infty_{n=1}\frac{\cos(n\theta)}{n^2}&=\frac{\theta^2}{4}-\frac{\pi\theta}{2}+\frac{\pi^2}{6}\\ \sum^\infty_{n=1}\frac{\sin(n\theta)}{n^3}&=\frac{\theta^3}{12}-\frac{\pi\theta^2}{4}+\frac{\pi^2\theta}{6}\\ \end{align} (which can be derived by considering $\Im\ln(1-e^{i\theta})$ and integrating), give us \begin{align} \Im{\rm Li}_3(e^{\pm\pi i/3}) =&\pm\frac{5\pi^3}{162}\\ \Im{\rm Li}_3(e^{2\pi i/3}) &=\frac{2\pi^3}{81}\\ {\rm Li}_2(e^{-\pi i/3}) &=\frac{\pi^2}{36}-i\sum^\infty_{n=1}\frac{\sin(n\pi/3)}{n^2}\\ &=\frac{\pi^2}{36}-\frac{i\sqrt{3}}{2}\sum^\infty_{n=0}\left[\frac{1}{(6n+1)^2}+\frac{1}{(6n+2)^2}-\frac{1}{(6n+4)^2}-\frac{1}{(6n+5)^2}\right]\\ &=\frac{\pi^2}{36}-\frac{i}{24\sqrt{3}}\left(\psi_1\left(\frac{1}{6}\right)+\psi_1\left(\frac{1}{3}\right)-\psi_1\left(\frac{2}{3}\right)-\psi_1\left(\frac{5}{6}\right)\right)\\ &=\frac{\pi^2}{36}-i\left(\frac{1}{2\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{\pi^2}{3\sqrt{3}}\right) \end{align} Furthermore, the dilogarithm reflection formula states $${\rm Li}_2(z)+{\rm Li}_2(1-z)=\frac{\pi^2}{6}-\ln{z}\ln(1-z)$$ Hence \begin{align} {\rm Li}_2(1-e^{2\pi i/3}) &=\frac{\pi^2}{6}-\frac{2\pi i}{3}\left(\frac{\ln{3}}{2}-\frac{\pi i}{6}\right)-\left(-\frac{\pi^2}{18}+i\sum^\infty_{n=1}\frac{\sin(2n\pi/3)}{n^2}\right)\\ &=\frac{\pi^2}{6}-\frac{2\pi i}{3}\left(\frac{\ln{3}}{2}-\frac{\pi i}{6}\right)-\left(-\frac{\pi^2}{18}+\frac{i\sqrt{3}}{2}\sum^\infty_{n=0}\left[\frac{1}{(3n+1)^2}-\frac{1}{(3n+2)^2}\right]\right)\\ &=\frac{\pi^2}{9}-i\left(\frac{1}{3\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{2\pi^2}{9\sqrt{3}}+\frac{\pi}{3}\ln{3}\right) \end{align}


Evaluation of $\mathcal{A}_3$

Similarly, we start with the evaluation of $\mathcal{B}_2$. \begin{align} \mathcal{B}_2 &=\int^\frac{\pi}{3}_\frac{\pi}{6}\ln^2{3}-4\ln{3}\ln(2\cos{x})+4x^2+4\operatorname{Re}\ln^2(1+e^{2ix})\ {\rm d}x\\ &=-\frac{\ln{3}}{3\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{7\pi^3}{162}+\frac{2\pi^2}{9\sqrt{3}}\ln{3}+\frac{\pi}{6}\ln^2{3}+8\Re\sum^\infty_{n=1}\frac{(-1)^{n}H_{n-1}}{n}\int^\frac{\pi}{3}_\frac{\pi}{6}e^{2inx}\ {\rm d}x\\ &=-\frac{\ln{3}}{3\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{7\pi^3}{162}+\frac{2\pi^2}{9\sqrt{3}}\ln{3}+\frac{\pi}{6}\ln^2{3}-4\sum^\infty_{n=1}\frac{1}{n^3}\left(\sin\left(\frac{2\pi n}{3}\right)-\sin\left(\frac{\pi n}{3}\right)\right)\\ &\ \ \ \ \ +4\Im\sum^\infty_{n=1}\frac{H_{n}}{n^2}\left(e^{2\pi in/3}-e^{\pi in/3}\right)\\ &=-\frac{\ln{3}}{3\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{11\pi^3}{162}+\frac{2\pi^2}{9\sqrt{3}}\ln{3}+\frac{\pi}{6}\ln^2{3}\\ &\ \ \ \ \ +4\Im\left[{\rm Li}_3(z)-{\rm Li}_3(1-z)+{\rm Li}_2(1-z)\ln(1-z)+\frac{1}{2}\ln{z}\ln^2(1-z)+\zeta(3)\right]^{e^{2\pi i/3}}_{e^{\pi i/3}} \end{align} where I used the generating function of $\dfrac{H_n}{n^2}$. Using results derived in the previous section, \begin{align} &\ \ \ \ \Im\left[{\rm Li}_3(z)-{\rm Li}_3(1-z)+{\rm Li}_2(1-z)\ln(1-z)+\frac{1}{2}\ln{z}\ln^2(1-z)+\zeta(3)\right]^{e^{2\pi i/3}}_{e^{\pi i/3}}\\ &=\color{brown}{\frac{2\pi^3}{81}}-\color{green}{\frac{5\pi^3}{162}}+\color{purple}{\left(-\frac{5\pi^3}{162}\right)}-\Im{\rm Li}_3(1-e^{2\pi i/3})\\ &\ \ \ \ +\Im\color{blue}{\left(\frac{\pi^2}{9}-i\left(\frac{1}{3\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{2\pi^2}{9\sqrt{3}}+\frac{\pi}{3}\ln{3}\right)\right)}\left(\frac{\ln{3}}{2}-\frac{\pi i}{6}\right)\\ &\ \ \ \ -\Im\color{red}{\left(\frac{\pi^2}{36}-i\left(\frac{1}{2\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{\pi^2}{3\sqrt{3}}\right)\right)}\left(-\frac{\pi i}{3}\right)+\frac{\pi^3}{108}+\frac{\pi}{12}\ln^2{3}\\ &=-\Im{\rm Li}_3(1-e^{2\pi i/3})-\frac{\ln{3}}{6\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{\pi^3}{27}+\frac{\pi^2}{9\sqrt{3}}\ln{3}-\frac{\pi}{12}\ln^2{3} \end{align} Therefore \begin{align} \mathcal{B}_2 &=-4\Im{\rm Li}_3(1-e^{2\pi i/3})-\frac{\ln{3}}{\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{13\pi^3}{162}+\frac{2\pi^2}{3\sqrt{3}}\ln{3}-\frac{\pi}{6}\ln^2{3} \end{align} and finally, \begin{align} \mathcal{A}_3 &=3\sqrt{3}\left(-4\Im{\rm Li}_3(1-e^{2\pi i/3})-\frac{\ln{3}}{\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{13\pi^3}{162}+\frac{2\pi^2}{3\sqrt{3}}\ln{3}-\frac{\pi}{6}\ln^2{3}\right)\\ &\ \ \ \ -6\left(-\frac{1}{3}\psi_1\left(\frac{1}{3}\right)+\frac{2\pi^2}{9}+\frac{\pi}{\sqrt{3}}\ln{3}+\frac{3}{2}\ln^2{3}-\frac{2\pi}{\sqrt{3}}-6\ln{3}+8\right)+\frac{3}{2}\ln^3{3}\\ &=\color{darkorange}{-12\sqrt{3}\Im{\rm Li}_3(1-e^{2\pi i/3})+(2-3\ln{3})\psi_1\left(\frac{1}{3}\right)+\frac{3}{2}\ln^3{3}-\left(\frac{\sqrt{3}\pi}{2}+9\right)\ln^2{3}}\\ &\ \ \ \ \color{darkorange}{+(2\pi^2-2\sqrt{3}\pi)\ln{3}-\left(\frac{13\sqrt{3}\pi^3}{54}+\frac{4\pi^2}{3}-4\sqrt{3}\pi-36\ln{3}+48\right)} \end{align}

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I found the antiderivative: $$\begin{align}\int\ln^3\!\left(1+x+x^2\right)dx&=\xi\,\sqrt3\,\Big[\alpha^3-6\alpha^2+24\alpha-48\Big]\\&-\beta\,\sqrt3\,\Big[4\beta^2-3\alpha\ln3+6\alpha-24+6\ln3\Big]\phantom{\Huge|}\\&+6\,\sqrt3\,\Im\Big[(\alpha-2-2\beta\,i)\,\operatorname{Li}_2(\gamma)-2\,\operatorname{Li}_3(\gamma)\Big]\phantom{\Huge|}\end{align}$$ where $$\begin{align}&\color{maroon}{\alpha=\ln\!\left(1+x+x^2\right)}\\&\color{orange}{\beta=\arctan(2\,\xi)}\phantom{\Huge|}\\&\color{green}{\gamma=\frac12-i\,\xi}\phantom{\Huge|}\\&\color{blue}{\xi=\frac{1+2x}{2\sqrt3}}\phantom{\Huge|}\end{align}$$

It is valid and continuous at least for $x\ge0$, so it is good for our purposes.

Unfortunately, I cannot demonstrate any systematic approach that leads to this result, I got it with a series of lucky guesses and applications of PSLQ algorithm to determine rational coefficients, and finally proved its correctness using differentiation.

It yields the conjectured result modulo several polylogarithm identities that I'm still trying to prove.

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  • $\begingroup$ if we use the fact that $\ln{(1+x+x^2)} = \ln{(x-a)}+\ln{(x-\bar{a})}$, then the resulting integrals have easily computable antiderivatives in terms of trilogarithm and logarithms but it's lengthy $\endgroup$ – Oussama Boussif Sep 11 '15 at 10:40

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