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

Question

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?

Answer 1

Here is a closed form solution for the integral

$$\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x=\frac{8\sqrt{3}}{243}\pi^3 \sim 1.768047624. $$

I have introduced a general technique, which is based on partial fraction combined with the use of dilogarithm function, to solve the integral

$$ \int _{a }^{b }\!{\frac {\ln \left( tx + u \right) }{m{x}^{2}+nx+p}}{dx}.$$

In your case, instead of using the dilogarithm function, we will use the polylogarithm function $\operatorname{Li}_{s}(z)$ and the whole problem boils down to evaluate integrals of the form

$$ \int_{0}^{1} \frac{\ln(x)^2}{x-\alpha}= -2 Li_{3}\left(\frac{1}{a}\right), $$

where

$$ \operatorname{Li}_{s}(z) = \sum_{k=1}^{\infty}\frac{z^k}{k^s}. $$

Answer 2

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.

Answer 3

$$\int_0^1 \dfrac{\ln^2(x)}{1+x+x^2} = \int_0^1 \dfrac{1-x}{1-x^3} \ln^2(x) dx = \int_0^1 (1-x)\ln^2(x) \left(\sum_{k=0}^{\infty} x^{3k}\right) dx$$ 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, $$f = \dfrac{13}{27} \zeta(3) + \dfrac{2 \pi^3}{81 \sqrt3}; \,\,\,\,\,\,\,\, g = \dfrac{13}{27} \zeta(3) - \dfrac{2 \pi^3}{81 \sqrt3}$$

Answer 4

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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}

Answer 5

For $\lambda>0$,

\begin{align}I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x\end{align}

Perform the change of variable $x=\lambda y$,

\begin{align}I(\lambda)&=\lambda\int_0^{\infty}\frac{\ln ^2 \left(\lambda y\right)}{\left(\lambda y\right)^2+\lambda^2y+\lambda ^2}\text{d}y\\ &=\frac{1}{\lambda}\int_0^{\infty}\frac{\ln ^2 \left(\lambda y\right)}{ y^2+y+1}\text{d}y\\ &=\frac{1}{\lambda}\int_0^{\infty}\frac{\ln ^2 \lambda }{ y^2+y+1}\text{d}y+\frac{1}{\lambda}\int_0^{\infty}\frac{\ln ^2 y}{ y^2+y+1}\text{d}y\\ &=\frac{\ln ^2 \lambda}{\lambda}\left[\frac{2}{\sqrt{3}}\arctan\left(\frac{2y+1}{\sqrt{3}}\right)\right]_0^\infty+\frac{1}{\lambda}\int_0^{\infty}\frac{\ln ^2 y}{ y^2+y+1}\text{d}y\\ &=\frac{\ln ^2 \lambda}{\lambda}\left(\frac{\pi}{\sqrt{3}}-\frac{\pi}{3\sqrt{3}}\right)+\frac{1}{\lambda}\int_0^{\infty}\frac{\ln ^2 y}{ y^2+y+1}\text{d}y\\ &=\frac{2\pi\ln ^2 \lambda}{3\lambda\sqrt{3}}+\frac{1}{\lambda}\int_0^{\infty}\frac{\ln ^2 y}{ y^2+y+1}\text{d}y\\ \end{align}

Consider,

\begin{align}L&=\int_0^\infty \frac{\ln^2 x}{x^2+x+1}\,dx\end{align}

\begin{align}M&=\int_0^\infty\int_0^\infty \frac{\ln^2(xy)}{(x^2+x+1)(y^2+y+1)}\,dx\,dy\\ &=2L\int_0^\infty \frac{\ln x}{x^2+x+1}\,dx\\ &=2L\left[\frac{2}{\sqrt{3}}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right]_0^\infty\\ &=2L\left(\frac{\pi}{\sqrt{3}}-\frac{\pi}{3\sqrt{3}}\right)\\ &=\frac{4\pi}{3\sqrt{3}}L \end{align}

On the other hand, perfom the change of variable $u=yx$,

\begin{align}M&=\int_0^\infty \int_0^\infty\frac{y\ln^2 u}{(u^2+uy+y^2)(y^2+y+1)}\,du\,dy\\ &=\int_0^\infty \left[\frac{(u+1)\ln\left(\frac{y^2+y+1}{y^2+uy+u^2}\right)}{2(u^3-1)}-\frac{\arctan\left(\frac{2y+u}{\sqrt{3}u}\right)+\arctan\left(\frac{2y+1}{\sqrt{3}}\right)}{\sqrt{3}(u^2+u+1)}\right]_{y=0}^{y=\infty}\ln^2 u\,du\\ &=-\frac{\pi}{\sqrt{3}}\int_0^\infty \frac{\ln^2 u}{u^2+u+1}\,du+\int_0^\infty\frac{(u+1)\ln^3 u}{u^3-1}\,du+\frac{\pi}{3\sqrt{3}}\int_0^\infty \frac{\ln^2 u}{u^2+u+1}\,du\\ &=\int_0^\infty\frac{(u+1)\ln^3 u}{u^3-1}\,du-\frac{2\pi}{3\sqrt{3}}\int_0^\infty \frac{\ln^2 u}{u^2+u+1}\,du\\ &=\int_0^\infty\frac{(u+1)\ln^3 u}{u^3-1}\,du-\frac{2\pi}{3\sqrt{3}}L \end{align}

Therefore,

\begin{align}L&=\frac{\sqrt{3}}{2\pi}\int_0^\infty \frac{(u+1)\ln^3 u}{u^3-1}\,du\\ &=\frac{\sqrt{3}}{\pi}\int_0^1 \frac{(u+1)\ln^3 u}{u^3-1}\,du\\ &=\frac{\sqrt{3}}{\pi}\left(\int_0^1\frac{\ln^3 u}{u-1}\,du-\int_0^1\frac{u^2\ln^3 u}{u^3-1}\,du\right)\\ \end{align}

In the latter integral perform the change of variable $v=u^3$,

\begin{align}L&=\frac{\sqrt{3}}{\pi}\left(1-\frac{1}{3^4}\right)\int_0^1\frac{\ln^3 u}{u-1}\,du\\ &=\frac{80\sqrt{3}}{81\pi}\int_0^1\frac{\ln^3 u}{u-1}\,du\\ &=-\frac{80\sqrt{3}}{81\pi}\int_0^1 \left(\sum_{n=0}^\infty u^n\right)\ln^3 u\,du\\ &=-\frac{80\sqrt{3}}{81\pi}\sum_{n=0}^\infty\left(\int_0^1 u^n\ln^3 u\,du\right)\\ &=-\frac{80\sqrt{3}}{81\pi}\times -6\sum_{n=0}^\infty \frac{1}{(n+1)^4}\\ &=\frac{160\sqrt{3}}{27\pi}\zeta(4) \end{align}

If you know that,

\begin{align}\zeta(4)=\frac{\pi^4}{90}\end{align}

Therefore,

\begin{align}L&=\frac{160\sqrt{3}}{27\pi}\times \frac{\pi^4}{90}\\ &=\boxed{\frac{16\sqrt{3}}{243}\pi^3} \end{align}

Thus,

For $\lambda>0$,

\begin{align}\boxed{I(\lambda)=\frac{2\pi\ln ^2 \lambda}{3\lambda\sqrt{3}}+\frac{16\sqrt{3}}{243\lambda}\pi^3}\\\end{align}

Answer 6

Rewrite \begin{align} \int_0^1 \dfrac{\ln^2 x }{1+x+x^2}\ dx&= \int_0^1 \dfrac{1-x}{1-x^3} \ln^2x\ dx\\ & = \int_0^1\sum_{k=0}^{\infty} x^{3k}\ (1-x) \ln^2x\ dx\\ &=\sum_{k=0}^{\infty}\left(\int_0^1 x^{3k}\ln^2x\ dx-\int_0^1 x^{3k+1}\ln^2x\ dx\right).\tag1 \end{align} Using $$ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots\tag2 $$ then $(1)$ turns out to be \begin{align} \int_0^1 \dfrac{\ln^2 x }{1+x+x^2}\ dx&= \sum_{k=0}^{\infty}\left[\frac{2}{(3k+1)^{3}}-\frac{2}{(3k+2)^{3}}\right]\\ &= \frac1{3^3}\sum_{k=0}^{\infty}\left[\frac{2}{\left(k+\frac13\right)^{3}}-\frac{2}{\left(k+\frac23\right)^{3}}\right].\tag3\\ \end{align} Consider series representation of polygamma function $$ \psi_n(z)=(-1)^{n+1}\sum_{k=0}^\infty\frac{n!}{(k+z)^{n+1}}\quad;\quad\text{for}\ n>0\tag4 $$ and its reflection formula $$ \psi_n(1-z)+(-1)^{n+1}\psi_n(z)=(-1)^{n}\pi\frac{d^n}{dz^n}\cot(\pi z).\tag5 $$ Using $(4)$ and $(5)$, then $(3)$ becomes \begin{align} \int_0^1 \dfrac{\ln^2 x }{1+x+x^2}\ dx&= \frac1{3^3}\bigg[\psi_2\left(\frac23\right)-\psi_2\left(\frac13\right)\bigg]\\ &=\frac\pi{27}\left.\frac{d^2}{dz^2}\cot(\pi z)\right|_{z=\frac13}\\ &=\large\color{blue}{\frac{8\sqrt3}{243}\pi^3}. \end{align}