Integral $\int_0^1 \frac{\ln(1+x)}{1+x^3}dx$

Solution 1:

The 'sister integral' approach works for quite a few other integrals, but I do not know how to proceed in this particular case as well (note, by the way, that you have used the geometric series outside its radius of convergence in your calculations), so here's a sketch of the somewhat laborious brute-force method:

After the first partial fraction decomposition we have $I = \frac{1}{6} \ln^2(2) + \frac{1}{3} K$, where $$ K = \int \limits_0^1 \frac{(2-x) \ln(1+x)}{1 - x + x^2} \, \mathrm{d} x \, .$$ Now we introduce the sixth root of unity $\alpha \equiv \mathrm{e}^{\mathrm{i} \pi/3} = \frac{1 + \sqrt{3} \mathrm{i}}{2}$. It has the useful properties $\overline{\alpha} = 1- \alpha = - \alpha^2$, $\frac{\alpha}{1+\alpha} = \frac{\mathrm{i} \overline{\alpha}}{\sqrt{3}}$, $\frac{\overline{\alpha}}{1+\alpha} = - \frac{\mathrm{i}}{\sqrt{3}}$ and appears when doing partial fractions once more: $$ \frac{2 - x}{1 - x + x^2} = \frac{- \alpha}{x - \alpha} + \frac{-\overline{\alpha}}{x - \overline{\alpha}} = 2 \operatorname{Re} \left[\frac{- \alpha}{x - \alpha}\right] \, , \, x \in \mathbb{R} \, .$$ Therefore, \begin{align} K &= 2 \operatorname{Re} \left[\alpha \int \limits_0^1 \frac{- \ln(1+x)}{x - \alpha} \, \mathrm{d} x \right] \stackrel{t = x - \alpha}{=} 2 \operatorname{Re} \left[\alpha \int \limits_{-\alpha}^{\overline{\alpha}} \frac{- \ln(1+\alpha) - \ln \left(1 + \frac{t}{1+\alpha}\right) }{t} \, \mathrm{d} t \right] \\ &\hspace{-8pt}\stackrel{s = \frac{-t}{1+\alpha}}{=} 2 \operatorname{Re} \left[\alpha \ln(1+\alpha) \left[\ln(-\alpha) - \ln(\overline{\alpha})\right] + \alpha \int \limits_{\frac{\alpha}{1 + \alpha}}^{-\frac{\overline{\alpha}}{1+\alpha}} \frac{- \ln(1-s)}{s} \, \mathrm{d} s \right] \\ &= \frac{\pi^2}{18} + \frac{\pi \ln(3)}{2 \sqrt{3}} + \operatorname{Re} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i}}{\sqrt{3}}\right) - \operatorname{Li}_2 \left(\frac{\mathrm{i \overline{\alpha}}}{\sqrt{3}}\right)\right] - \sqrt{3} \operatorname{Im} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i}}{\sqrt{3}}\right) - \operatorname{Li}_2 \left(\frac{\mathrm{i \overline{\alpha}}}{\sqrt{3}}\right)\right] \, . \end{align} The dilogarithm values can now be simplified using the various functional equations. We obtain \begin{align} \operatorname{Re} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i}}{\sqrt{3}}\right) \right] &= \frac{1}{2} \operatorname{Li}_2 \left(\frac{1}{3}\right) - \frac{\pi^2}{24} + \frac{1}{8} \ln^2(3) \\ \operatorname{Re} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i \overline{\alpha}}}{\sqrt{3}}\right) \right] &= \frac{5 \pi^2}{72} - \frac{1}{8} \ln^2 (3) \, . \end{align} The imaginary parts are a bit harder to compute (related questions are found here and here), but a reasonably nice expression in terms of the trigamma function can be derived: $$ \operatorname{Im} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i}}{\sqrt{3}}\right) - \operatorname{Li}_2 \left(\frac{\mathrm{i \overline{\alpha}}}{\sqrt{3}}\right)\right] = - \frac{\pi^2}{18 \sqrt{3}} + \frac{\operatorname{\psi}_1 \left(\frac{1}{3}\right)}{12 \sqrt{3}} \, .$$ Thus we arrive at $$ K = \frac{1}{4} \ln^2 (3) + \frac{\pi \ln(3)}{2 \sqrt{3}} + \frac{1}{2} \operatorname{Li}_2 \left(\frac{1}{3}\right) - \frac{1}{12} \operatorname{\psi}_1 \left(\frac{1}{3}\right) $$ and $$ \boxed{I = \frac{1}{6} \ln^2 (2) + \frac{1}{12} \ln^2 (3) + \frac{\pi \ln(3)}{6 \sqrt{3}} + \frac{1}{6} \operatorname{Li}_2 \left(\frac{1}{3}\right) - \frac{1}{36}\operatorname{\psi}_1 \left(\frac{1}{3}\right)} \, .$$ It is of course up to you whether you consider this a nice result, but I have no idea how to simplify it any further.

Solution 2:

I'm not sure if this works but here's an attempt

Note that: (see here) $$\frac1{1+x^3}=\frac1{(x-e_0)(x-e_1)(x-e_2)}=-\frac13\left(\frac{e_0}{x-e_0}+\frac{e_1}{x-e_1}+\frac{e_2}{x-e_2}\right)$$ where $e_k=e^{\frac{i\pi}3(2k+1)}$. So $$I=-\frac13\sum_{k=0}^{2}e_k\int_0^1\frac{\log(x+1)}{x-e_k}dx=-\frac13\sum_{k=0}^2 e_kj_k$$ First notice that $x-e_1=x+1$ so $$j_1=\int_0^1\frac{\log(1+x)}{1+x}dx=\frac{\log^2 2}2$$ Then we see that $$j_\pm=\int_0^1\frac{\log(1+x)}{x-\frac12\mp \frac{i\sqrt3}2}dx$$ Then as commented by @nospoon, $$\int_0^1\frac{\log(1+x)}{a+x}dx=\log(2)\log\left(\frac{a+1}{a-1}\right)+\mathrm{Li}_2\left(\frac2{1-a}\right)-\mathrm{Li}_2\left(\frac1{1-a}\right)$$ So $$j_\pm=\log(2)\log\left(\pm\frac{i}{\sqrt3}\right)+\mathrm{Li}_2\left(-1\pm\frac{i}{\sqrt3}\right)-\mathrm{Li}_2\left(-\frac12\pm\frac{i}{2\sqrt3}\right)$$ $$j_\pm=\frac{\pm i\pi-\log3}2\log2+\mathrm{Li}_2\left(-1\pm\frac{i}{\sqrt3}\right)-\mathrm{Li}_2\left(-\frac12\pm\frac{i}{2\sqrt3}\right)$$ Where $j_+=j_0$ and $j_-=j_2$. So we have the monstrous result $$\begin{align} \log^2(2)-6I=&(1+i\sqrt3)\left[\frac{i\pi-\log3}2\log2+\mathrm{Li}_2\left(-1+\frac{i}{\sqrt3}\right)-\mathrm{Li}_2\left(-\frac12+\frac{i}{2\sqrt3}\right)\right]\\ +&(1-i\sqrt3)\left[\frac{-i\pi-\log3}2\log2+\mathrm{Li}_2\left(-1-\frac{i}{\sqrt3}\right)-\mathrm{Li}_2\left(-\frac12-\frac{i}{2\sqrt3}\right)\right]\\ =&(1+i\sqrt3)\left[\mathrm{Li}_2\left(-1+\frac{i}{\sqrt3}\right)-\mathrm{Li}_2\left(-\frac12+\frac{i}{2\sqrt3}\right)\right]\\ +&(1-i\sqrt3)\left[\mathrm{Li}_2\left(-1-\frac{i}{\sqrt3}\right)-\mathrm{Li}_2\left(-\frac12-\frac{i}{2\sqrt3}\right)\right]-\log(3)\log(2) \end{align}$$ I do not know how to simplify the $\mathrm{Li}_2$ terms but I'm sure others would be able to.