How to proceed in $I=\int_0^{\frac{\pi}{3}}\left(\ln\left(\frac{2\tan x}{\tan x +\sqrt{3}}\right)\right)^2\,dx$?

Question

$$I=\int_0^{\frac{\pi}{3}}\left(\ln\left(\frac{2\tan x}{\tan x +\sqrt{3}}\right)\right)^2dx$$

using $f(x) = f(a-x)$

$$I=\int_0^{\frac{\pi}{3}}\left(\ln\left(\frac{2\tan \left(\frac{\pi}{3}-x\right)}{\tan \left(\frac{\pi}{3}-x\right) +\sqrt{3}}\right)\right)^2 dx$$

using $\tan \left(\frac{\pi}{3}-x\right) = \frac{\tan\left(\frac{\pi}{3}\right)-\tan\left(x\right)}{1+\tan\left(\frac{\pi}{3}\right)\tan\left(x\right)}$

using $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$

$$I=\int_0^{\frac{\pi}{3}}\left(\ln\left(\frac{\sqrt3-\tan x}{\sqrt3+\tan x}\right)\right)^2 dx$$

Substituting $\tan (x) = \sqrt3 t$;

$$I=\sqrt3 \int_0^1 \ln \left(\frac{1-t}{1+t}\right)^2\frac{1}{1+3t^2} dt$$

Using this substitution $\frac{1-t}{1+t} = x$ as our bounds are $0$ and $1$;

$$I=\frac{\sqrt3}{2}\int_0^1 \frac{\ln^2(x)}{x^2-x+1} dx$$

I hit a deadend here and I am not able to find anything on this integral that can solve this.

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Edit :

Trying @Vaskara_GRek_O comment,

$$I=\frac{\sqrt3}{2}\int_0^1 \frac{\ln^2(x)}{x^2-x+1} \,dx$$

$$u= 1-x\to -\,du=\,dx \implies [0,1]\to[1,0]$$

$$I=\frac{\sqrt3}{2}\int_0^1 \frac{\ln^2(1-u)}{u^2-u+1} \,du$$

$$\ln^2(1-u)=2\sum_{n=1}^\infty \frac{H_n}{n+1}u^n$$

$$I=\sqrt3 \int_0^1 \frac{1}{u^2-u+1} \sum_{n=1}^\infty \frac{H_n}{n+1}u^n \,du$$

$$I=\sqrt3 \sum_{n=1}^\infty \frac{H_n}{n+1} \int_0^1 \frac{u^n}{u^2-u+1} \,du$$

$$\int_0^1 \frac{u^n}{u^2-u+1} \,du=\frac{1}{n+1}\;_3F_2\left(1,1,1+n;1+\frac{n}{2},\frac{3+n}{2},\frac{1}{4}\right)$$

$$I=\sqrt3 \sum_{n=1}^\infty \frac{H_n}{(n+1)^2} \;_3F_2\left(1,1,1+n;1+\frac{n}{2},\frac{3+n}{2},\frac{1}{4}\right)$$

How does one even compute this scary sum?

Taken from here

Answer 1

Note that \begin{align} K=&\int_0^1\frac{\ln^2 x}{x^2-x+1}dx\\ =&\int_0^1\frac{\ln^2 x}{x^3+1}dx + \int_0^1\frac{x\ln^2 x}{x^3+1}\overset{x\to 1/x}{dx} =\int_0^\infty\frac{\ln^2x}{x^3+1}dx\\ =& \ \frac{d^2}{da^2} \bigg(\int_0^\infty\frac{x^{a-1}}{x^3+1}dx\bigg)_{a=1} = \frac{d^2}{da^2} \left( \frac\pi3 \csc\frac{\pi a}3\right) _{a=1}=\frac{10\pi^3}{81\sqrt3} \end{align} Thus, $\int_0^{\frac{\pi}{3}}\ln^2\frac{2\tan x}{\tan x +\sqrt{3}}dx= \frac{\sqrt3}{2}K= \frac{5\pi^3}{81} $.

Answer 2

Rewrite $$I=\int\frac{\log ^2(x)}{x^2-x+1}\,dx=\int\frac{\log ^2(x)}{(x-a)(x-b)}\,dx$$where $$a=\frac{1+i \sqrt{3}}{2}\qquad \text{and} \qquad b=\frac{1-i \sqrt{3}}{2} $$ Use partial fraction decomposition $$I=\frac 1{a-b}\int \Big( \frac 1{x-a}-\frac 1{x-b}\Big)\log ^2(x)\,dx$$

For each integral, use twice integration by parts $$J=\int \frac{\log ^2(x)}{x-c}\,dx=-2 \text{Li}_3\left(\frac{x}{c}\right)+ 2 \log (x) \text{Li}_2\left(\frac{x}{c}\right)+ \log ^2(x) \log\left(-\frac{x-c}{c}\right)$$

Using the bounds, the result simplifies a lot.

I am sure that you can take it from here.

Answer 3

Noticing that $$ I=\frac{\sqrt{3}}{2} \int_0^1 \frac{\ln ^2 x}{x^2-x+1} d x= \frac{\sqrt{3}}{2} I^{\prime \prime}(0) $$ where $$ \begin{aligned} I(a)&= \int_0^1 \frac{x^a}{x^2-x+1} d x \\ &= \int_0^1 \frac{(x+1) x^a}{x^3+1} d x \\ &= \int_0^1 \frac{x^{a+1}}{x^3+1} d x+\int_0^1 \frac{x^a}{x^3+1} d x \\ &= \frac{1}{6}\left[\psi\left(\frac{a+5}{6}\right)-\psi\left(\frac{a+2}{6}\right)+\psi\left(\frac{a+4}{6}\right)-\psi\left(\frac{a+1}{6}\right)\right] \\&=\frac{1}{6}\left[\left(\psi\left(\frac{a+5}{6}\right)-\psi \left(\frac{a+1}{6}\right)\right)-\left(\psi\left(\frac{a+2}{6}\right)-\psi\left(\frac{a+4}{6}\right)\right)\right] \\ \end{aligned} $$ Differentiating w.r.t. $a$ twice at $a=0$ yields $$ \begin{aligned} I^{\prime \prime}(0)= & \frac{1}{216}\left[\left(\psi^{\prime \prime}\left(\frac{5}{6}\right)-\psi^{\prime \prime}\left(\frac{1}{6}\right)\right)+\left(\psi^{\prime \prime}\left(\frac{4}{6}\right)-\psi^{\prime \prime}\left(\frac{2}{6}\right)\right)\right] \\ = & \frac{1}{216}\left(8 \sqrt{3} \pi^3+\frac{8 \pi^3}{2 \sqrt{3}}\right)\\=&\frac{10 \pi^3}{81 \sqrt{3}} \end{aligned} $$ Plugging back gives $$\boxed{\int_0^{\frac{\pi}{3}}\left(\ln\left(\frac{2\tan x}{\tan x +\sqrt{3}}\right)\right)^2dx =\frac{5\pi^3}{81}} $$

Answer 4

In fact $$\begin{eqnarray} I&=&\frac{\sqrt3}{2}\int_0^1 \frac{\ln^2(x)}{x^2-x+1} dx=\frac{\sqrt3}{2}\int_0^1 \frac{(1+x)\ln^2(x)}{1+x^3} dx\\ &=&\frac{\sqrt3}{2}\int_0^1 \sum_{n=0}^\infty (-1)^n(1+x)x^{3n}\ln^2(x)dx\\ &=&\frac{\sqrt3}{2}\sum_{n=0}^\infty(-1)^n\int_0^1 (1+x)x^{3n}\ln^2(x)dx\\ &=&\sqrt3\sum_{n=0}^\infty(-1)^n\bigg(\frac1{(3n+1)^3}+\frac1{(3n+2)^3}\bigg)\\ &=&\sqrt3\sum_{n=0}^\infty(-1)^n\bigg(\frac1{(3n+1)^3}+\frac1{(3n+2)^3}+\frac1{(3n+3)^3}\bigg)-\sqrt3\sum_{n=0}^\infty(-1)^n\frac1{(3n+3)^3}\\ &=&\sqrt3\sum_{n=0}^\infty(-1)^n\frac1{(n+1)^3}-\frac{\sqrt3}{27}\sum_{n=0}^\infty(-1)^n\frac1{(n+1)^3}\\ &=&\frac{26\sqrt3}{27}\sum_{n=0}^\infty(-1)^n\frac1{(n+1)^3}=\frac{26\sqrt3}{27}\cdot\frac{3\zeta(3)}{4}=\frac{13\sqrt3}{36}\zeta(3). \end{eqnarray}$$ Here $$ \sum_{n=0}^\infty(-1)^n\frac1{(n+1)^3}=\frac{3\zeta(3)}{4} $$ is used.

Answer 5

\begin{align}J&=\frac{\sqrt{3}}{2}\underbrace{\int_0^1\frac{\ln^2 x}{x^2-x+1}dx}_{=K}\\ K&\overset{u=\frac1x}=\int_1^\infty\frac{\ln^2u}{u^2-u+1}du=\frac{1}{2}\int_0^\infty\frac{\ln^2u}{u^2-u+1}du\\ Q&=\int_0^\infty\int_0^\infty\frac{\ln^2 (xy)}{(x^2-x+1)(y^2-y+1)}dxdy\\ &\overset{u(x)=xy}=\int_0^\infty\int_0^\infty\frac{\ln^2 u}{(y^2-y+1)(y^2-uy+u^2)}dudy\\ &=\int_0^\infty\left[\frac{(u+1)\left(\frac{y^2-y+1}{y^2-uy+u^2}\right)}{2(u^3-1)}+\frac{\arctan\left(\frac{2y-u}{u\sqrt{3}}\right)}{(u^2+u+1)\sqrt{3}}+\frac{\arctan\left(\frac{2y-1}{\sqrt{3}}\right)}{(u^2+u+1)\sqrt{3}}\right]_0^\infty\ln^2u\, du\\ &=\underbrace{\int_0^\infty\frac{(u+1)\ln^3u}{u^3-1}du}_{=A}+\frac{4\pi}{3\sqrt{3}}\underbrace{\int_0^\infty\frac{\ln^2u}{u^2+u+1}du}_{=B}\\ A&=2\int_0^1 \frac{(u+1)\ln^3u}{u^3-1}du=2\int_0^1\frac{\ln^3u}{u-1}du-2\underbrace{\int_0^1\frac{u^2\ln^3u}{u^3-1}du}_{z=u^3}\\&=\frac{160}{81}\int_0^1\frac{\ln^3u}{u-1}du=\boxed{\frac{32\pi^4}{243}}\\ B&=\int_0^\infty\frac{\ln^2u}{u^2-u+1}du-\underbrace{\int_0^\infty\frac{2u\ln^2u}{u^4+u^2+1}du}_{z=u^2}=2K-\frac{1}{4}B=\boxed{\frac{8}{5}K}\\ Q&=\boxed{\frac{32\pi^4}{243}+\frac{32K\pi}{15\sqrt{3}}} \end{align} On the other hand, \begin{align}Q&=4K\left(\int_0^\infty\frac{1}{x^2-x+1}dx\right)=\frac{16K\pi}{3\sqrt{3}}\end{align} Therefore, \begin{align}K&=\frac{10\pi^3\sqrt{3}}{243}\\ J&=\boxed{\frac{5\pi^3}{81}} \end{align}

Answer 6

$$I=\int_0^{\frac{\pi}{3}}\left(\ln\left(\frac{2\tan x}{\tan x +\sqrt{3}}\right)\right)^2dx$$

Put $\tan x = \sqrt3 t$,

$$I=\sqrt3\int_0^1\frac{1}{1+3t^2}\ln^2\left(\frac{2t}{1+t}\right)\,dt$$

Put $\frac{2t}{1+t}=\frac{1}{u}$,

$$I=\frac{\sqrt3}{2}\int_1^\infty\frac{\ln^2(u)}{u^2-u+1}\,du$$

Using, $$\int_0^\infty=\int_0^1+\int_1^\infty\implies \int_1^\infty=\int_0^\infty-\int_0^1$$

$$I=\frac{\sqrt3}{2}\int_0^\infty\frac{\ln^2(u)}{u^2-u+1}\,du-\frac{\sqrt3}{2}\int_0^1\frac{\ln^2(u)}{u^2-u+1}\,du$$

$$I=\frac{\sqrt3}{4}\int_0^\infty\frac{\ln^2(u)}{u^2-u+1}\,du$$

Using the already derived Mellin Transform,

$$\color{red}{F(s,\alpha)=\int_0^\infty \frac{x^{s-1}}{x^2+2x\cos\alpha +1}\,dx=\frac{\pi\sin((1-s)\alpha)}{\sin\alpha\sin(\pi s)}}$$

$$\boxed{\color{blue}{I=\frac{\sqrt3}{4}\lim_{s\to 1}\frac{\partial^2}{\partial s^2}F\left(s,\frac{2\pi}{3}\right)=\frac{5\pi^3}{81}}}$$