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How can i evaluate $$\int _0^1\frac{\ln \left(1-x\right)\ln \left(1-x^4\right)}{x}\:dx$$ WolframAlpha offers no closed form and it numerically approximates it to $1.07801$.

I'd appreciate any hints.

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There is closed form for the antiderivative but it would require pages to write (have a look here).

Using the bounds $$\int _0^1\frac{\ln \left(1-x\right)\ln \left(1-x^4\right)}{x}\:dx=\frac{67 }{32}\zeta (3)-\frac{\pi }{2}C$$

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$$I=\int_0^1\frac{\ln(1-x)\ln(1-x^4)}{x}dx=-\sum_{n=1}^\infty\frac1n\int_0^1 x^{4n-1}\ln(1-x)dx=\sum_{n=1}^\infty\frac{H_{4n}}{4n^2}$$

use $$\sum_{n=1}^\infty f(2n)=\frac12\sum_{n=1}^\infty f(n)+\frac12\sum_{n=1}^\infty (-1)^nf(n)\tag1$$

$$I=\frac12\sum_{n=1}^\infty\frac{H_{2n}}{n^2}+\frac12\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^2}=\frac12S_1+\frac12S_2$$

use (1) again for the $S_1$ and use $\sum_{n=1}^\infty (-1)^nf(2n)=\Re\sum_{n=1}^\infty i^n f(n)$ for $S_2$.

Hint: we need the identity to calculate $S_2$.

$$\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$

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