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As title . I tried to work it out $$\int_0^1 \frac{1-2x}{x^2-x+1} \ln (1-x) dx$$

But could not find a solution.

Wolfram Alpha gives a solution using $Li(c)$ , but I’m not familiar with the special function, couldn’t understand it.

J.Dmaths
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athos
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    If WA uses the "logarithmic integral" function, Li, then most likely there is no way to work out the answer without using that function. You just have to bite the bullet, and learn about the logarithmic integral (or forget about it, and go on to the next problem). – Gerry Myerson Feb 12 '24 at 08:20

2 Answers2

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\begin{align} &\int_0^1 \frac{1-2x}{x^2-x+1} \ln (1-x)\overset{ibp}{ dx}\\ =& -\int_0^1\frac{\ln(1-x+x^2)}{1-x}\overset{x\to 1-x}{dx} =-\int_0^1\frac{\ln(1-x+x^2)}{x}{dx}\\ =& \int_0^1\frac{\ln(1+x)}{x}{dx}-\int_0^1\frac{\ln(1+x^3)}{x}\overset{x^3\to x}{dx}\\ =& \ \frac23\int_0^1\frac{\ln(1+x)}{x}{dx}=\frac{\pi^2}{18} \end{align} where $\int_0^1\frac{\ln(1+x)}{x}{dx}= \frac{\pi^2}{12}$.

Quanto
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If you cannot use polylogarithms yet, use a series expansion around $x=1$

$$\frac{1-2 x}{x^2-x+1}= -\sum_{n=0}^\infty \left(\sqrt{3} \sin \left(\frac{2 \pi n}{3}\right)+\cos \left(\frac{2 \pi n}{3}\right)\right)\,(x-1)^n$$ Using one integration by parts $$\int (x-1)^n \log(1-x)\,dx=\frac{(x-1)^{n+1} ((n+1) \log (1-x)-1)}{(n+1)^2}$$ $$\int_0^1 (x-1)^n \log(1-x)\,dx=-\frac{(-1)^n}{(n+1)^2}$$ which makes that your integral is $$\int_0^1 \frac{1-2x}{x^2-x+1} \log (1-x)\, dx=\sum_{n=0}^\infty (-1)^n \, \frac{\sqrt{3} \sin \left(\frac{2 \pi n}{3}\right)+\cos \left(\frac{2 \pi n}{3}\right) } {(n+1)^2}$$ Using the fist $100$ terms, you get $0.548211$ while the exact solution is $0.548311$

Claude Leibovici
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