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

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).

• 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. – Lada Dudnikova Apr 19 at 19:58
• 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 $0$. – Lada Dudnikova Apr 19 at 20:13
• \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 – FDP Apr 21 at 9:31
• Hint: $$\displaystyle \int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2\ln\left(\frac{a+1}{a-1}\right)+\operatorname{Li_2}\left(\frac2{1-a}\right)-\operatorname{Li_2}\left(\frac1{1-a}\right)$$ – nospoon Apr 24 at 21:47

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.
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.