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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?

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I think you misplaced the $\lambda$'s - check your question again. –  nbubis Feb 4 '13 at 14:45
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@nbubis: It looks fine to me. Note that the substitution $x=\frac{1}{u}$ yields $$\int_{0}^{\infty}\frac{\left(\log x\right)^{2}}{x^{2}+\lambda x+\lambda^{2}}dx=\int_{0}^{\infty}\frac{\left(\log u\right)^{2}}{1+\lambda u+\lambda u^{2}}du. $$ –  Eric Naslund Feb 4 '13 at 15:00
    
Did you use $(\ln x-\ln \lambda)^2=\ln^2x+\ln^2\lambda$ for the third equality? Not true. –  1015 Feb 4 '13 at 15:05
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@julien : it is true because $\int_0^{\infty}\frac{\ln x}{x^2+x+1}\text{d}x=0$ :) –  Ryan Feb 4 '13 at 15:16
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In fact, $\int_0^\infty\frac{\log(x)}{x^2+2ax+1}\mathrm{d}x=0$ for $|a|\lt1$. However, a mention in the proof would be less confusing. –  robjohn Feb 5 '13 at 8:23

5 Answers 5

up vote 2 down vote accepted

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

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I wonder if it's useful to use the integral representation of the Hurwitz zeta function for the 2 sums. (+1) –  Chris's sis Feb 5 '13 at 8:21
    
@Chris'ssister I haven't used Hurwitz zeta function frequently. I guess Hurwitz zeta function is equivalent to the Polylogarithm approach. Would using Hurwitz zeta function simplify things further? –  user17762 Feb 5 '13 at 8:24
    
I didn't check that yet, but it was my first thought that came to my mind when I firstly saw the sums. –  Chris's sis Feb 5 '13 at 8:27

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

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

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Note: I tried Maple 17 and Mathematica Wolfram alpha to evaluate $ \int_0^1\frac{\ln^3x}{x^2+x+1}\text{d}x $ but only numerical answers. –  Mhenni Benghorbal Aug 21 '13 at 18:54

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}

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@Chinny84 Sorry, I made a typo. I forgot to include $$ \sum_{k=0}^\infty x^{3k}=\frac1{1-x^3}. $$ –  Tunk-Fey Jul 31 at 13:06
    
I thought as much, but I didn't fell terribly confident with my conclusion. +1 –  Chinny84 Jul 31 at 13:08
    
@Chinny84 Thanks for your consideration. Really appreciate it. I am very lucky you were not too harsh to me (you could downvote my answer but you didn't, thank you). $\ddot\smile$ –  Tunk-Fey Jul 31 at 13:12
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I don't down vote unless it's completely wrong, and I followed the rest of the beautiful argument well. Also, these types of manipulation of definite integrals with series expansions I have not come across before joint MSE, so my toolbox has grown by answers such as yourself and co, so I was just trying to understand more so rather than pick up the typo(which I thought was a possibility also). :) –  Chinny84 Jul 31 at 13:47

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.

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Nice solution thx! Anyone knows a method on zeta function? –  Ryan Feb 4 '13 at 15:17
    
@Ryan: I tried playing around with this question once before, and I couldn't come up with anything other than the above solution. It is a tricky integral. –  Eric Naslund Feb 4 '13 at 23:35

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$

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

\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}

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