A NICE PAIR OF LOGARITHMIC INTEGRALS-PART I

Let´s evaluate today the pair of integrals


\int_0^1 \frac{\ln^2\left(x\right)}{x^2-x +1}\,dx=\frac{10\pi^3}{81 \sqrt{3}}


\int_0^1 \frac{\ln^2\left(x\right)}{x^2+x +1}\,dx=\frac{8\pi^3}{81 \sqrt{3}}


Consider the following integral:


\begin{aligned} I&=\int_0^1 \frac{\ln^2\left(x\right)}{x^2+2x \cos(\theta)+1}\,dx\\ &=\int_0^1 \frac{\ln^2\left(x\right)}{x^2+\left(e^{i \theta}+e^{-i \theta}\right)x+1}\,dx\\ &=\int_0^1 \frac{\ln^2\left(x\right)}{(1+xe^{-i \theta})(1+xe^{i \theta})}\,dx\\ &=\int_0^1 \ln^2\left(x\right) \left( \left(\sum_{k=0}^\infty (-1)^k e^{-i \theta n}x^k\right) \cdot \left(\sum_{k=0}^\infty (-1)^k e^{i \theta n}x^k\right)\right)\,dx\\ &=\int_0^1 \ln^2\left(x\right)\left(\frac{1}{\sin\left(\theta \right)} \sum_{k=1}^\infty(-1)^{k-1} \sin(k \theta) x^{k-1}\right)\,dx\\ &=\frac{1}{\sin\left(\theta \right)} \sum_{k=1}^\infty(-1)^{k-1} \sin(k \theta)\int_0^1 \ln^2\left(x\right)\right)x^{k-1}\,dx\\ &=\frac{1}{\sin\left(\theta \right)} \sum_{k=1}^\infty(-1)^{k-1} \sin(k \theta)\left( \frac{(-1)^2 2!}{k^3}\right)\\ &=\frac{2}{\sin\left(\theta \right)} \sum_{k=1}^\infty \frac{(-1)^{k-1} \sin(k \theta)}{k^3}\\ &=\frac{2}{\sin\left(\theta \right)}\left(\frac{\pi^2}{12}x-\frac{x^3}{12} \right) \qquad \blacksquare \end{aligned}

If we set    \theta=\frac{2 \pi}{3}

 

\begin{aligned} I&=\int_0^1 \frac{\ln^2\left(x\right)}{x^2-x +1}\,dx\\ &=\frac{4}{\sqrt{3}}\left(\frac{\pi^2}{12}\left(\frac{2 \pi}{3} \right)-\frac{1}{12} \left(\frac{2 \pi}{3} \right)^3\right)\\ &=\frac{\pi^3}{3 \sqrt{3}}\left({\frac{2}{3}-\frac{8 }{27} \right) \\ &=\frac{10\pi^3}{81 \sqrt{3}}\qquad \blacksquare \end{aligned}

If we set \theta=\frac{ \pi}{3}

\begin{aligned} I&=\int_0^1 \frac{\ln^2\left(x\right)}{x^2+x +1}\,dx\\ &=\frac{4}{\sqrt{3}}\left(\frac{\pi^2}{12}\left(\frac{ \pi}{3} \right)-\frac{1}{12} \left(\frac{ \pi}{3} \right)^3\right)\\ &=\frac{\pi^3}{3 \sqrt{3}}\left({\frac{1}{3}-\frac{1 }{27} \right) \\ &=\frac{8\pi^3}{81 \sqrt{3}}\qquad \blacksquare \end{aligned}




Appendix

\begin{aligned} P&= \left(\sum_{n=0}^\infty (-1)^ne^{-i \theta n}x^n\right) \cdot \left(\sum_{n=0}^\infty (-1)^ne^{i \theta n}x^n\right)\\ &=\left(\sum_{n=0}^\infty (-e^{-i \theta })^nx^n\right) \cdot \left(\sum_{n=0}^\infty (-e^{i \theta })^nx^n\right) \\ \end{aligned}


Taking the Cauchy product, We have that  a_k=(-e^{-i \theta })^k  and  b_k=(-e^{i \theta })^k ,   we therefore obtain


\begin{aligned} P&= \left(\sum_{n=0}^\infty (-1)^ne^{-i \theta n}x^n\right) \cdot \left(\sum_{n=0}^\infty (-1)^ne^{i \theta n}x^n\right)\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n (-e^{-i \theta })^k(-e^{i \theta })^{n-k} \right)x^n\\ &=\sum_{n=0}^\infty(-1)^n\left(\sum_{k=0}^n e^{i n \theta }e^{-2i k \theta } \right)x^n\\ &=\sum_{n=0}^\infty(-1)^n e^{i n \theta }\left(\sum_{k=0}^n e^{-2i k \theta } \right)x^n\\ &=\sum_{n=0}^\infty(-1)^n e^{i n \theta }\left(\frac{1-e^{-2i (n+1) \theta }}{1-e^{-2i \theta }} \right)x^n \qquad \left(\text{geometric sum} \right)\\ &=\sum_{n=0}^\infty(-1)^n e^{i n \theta }\frac{e^{i \theta}}{e^{i \theta}}\left(\frac{1-e^{-2i (n+1) \theta }}{1-e^{-2i \theta }} \right)x^n \\ &=\sum_{n=0}^\infty(-1)^n\left(\frac{e^{i (n+1)\theta}-e^{-i (n+1) \theta }}{e^{i \theta}-e^{-i \theta }} \right)x^n \\ &=\sum_{n=0}^\infty(-1)^n\left(\frac{\sin\left((n+1)\theta \right)}{\sin\left(\theta \right)} \right)x^n \\ &=\sum_{n=1}^\infty(-1)^{n-1}\left(\frac{\sin\left(n\theta \right)}{\sin\left(\theta \right)} \right)x^{n-1}\\ &=\frac{1}{\sin\left(\theta \right)} \sum_{n=1}^\infty(-1)^{n-1} \sin(n \theta) x^{n-1}\qquad \blacksquare\\ \end{aligned}

From this post we know that


\sum_{k=1}^\infty \frac{(-1)^{k-1} \sin(k \theta)}{k}=\frac{x}{2}(A.1)


Integrating both sides of (A.1) from 0 to x we obtain


-\sum_{k=1}^\infty \frac{(-1)^{k-1} \cos(k \theta)}{k^2}=\frac{x^2}{4}+C(A.2)


To find the constant, set x=0, then


\begin{aligned} -\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^2}=C\\ -\frac{\pi^2}{12}=C \end{aligned}

And we conclude that


\sum_{k=1}^\infty \frac{(-1)^{k-1} \cos(k \theta)}{k^2}=\frac{\pi^2}{12}-\frac{x^2}{4}(A.3)


Integrating both sides of (A.3) from 0 to x we obtain


\sum_{k=1}^\infty \frac{(-1)^{k-1} \sin(k \theta)}{k^3}=\frac{\pi^2}{12}x-\frac{x^3}{12}+C(A.4)


To find the constant, set x=\pi in (A.4), then


\begin{aligned} 0=\frac{\pi^3}{12}-\frac{\pi^3}{12}+C\\ C=0 \end{aligned}

And we obtain


\boxed{\sum_{k=1}^\infty \frac{(-1)^{k-1} \sin(k \theta)}{k^3}=\frac{\pi^2}{12}x-\frac{x^3}{12}}(A.5)

\int_{0}^{1} x^{m} \ln ^{n}(x) d x=\frac{(-1)^{n} n !}{(m+1)^{n+1}}



\begin{aligned} I&=\int_{0}^{1} x^{m} \ln ^{n}(x) d x\\ &=\int_{0}^{\infty} e^{-m x} \ln ^{n}\left(e^{-x}\right) e^{-x} d x \qquad ( x \to e^{-x}) \\ &=\int_{0}^{\infty} e^{-(m+1)x}(-1)^{n} x^{n} d x\\ &=\frac{(-1)^{n}}{(m+1)^{n+1}} \int_{0}^{\infty} e^{-x} x^{n} d x \qquad ((m+1) x \to x)\\ &=\frac{(-1)^{n} n!}{(m+1)^{n+1}} \qquad \blacksquare \end{aligned}

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