Earlier today I saw this integral around here and gave it a try without success, unfortunately it got taken down so it didn't receive to much attention, but I think it's a nice integral (although it seems quite hard) and finding a closed form it's worth trying.

Evaluate $$I=\int_0^1 \frac{\ln(1+x)}{1+x^3}dx$$

I tried to work with it's sister integral:$$J=\int_0^1 \frac{\ln(1-x)}{1+x^3}dx\Rightarrow I-J=-\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^3}dx$$ Via the substitution $\frac{1-x}{1+x}=t$ we get: $$I-J=-\int_0^1 \frac{(1+t)\ln t}{1+3t^2}dt=-\sum_{n=0}^\infty (-3)^n \int_0^1 t^{2n}\ln t dt -\sum_{n=0}^\infty (-3)^n \int_0^1 t^{2n+1}\ln t dt $$ $$\int_0^1 x^k dx=\frac{1}{k+1}\overset{\frac{d}{dk}}\Rightarrow \int_0^1 x^k\ln x dx=-\frac{1}{(k+1)^2}$$ $$\Rightarrow I-J=\sum_{n=0}^\infty \frac{(-3)^n }{(2n+1)^2}+\sum_{n=0}^\infty \frac{(-3)^n }{(2n+2)^2}=\sum_{n=0}^\infty \frac{(-3)^n }{(2n+1)^2}+\frac{\operatorname{Li_2 (-3)}}{12}$$ But the first sum is quite ugly looking, so I don't think it's a great approach to the integral and I'm struggling for $I+J$ too.

I remember that OP tried to do partial fractions such as: $$I=\frac13 \int_0^1 \frac{\ln(1+x)}{1+x}dx-\frac13 \int_0^1 \frac{(x-2)\ln(1+x)}{x^2-x+1}dx$$ And he applied Feynman's trick for the second integral, yet the computation are unbearable and it didn't even spit out the correct result (hopefully there can be a nice closed form and I would like to see one).

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    $\begingroup$ This integral can be done with some complex integration. You can go from $1$ to $0$ and then go around 0 counterclockwise and go from $0$ to $1$. The function will be changed since $\ln(x)$ is multivalued function. Alas, I'm not fluent in this, and can't provide the details at once, I just saw this calculating on the seminars. $\endgroup$ – Lada Dudnikova Apr 19 at 19:58
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    $\begingroup$ sorry, the method will better work for $J$ because $\ln(1-x)$ has a branch point at $1$, the denominator doesn't change at $1$ and you ought not do the change of the beginning and the end of your integration path to $-1$ and $0$. $\endgroup$ – Lada Dudnikova Apr 19 at 20:13
  • $\begingroup$ \begin{align}\text{I}=-\int_{\frac{1}{2}}^1\frac{x\ln x}{1-3x+3x^2}\,dx\end{align} and ask Wolfram: integrate -x*log(x)/(1-3*x+3*x^2),x $\endgroup$ – FDP Apr 21 at 9:31
  • $\begingroup$ Hint: $$\displaystyle \int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2\ln\left(\frac{a+1}{a-1}\right)+\operatorname{Li_2}\left(\frac2{1-a}\right)-\operatorname{Li_2}\left(\frac1{1-a}\right)$$ $\endgroup$ – nospoon Apr 24 at 21:47

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.

  • $\begingroup$ Well, this is surprisingly nice, thank you! I was about to give up on this integral. $\endgroup$ – カカロット Apr 25 at 18:13
  • $\begingroup$ \begin{align}\ln^2 (2),\ln^2 (3),\frac{\pi \ln(3)}{6 \sqrt{3}} ,\operatorname{Li}_2 \left(\frac{1}{3}\right),\operatorname{\psi}_1 \left(\frac{1}{3}\right)\end{align} seem to be rationally independent. $\endgroup$ – FDP Apr 25 at 23:25
  • $\begingroup$ I'm not sure how to feel about this result... However, nicely done! (+1) $\endgroup$ – mrtaurho Apr 27 at 20:55

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.


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