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I am trying to prove the below closed form $$\int_0^1 \frac {\ln(1-x)\ln(1-x^3)}{x} dx=\frac 53 \zeta(3)-\frac π3 \operatorname{Cl}_2(\frac {2\pi}{3}) $$ where $\operatorname{Cl}_2$ denotes the Clausen function.

You can easily break it into sum of two integrals $$\int_0^1 \frac {\ln^2(1-x)}{x} dx+\int_0^1\frac {\ln(1-x)\ln(x^2+x+1)}{x} dx$$ The first one is trivial as you can see. But the second one I am struggling with I tried to use the series representation of $\ln(1-x)$ and then use IBP but the resulting series is not less hard that the integral. Any ideas to do it ?

user170231
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Gauss
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2 Answers2

5

$$\int_0^1\frac{\log(1-x)\log(1-x^3)}{x}dx=-\sum_{n=1}^\infty\frac1n\int_0^1x^{3n-1}\log(1-x)dx $$

From this post we know that, $$\int_0^1x^{n-1}\log(1-x)dx=-\frac{H_n}{n}\\\therefore\stackrel{n\space\mapsto3n}{\implies}\int_0^1x^{3n-1}\log(1-x)dx=-\frac{H_{3n}}{3n}$$ And , thus : $$I=3\sum_{n=1}^\infty\frac{H_{3n}}{9n^2}=\frac23\left[-\frac12\frac{H_1}{1}-\frac12\frac{H_2}{2^2}+\frac{H_3}{3^2}-\dots +\frac12\frac{H_1}{1}+\frac{H_2}{2^2}+\dots\right ]\\=\frac23\sum_{n=1}^\infty\frac{H_n}{n^2}\cos\left(\frac23\pi n\right)+\frac13\sum_n^\infty\frac{H_n}{n^2}=\frac23\left( \zeta(3)+\operatorname{Re}\sum_{n=1}^\infty\frac{H_n}{n^2}\left(\exp(\frac23\pi i) \right)\right)$$

Now , from this post and some simplification we get: $$\sum_{n=1}^\infty\frac{H_{3n}}{9n^2}=\frac59\zeta(3)+\frac{2\pi^3}{81\sqrt3}-\frac{\pi}{27\sqrt3}\psi^{{I}}\left(\frac13\right)$$

The resault follows.

Antony Theo.
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See the answer as an alternative. Ths general form is given by $$\sum_{n=1}^{\infty} \frac{H_{\alpha n}}{n^{2}}=\frac{1}{2}\left(\alpha^{2}+\frac{3}{\alpha}\right) \zeta(3)-\frac{\pi}{2 \alpha} \sum_{k=1}^{\alpha-1} \cot \left(\frac{\pi k}{\alpha}\right) \psi^{(1)}\left(\frac{k}{\alpha}\right), $$

$${\sum_{n=1}^{\infty} \frac{H_{\alpha n}}{n^{4}}=\frac{1}{2}\left(\alpha^{4}+\frac{5}{\alpha}\right) \zeta(5)-\frac{\pi^2\alpha^2}{6}\zeta(3) -\frac{\pi}{12 \alpha} \sum_{k=1}^{\alpha-1} \cot \left(\frac{\pi k}{\alpha}\right) \psi^{(3)}\left(\frac{k}{\alpha}\right)}, $$ which explains $$ \begin{aligned} &\int_{0}^{1} \frac{\ln^2(1-x)}{x} \text{d}x=2\zeta(3),\\ &\int_{0}^{1} \frac{\ln(1-x)\ln(1+x)}{x} \text{d}x=-\frac{5}{8}\zeta(3),\\ &\int_{0}^{1} \frac{\ln(1-x)\ln(1+x^2)}{x} \text{d}x=\frac{23}{32}\zeta(3)-\frac{\pi}{2}G, \\ &\int_{0}^{1} \frac{\ln(1-x)\ln(1+x+x^2)}{x}\text{d}x=-\frac{1}{3} \zeta(3)- \frac{\pi}{18\sqrt{3} } \left [ \psi^{(1)}\left ( \frac{1}{3} \right )-\psi^{(1)}\left ( \frac{2}{3} \right )\right ],\\ \\ & \int_{0}^{1} \frac{\text{Li}_2^2(x)}{x}\text{d}x=\frac{\pi^2}{3}\zeta(3)-3\zeta(5),\\ &\int_{0}^{1} \frac{\text{Li}_2(x)\text{Li}_2(x^2)}{x}\text{d}x=\frac{\pi^2}{4}\zeta(3)-\frac{37}{16}\zeta(5),\\ &\int_{0}^{1} \frac{\text{Li}_2(x)\text{Li}_2(x^3)}{x}\text{d}x=\frac{2\pi^2}{9}\zeta(3)-\frac{124}{27}\zeta(5) +\frac{\pi}{324\sqrt{3} } \left [ \psi^{(3)}\left ( \frac{1}{3} \right )-\psi^{(3)}\left ( \frac{2}{3} \right )\right ],\\ &\int_{0}^{1} \frac{\text{Li}_2(x)\text{Li}_2(x^4)}{x}\text{d}x=\frac{5\pi^2}{24}\zeta(3)-\frac{1029}{128}\zeta(5)+2\pi\beta(4). \end{aligned} $$

Setness Ramesory
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