Conjecture $\int_0^1\frac{\ln^2\left(1+x+x^2\right)}x dx\stackrel?=\frac{2\pi}{9\sqrt3}\psi^{(1)}(\tfrac13)-\frac{4\pi^3}{27\sqrt3}-\frac23\zeta(3)$

I'm interested in the following definite integral: $$I=\int_0^1\frac{\ln^2\!\left(1+x+x^2\right)}x\,dx.\tag1$$ The corresponding antiderivative can be evaluated with Mathematica, but even after simplification is quite clumsy. It matches results of numerical integration, and its correctness can potentially be verified by hand using differentiation. So, we are assured that a closed form exists for $I$, albeit complicated one.

My program for a numerical search for closed forms found a much simpler candidate:

$$I\stackrel{\color{gray}?}=\frac{2\pi}{9\sqrt3}\psi^{\small(1)}\!\left(\tfrac13\right)-\frac{4\pi^3}{27\sqrt3}-\frac23\zeta(3).\tag2$$

Note that the trigamma value here can be expressed in terms of the dilogarithm of complex argument (see formula $(5)$ here) or of the $2^{nd}$ order harmonic number of fractional argument: \begin{align}\psi^{\small(1)}\!\left(\tfrac13\right)&=\frac{2\pi^2}3+2\sqrt3\,\Im\,\operatorname{Li}_2\!\left[(-1)^{\small1/3}\right],\tag3\\\psi^{\small(1)}\!\left(\tfrac13\right)&=\frac{\pi^2}6+9-H^{\small(2)}_{\small1/3}.\tag4\end{align} Can we prove $(2)$, preferably not going through the huge intermediate antiderivative?

One possible direction that I thought of is to factor the polynomial under the logarithm: $$I=\int_0^1\Big[\ln\!\left(x+(-1)^{\small1/3}\right)+\ln\!\left(x-(-1)^{\small2/3}\right)\Big]^2x^{-1}\,dx.\tag5$$ After expanding the square brackets, Mathematica can find a simpler antiderivative for it. Can we reach $(2)$ following this direction manually?

• As a bonus point, we can also try to prove $$\int_0^1\frac{\ln^2\!\left(1+x+x^2\right)}{x^2}\,dx=\frac{4\pi^2}9-\frac32 \ln^2 3+\frac\pi{\sqrt3}\ln3-\frac13\psi^{\small(1)}\!\left(\tfrac13\right). \tag{\diamond}$$ – Vladimir Reshetnikov Sep 9 '15 at 0:03
• Bonus point is easier. Integrating by parts kills one of the logarithms; Any expression of the form $R_1(x)\ln R_2(x)$ admits an easily computable antiderivative expressed in terms of dilogarithms (factor $R_2$ and decompose $R_1$ into partial fractions). – Start wearing purple Sep 9 '15 at 0:37
• I knew this was Vladimir before I opened the post... I'm always impressed with the formulas you come up with! – Bruno Joyal Sep 9 '15 at 2:01
• You may want to consider Catalan's constant $K = \rm{Cl}_2\big(\tfrac\pi2\big)$ and its cubic counterpart Gieseking's constant $\color{red}\kappa =\rm{Cl}_2\big(\tfrac\pi3\big)$ as having equal status. Hence, many integrals would have a closed-form like, $$\int_0^1 \frac{\ln^2(x^2+x+1)}{x}dx =\frac{4\pi\,\color{red}\kappa}9-\frac{2\zeta(3)}3$$ $$\int_1^\infty \frac{\ln(x)}{x^2+x+1}dx =\frac{4\sqrt3\,\color{red}\kappa}9$$ etc. More can be found in this list. – Tito Piezas III Jul 1 at 17:03

1. Replace $$\ln(1+x+x^2)$$ by $$\ln(1-x^3)-\ln(1-x)$$.

2. Two of the resulting integrals are easy to compute: $$\int_0^1\frac{\ln^2\left(1-x^3\right)dx}{x}=\frac23\zeta(3),\quad \int_0^1\frac{\ln^2\left(1-x\right)dx}{x}=2\zeta(3).$$

3. Mathematica computes and fullsimplifies the remaining nontrivial integral $$\int_0^1\frac{\ln\left(1-x\right)\ln\left(1-x^3\right)dx}{x}$$ to a one-line expression containing a sum of two trilogarithms $$\operatorname{Li}_3(z_1)+\operatorname{Li}_3(z_2)$$.

4. It so happens that $$z_1+z_2=1$$, hence thanks to Landen's identity the above sum is equal to $$-\operatorname{Li}_3\left(\frac{z_1}{z_1-1}\right)+\zeta\left(3\right)+\text{elementary}.$$

5. Finally, it so happens that $$\frac{z_1}{z_1-1}=e^{2\pi i/3}$$ and the corresponding trilogarithmic value is known in terms of $$\zeta(3)$$.

• Thanks! 4. I would hesitate to call the term $\zeta(3)$ elementary. I believe, it's still an open question. – Vladimir Reshetnikov Sep 9 '15 at 1:55
$$\frac{1}{2}\int_1^3 \frac{\log ^2(x)}{x-1} \ dx+\frac{1}{2}\int_1^3\frac{\log ^2(x)}{ (x-1) \sqrt{4 x-3}} \ dx,$$