3

Question $$I=\int^1_0\frac{\ln(1+x)\ln(1+x^3)}{x}\mathrm{d}x$$

My approach so far

Let $$f(z)=\frac{\ln(1+z)\ln(1+z^3)}{z},\ I=\int^1_0f(z)\ \mathrm{d}z$$The main idea is to integrate $f(z)$ along three paths and obtain an algebraic equation with $I$.

The first path is simply the interval $[-1,1]$. Since the contributions from the indentations around the branch points tend to $0$, we may change the path to $e^{\pi i[0,1]}$. \begin{align} \int^1_{-1}f(z)\ \mathrm{d}z &=I-\int^1_0\frac{\ln(1-x)\ln(1-x^3)}{x}\mathrm{d}x\\ &=\mathrm{Im}\int^\pi_0\mathrm{Log}(1+e^{i\theta})\mathrm{Log}(1+e^{3i\theta})\ \mathrm{d}\theta\\ &=\frac{3}{2}\int^\frac{\pi}{3}_0\theta\ln\left(2\cos\frac{\theta}{2}\right)\mathrm{d}\theta+\frac{1}{2}\int^\frac{\pi}{3}_0\theta\ln\left(2\cos\frac{3\theta}{2}\right)\mathrm{d}\theta+\int^\pi_{\frac{\pi}{3}}\left(\frac{3}{2}\theta-\pi\right)\ln\left(2\cos\frac{\theta}{2}\right)\mathrm{d}\theta \end{align}The second path: \begin{align} \mathrm{Re}\int_{e^{\pi i/3}[0,1]}f(z)\ \mathrm{d}z &=\mathrm{Re}\int^1_0\frac{\ln(1+xe^{\pi i/3})\ln(1-x^3)}{x}\mathrm{d}x\\ &=\frac{1}{2}\int^1_0\frac{\ln(1+x+x^2)\ln(1-x^3)}{x}\mathrm{d}x\\ &=\frac{\zeta(3)}{3}-\frac{1}{2}\int^1_0\frac{\ln(1-x)\ln(1-x^3)}{x}\mathrm{d}x\\ &=\frac{\zeta(3)}{3}+\frac{1}{2}\int^1_{-1}f(z)\ \mathrm{d}z-\frac{1}{2}I \end{align}

The third path: \begin{align} \mathrm{Re}\int_{e^{\pi i[1/3,0]}}f(z)\ \mathrm{d}z &=\mathrm{Im}\int^\frac{\pi}{3}_0\mathrm{Log}(1+e^{i\theta})\mathrm{Log}(1+e^{3i\theta})\ \mathrm{d}\theta\\ &=\frac{3}{2}\int^\frac{\pi}{3}_0\theta\ln\left(2\cos\frac{\theta}{2}\right)\mathrm{d}\theta+\frac{1}{2}\int^\frac{\pi}{3}_0\theta\ln\left(2\cos\frac{3\theta}{2}\right)\mathrm{d}\theta \end{align}-

Martin.s
  • 1
  • 1
  • 22

1 Answers1

6

Let $\gamma=[0,1]\cup e^{\pi i[0,1/3]}\cup e^{\pi i/3}[1,0]$. By Cauchy's integral theorem, $$\int_\gamma f(z)\ \mathrm{d}z=0$$Therefore, $$I=\frac{2}{9}\zeta(3)+\color{red}{\frac{3}{2}\int^\frac{\pi}{3}_0\theta\ln\left(2\cos\frac{\theta}{2}\right)\mathrm{d}\theta}+\color{blue}{\frac{1}{2}\int^\frac{\pi}{3}_0\theta\ln\left(2\cos\frac{3\theta}{2}\right)\mathrm{d}\theta}+\color{green}{\frac{1}{3}\int^\pi_{\frac{\pi}{3}}\left(\frac{3}{2}\theta-\pi\right)\ln\left(2\cos\frac{\theta}{2}\right)\mathrm{d}\theta}$$The red integral: \begin{align} \color{red}{\frac{3}{2}\int^\frac{\pi}{3}_0\theta\ln\left(2\cos\frac{\theta}{2}\right)\mathrm{d}\theta} &=-\frac{3}{2}\mathrm{Re}\int^{e^{\pi i/3}}_1\frac{\ln{z}\ln(1+z)}{z}\mathrm{d}z\\ &=-\frac{3}{2}\mathrm{Re}\left[\mathrm{Li}_3(-z)-\mathrm{Li}_2(-z)\ln{z}\phantom{\frac{}{}}\right]^{e^{\pi i/3}}_1\\ &\color{red}{=-\frac{9}{8}\zeta(3)-\frac{3}{2}\mathrm{Cl}_3\left(\frac{2\pi}{3}\right)+\frac{\pi}{2}\mathrm{Cl}_2\left(\frac{2\pi}{3}\right)}\\ \end{align}The blue integral: \begin{align} \color{blue}{\frac{1}{2}\int^\frac{\pi}{3}_0\theta\ln\left(2\cos\frac{3\theta}{2}\right)\mathrm{d}\theta} &=\frac{1}{18}\mathrm{Re}\int^1_{-1}\frac{\ln{z}\ln(1+z)}{z}\mathrm{d}z\\ &=\frac{1}{18}\mathrm{Re}\left[\mathrm{Li}_3(-z)-\mathrm{Li}_2(-z)\ln{z}\phantom{\frac{}{}}\right]^1_{-1}\\ &\color{blue}{=-\frac{7}{72}\zeta(3)} \end{align}The green integral: \begin{align} \color{green}{\frac{1}{3}\int^\pi_{\frac{\pi}{3}}\left(\frac{3}{2}\theta-\pi\right)\ln\left(2\cos\frac{\theta}{2}\right)\mathrm{d}\theta} &=2\mathrm{Re}\int^\frac{\pi}{2}_\frac{\pi}{6}\theta\ln(1+e^{2i\theta})\ \mathrm{d}\theta-\frac{2\pi}{3}\mathrm{Re}\int^\frac{\pi}{2}_\frac{\pi}{6}\ln(1+e^{2i\theta})\ \mathrm{d}\theta\\ &=\frac{1}{2}\mathrm{Re}\int^{e^{\pi i/3}}_{-1}\frac{\ln{z}\ln(1+z)}{z}\mathrm{d}z+\frac{\pi}{3}\mathrm{Im}\int^{e^{\pi i/3}}_{-1}\frac{\ln(1+z)}{z}\mathrm{d}z\\ &=\frac{1}{2}\mathrm{Re}\left[\mathrm{Li}_3(-z)-\mathrm{Li}_2(-z)\ln{z}\phantom{\frac{}{}}\right]^{e^{\pi i/3}}_{-1}-\frac{\pi}{3}\mathrm{Im}\ \mathrm{Li}_2(-e^{\pi i/3})\\ &\color{green}{=-\frac{1}{2}\zeta(3)+\frac{1}{2}\mathrm{Cl}_3\left(\frac{2\pi}{3}\right)+\frac{\pi}{6}\mathrm{Cl}_2\left(\frac{2\pi}{3}\right)} \end{align}Finally, $$I=\int^1_0\frac{\ln(1+x)\ln(1+x^3)}{x}\mathrm{d}x=\boxed{\displaystyle-\frac{3}{2}\zeta(3)-\mathrm{Cl}_3\left(\frac{2\pi}{3}\right)+\frac{2\pi}{3}\mathrm{Cl}_2\left(\frac{2\pi}{3}\right)}$$