Definite integral, quotient of logarithm and polynomial: $I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x$

I was thinking this integral : $$I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x$$ What I do is use a Reciprocal subsitution, easy to show that: $$I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{\lambda ^2x^2+\lambda x+1}\text{d}x =\frac{1}{\lambda}\int_0^{\infty}\frac{(\ln x-\ln \lambda)^2}{x^2+x+1}\text{d}x \\ =\frac{\ln ^2\lambda}{\lambda}\int_0^{\infty}\frac{1}{x^2+x+1}\text{d}x+\frac{1}{\lambda}\int_0^{\infty}\frac{\ln ^2x}{x^2+x+1}\text{d}x \\ =\frac{\ln ^2\lambda}{\lambda}\frac{2\pi}{3\sqrt{3}}+\frac{2}{\lambda}\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x$$ But I hav no idea on process the remaining integral:( Anyone knows how to solve it?

THX guys! I came with another method might work for this: Recall the handy GF $$\frac{1}{x^2+x+1}=\frac{2}{\sqrt{3}}\sum_{k=0}^{\infty}\sin \left(\frac{2\pi}{3}\left(k+1\right)\right)x^k$$ Then we have : $$\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x=\frac{2}{\sqrt{3}}\sum_{k=0}^{\infty}\frac{2}{\left(k+1\right)^3}\sin \left(\frac{2\pi}{3}\left(k+1\right)\right)x^k$$ With $$\sum_{k=1}^{\infty}\frac{\sin (kx)}{k^3}=\frac{\pi ^2}{6}x-\frac{\pi}{4}x^2+\frac{x^3}{12}$$ What can we arrived?

Solution 1:

$$\int_0^1 \dfrac{\ln^2(x)}{1+x+x^2} = \int_0^1 \dfrac{1-x}{1-x^3} \ln^2(x) dx = \int_0^1 (1-x)\ln^2(x) \left(\sum_{k=0}^{\infty} x^{3k}\right) dx$$ Now note that $$\underbrace{\int_0^1 x^n \ln^2(x)dx = \int_{-\infty}^0 e^{nt} t^2 e^t dt}_{x \mapsto e^t} = \int_0^{\infty} t^2 e^{-(n+1)t}dt = \dfrac2{(1+n)^3}$$ Hence,$$\int_0^1 \dfrac{\ln^2(x)}{1+x+x^2} dx = \sum_{k=0}^{\infty} \left(\dfrac2{(1+3k)^3} - \dfrac2{(2+3k)^3} \right) = \dfrac{8 \pi^3}{81 \sqrt3}$$ Let us call $$\sum_{k=0}^{\infty} \dfrac1{(1+3k)^3} =f$$ and $$\sum_{k=0}^{\infty} \dfrac1{(2+3k)^3} =g$$ We are interested in $f-g$.

We have $$\text{Li}_3(\omega) = \sum_{k=1}^{\infty} \dfrac{\omega^k}{k^3} = \omega f + \omega^2 g + \dfrac{\zeta(3)}{27}$$ $$\text{Li}_3(\omega^2) = \sum_{k=1}^{\infty} \dfrac{\omega^{2k}}{k^3} = \omega^2 f + \omega g + \dfrac{\zeta(3)}{27}$$ where $\text{Li}_s(x)$ is the polylogarithm function defined as $$\text{Li}_s(x) = \sum_{k=0}^{\infty} \dfrac{x^k}{k^s}$$ Polylgarithm function satisfies a nice identity namely $$\text{Li}_n(e^{2 \pi ix}) + (-1)^n \text{Li}_n(e^{-2 \pi ix}) = - \dfrac{(2\pi i)^n}{n!}B_n(x)$$ where $B_n(x)$ are Bernoulli polynomials. Take $n=3$ and $x = 1/3$ to get that $$\text{Li}_3(\omega) - \text{Li}_3(\omega^2) = - \dfrac{(2\pi i)^3}{3!}B_3(1/3) = - \dfrac{(2\pi i)^3}{3!} \dfrac1{27} = \dfrac{8 \pi^3}{6 \times 27}i = \dfrac{4 \pi^3}{81}i$$ We also have that $$\text{Li}_3(\omega) - \text{Li}_3(\omega^2) = (\omega-\omega^2)(f-g) = \sqrt{3}i(f-g)$$ Hence, we get that $$f-g = \dfrac{4 \pi^3}{81 \sqrt3}$$ We can even get the values of $f$ and $g$ in terms of $\zeta(3)$. Note that $$f+g + \dfrac{\zeta(3)}{27} = \zeta(3) = \text{Li}_3(1)$$ Hence, $$f = \dfrac{13}{27} \zeta(3) + \dfrac{2 \pi^3}{81 \sqrt3}; \,\,\,\,\,\,\,\, g = \dfrac{13}{27} \zeta(3) - \dfrac{2 \pi^3}{81 \sqrt3}$$

Solution 2:

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${\tt\large\mbox{Just another one:}}$

\begin{align} &\color{#66f}{\large\int_{0}^{1}{\ln^{2}\pars{x} \over 1 + x + x^{2}}\,\dd x} =\lim_{\mu \to 0}\,\partiald[2]{}{\mu} \int_{0}^{1}{\pars{1 - x}x^{\mu} \over 1 - x^{3}}\,\dd x \\[3mm]&=\lim_{\mu \to 0}\,\partiald[2]{}{\mu}\int_{0}^{1} {x^{\mu/3} - x^{\pars{\mu + 1}/3} \over 1 - x}\,{1 \over 3}\,x^{-2/3}\dd x ={1 \over 3}\,\lim_{\mu \to 0}\,\partiald[2]{}{\mu}\int_{0}^{1} {x^{\pars{\mu - 2}/3} - x^{\pars{\mu - 1}/3} \over 1 - x}\,\dd x \\[3mm]&={1 \over 3}\,\lim_{\mu \to 0}\,\partiald[2]{}{\mu}\bracks{% \int_{0}^{1}{1 - x^{\pars{\mu - 1}/3} \over 1 - x}\,\dd x -\int_{0}^{1}{1 - x^{\pars{\mu - 2}/3} \over 1 - x}\,\dd x} \\[3mm]&={1 \over 3}\,\lim_{\mu \to 0}\,\partiald[2]{}{\mu} \bracks{\Psi\pars{\mu + 2 \over 3} - \Psi\pars{\mu + 1 \over 3}} ={1 \over 27}\bracks{\Psi''\pars{2 \over 3} - \Psi''\pars{1 \over 3}} \\[3mm]&={1 \over 27}\,\bracks{\pi\,\totald[2]{\cot\pars{\pi z}}{z}}_{z\ =\ ^1/_3}\ =\ {2\pi^{3} \over 27}\,\color{#c00000}{\cot\pars{\pi \over 3}} \color{#f0f}{\csc^{2}\pars{\pi \over 3}} ={2\pi^{3} \over 27}\,\color{#c00000}{{1 \over \root{3}}}\,\color{#f0f}{\pars{2 \over \root{3}}^{2}} \\[3mm]&=\color{#66f}{\large{8\root{3} \over 243}\,\pi^{3}} \approx 1.7680 \end{align}

Solution 3:

First, let me say that calculating this integral is not so easy, and according to my notes it evaluates to $\frac{16\pi^{3}}{81\sqrt{3}}.$

Hint: To evaluate $$\int_0^\infty \frac{(\log x)^2}{x^2+x+1}dx,$$ we take advantage of the fact that $$(a+1)^3-(a-1)^3 = 6a^2+2$$ and let $$f(x)=\frac{\left(\log x\right)^{3}}{x^{2}-x+1},$$ with a negative sign in the denominator. Consider the integral $\oint_{\mathcal{C}}f(x)dx$ where $\mathcal{C}$ is the counter clockwise oriented keyhole contour which is a circle of radius $R$, the circle of radius $\epsilon,$ and the two branches which extend to negative infinity on the upper and lower half plane. It is easy to see that as $R\rightarrow\infty,$ and as $\epsilon\rightarrow0,$ the two circles contribution goes to zero. In the limit, the integral on the branches becomes $$\int_{0}^{\infty}\frac{\left(\log x+\pi i\right)^{3}}{x^{2}+x+1}dx-\int_{0}^{\infty}\frac{\left(\log x-\pi i\right)^{3}}{x^{2}+x+1}dx.$$

I think you can solve it from here.