# A generalized integral need help

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
@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
@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
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
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$$\int_0^1 \dfrac{\ln^2(x)}{1+x+x^2} = \int_0^1 \dfrac{1-x}{1-x^3} \ln^2(x) dx = \int_0^1 (1-x)\ln^2(x) \left(\sum_{k=0}^{\infty} x^{3k}\right) dx$$ 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
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 $$\int_{0}^{\infty}\frac{\left(\log x+\pi i\right)^{3}}{x^{2}+x+1}dx-\int_{0}^{\infty}\frac{\left(\log x-\pi i\right)^{3}}{x^{2}+x+1}dx.$$