Seeking alternative methods for $\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$

Question

I've solved this one by first tackling, $$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$ But i'd like to know other ways to solve it since the way i did it was a bit lengthy and not that straightforward.

Answer 1

$$\int_0^1\frac{\ln(x^2-x+1)}{x(1-x)}\ dx=\underbrace{\int_0^1\frac{\ln(x^2-x+1)}{1-x}\ dx}_{x\to 1-x}+\int_0^1\frac{\ln(x^2-x+1)}{x}\ dx$$

$$=2\int_0^1\frac{\ln(x^2-x+1)}{x}\ dx=2\underbrace{\int_0^1\frac{\ln(x^3+1)}{x}\ dx}_{x^3\to x}-2\int_0^1\frac{\ln(1+x)}{x}\ dx$$

$$=-\frac43\int_0^1\frac{\ln(1+x)}{x}\ dx=-\frac43\cdot\frac12\zeta(2)=-\frac{\pi^2}{9}$$

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A different way to calculate the last integral is to use the identity

$$\sum_{n=1}^{\infty}\frac{x^n}{n}\cos(an)=-\frac12\ln(1-2x\cos(a)+x^2)$$

Set $a=\frac{\pi}{3}$ we get

$$\ln(1-x+x^2)=-2\sum_{n=1}^\infty \frac{x^n}{n}\cos(n\pi/3)$$

so

$$\int_0^1\frac{\ln(1-x+x^2)}{x}\ dx=-2\sum_{n=1}^\infty \frac{\cos(n\pi/3)}{n}\int_0^1 x^{n-1}\ dx$$ $$=2\sum_{n=1}^\infty\frac{\cos(n\pi/3)}{n^2}=-\frac{\pi^2}{18}$$

where the last result follows from integrating both sides of the common identity

$$\sum_{n=1}^\infty \frac{\sin(nx)}{n}=\frac{\pi-x}{2}$$

Answer 2

On the path of Dennis Orton... \begin{align}J&=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx\\ &=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x}\:dx+\int _0^1\frac{\ln \left(t^2-t+1\right)}{1-t}\:dt\\ &\overset{x=1-t}=2\int _0^1\frac{\ln \left(x^2-x+1\right)}{x}\:dx\\ &=2\left(\int _0^1\frac{\ln \left(\frac{1+x^3}{1+x}\right)}{x}\:dx\right)\\ &=2\left(\int _0^1\frac{\ln \left(1+t^3\right)}{t}\:dt-\int _0^1\frac{\ln \left(1+x\right)}{x}\:dx\right)\\ &\overset{x=t^3}=\frac{2}{3}\int _0^1\frac{\ln \left(1+x\right)}{x}\:dt-2\int _0^1\frac{\ln \left(1+x\right)}{x}\:dx\\ &=-\frac{4}{3}\int _0^1\frac{\ln \left(1+x\right)}{x}\:dx\\ &=-\frac{4}{3}\left(\int_0^1\frac{\ln \left(1-t^2\right)}{t}\:dt-\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx\right)\\ &\overset{x=t^2}=-\frac{4}{3}\left(\frac{1}{2} \int _0^1\frac{\ln \left(1-x\right)}{x}\:dx-\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx\right)\\ &=\frac{2}{3}\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx\\ &=\frac{2}{3}\left(-\int_0^1 \left(\sum_{n=1}^\infty\frac{x^{n-1}}{n}\right)\,dx\right)\\ &=-\frac{2}{3}\sum_{n=1}^\infty\left(\int_0^1 \frac{x^{n-1}}{n}\,dx\right)\\ &=-\frac{2}{3}\sum_{n=1}^\infty\frac{1}{n^2}\\ &=-\frac{2}{3}\times\frac{\pi^2}{6}\\ &=\boxed{-\frac{\pi^2}{9}} \end{align} NB: I assume $\displaystyle \sum_{n=1}^\infty\frac{1}{n^2}=\zeta(2)=\frac{\pi^2}{6}$

PS: Sorry, i didn't see the soluton of Ali Shather

Answer 3

My approach. $$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$ $$=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx\:+\int _1^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$ Let $\displaystyle x=\frac{1}{t}$ for the $2$nd integral. $$\int _0^1\frac{\ln \left(t^2-t+1\right)}{t\left(1-t\right)}\:dt\:+\int _0^1\frac{\ln \left(t^2-t+1\right)}{t-1}\:dt-2\int _0^1\frac{\ln \left(t\right)}{t-1}\:dt$$ $$=\int _0^1\left(\frac{1}{t\left(1-t\right)}+\frac{1}{t-1}\right)\ln \left(t^2-t+1\right)\:dt\:-2\sum _{k=0}^{\infty }\frac{1}{\left(k+1\right)^2}$$ $$=\int _0^1\frac{\ln \left(t^2-t+1\right)}{t}\:dt\:-\frac{\pi ^2}{3}$$ $$\int _0^1\frac{\ln \left(t^3+1\right)}{t}\:dt-\int _0^1\frac{\ln \left(t+1\right)}{t}\:dt-\frac{\pi ^2}{3}$$ $$\int _0^1\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k}t^{3k-1}\:dt-\int _0^1\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k}t^{k-1}\:dt-\frac{\pi ^2}{3}$$ $$\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{3k^2}+\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+2}}{k^2}-\frac{\pi ^2}{3}$$ $$\frac{\pi ^2}{36}-\frac{\pi ^2}{12}-\frac{\pi ^2}{3}=-\frac{7\pi ^2}{18}$$ So, $$\boxed{\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx=-\frac{7\pi ^2}{18}}$$ In order to find the desired integral i used this previous expression. $$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx=\int _0^1\frac{\ln \left(t^2-t+1\right)}{t}\:dt\:-\frac{\pi ^2}{3}$$ And let $\displaystyle t=1-u$ $$-\frac{7\pi ^2}{18}=-\int _0^1\frac{\ln \left(u^2-u+1\right)}{u-1}\:du\:-\frac{\pi ^2}{3}$$ $$\boxed{\int _0^1\frac{\ln \left(u^2-u+1\right)}{u-1}\:du\:=\frac{\pi ^2}{18}}$$ Notice that on the $3$rd line all we have to do is to put in the result we just found and we'll be done. $$-\frac{7\pi ^2}{18}=\int _0^1\frac{\ln \:\left(t^2-t+1\right)}{t\left(1-t\right)}\:dt\:+\frac{\pi ^2}{18}-\frac{\pi ^2}{3}$$ And finally. $$\boxed{\int _0^1\frac{\ln \:\left(t^2-t+1\right)}{t\left(1-t\right)}\:dt\:=-\frac{\pi ^2}{9}}$$

Answer 4

I don't know if you like this or not. Let $$I(\alpha)=\int _0^1\frac{\ln \left[4\sin^2(\alpha)(x^2-x)+1\right]}{x\left(1-x\right)}\:dx, \alpha\in[0,\pi/6]$$ Then $I(0)=0, I(\frac{\pi}{6})=I$. Since \begin{eqnarray} I'(\alpha)&=&-\int _0^1\frac{4\sin(2\alpha)}{4\sin^2(\alpha)(x^2-x)+1}\,dx\\ &=&-\int _0^1\frac{2\cot(\alpha)}{x^2-x+\frac1{4\sin^2(\alpha)}}\,dx\\ &=&-\int _0^1\frac{2\cot(\alpha)}{(x-\frac12)^2+\frac1{4}\cot^2(\alpha)}\,dx\\ &=&-8\alpha. \end{eqnarray} So $$ I=\int_0^{\pi/6}I'(\alpha)\;d\alpha=-\int_0^{\pi/6}8\alpha\;d\alpha=-\frac{\pi^2}{9}. $$

Answer 5

Expanding in series and integrating gives $$\newcommand{\Li}{\operatorname{Li}} \int_0^1\log(1+\alpha x)\,\frac{\mathrm{d}x}x=-\Li_2(-\alpha)\tag1 $$ Setting $\omega=e^{i2\pi/3}$, we get $$ \begin{align} \int_0^1\frac{\log\left(x^2-x+1\right)}{x(1-x)}\,\mathrm{d}x &=\int_0^1\log\left(1-x+x^2\right)\left(\frac1x+\frac1{1-x}\right)\mathrm{d}x\tag2\\ &=2\int_0^1\log\left(1-x+x^2\right)\frac{\mathrm{d}x}x\tag3\\ &=2\int_0^1\left(\log\left(1+\omega x\right)+\log\left(1+\omega^2x\right)\right)\frac{\mathrm{d}x}x\tag4\\[6pt] &=-2\left(\Li(e^{i\pi/3})+\Li\left(e^{-i\pi/3}\right)\right)\tag5\\[6pt] &=-2\left(\frac{\pi^2}3-\frac{5\pi^2}{18}\right)\tag6\\ &=-\frac{\pi^2}9\tag7 \end{align} $$ Explanation:
$(2)$: partial fractions
$(3)$: substitute $x\mapsto1-x$ to get $\frac1{1-x}\mapsto\frac1x$
$(4)$: factor $1-x+x^2$
$(5)$: apply $(1)$
$(6)$: apply $(14)$ from this answer
$(7)$: simplfy

Answer 6

Here is another way to do it :

Let's start by using the substitution $ \left\lbrace\begin{aligned}t&=\frac{1-\sqrt{1-x}}{2}\\ \mathrm{d}t&=\frac{\mathrm{d}x}{4\sqrt{1-x}}\end{aligned}\right. $, we have : $$ \int_{0}^{\frac{1}{2}}{\frac{\ln{\left(1-t+t^{2}\right)}}{t\left(1-t\right)}\,\mathrm{d}x}=\int_{0}^{1}{\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x} $$

By using the substitution $ \left\lbrace\begin{aligned}t&=\frac{1+\sqrt{1-x}}{2}\\ \mathrm{d}t&=-\frac{\mathrm{d}x}{4\sqrt{1-x}}\end{aligned}\right. $, we have : $$ \int_{\frac{1}{2}}^{1}{\frac{\ln{\left(1-t+t^{2}\right)}}{t\left(1-t\right)}\,\mathrm{d}x}=\int_{0}^{1}{\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x} $$

Thus : $$ \int_{0}^{1}{\frac{\ln{\left(1-t+t^{2}\right)}}{t\left(1-t\right)}\,\mathrm{d}x}=2\int_{0}^{1}{\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x} $$

Let's work now on $ \int_{0}^{1}{\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x} $ shall we :

Substitute : $ \left\lbrace\begin{aligned} u &=\sqrt{1-x} \\ \mathrm{d}u &=-\frac{\mathrm{d}x}{2\sqrt{1-x}} \end{aligned}\right. $, we get :

$ \displaystyle\int_{0}^{1}{\displaystyle\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x}=2\displaystyle\int_{0}^{1}{\displaystyle\frac{\ln{\left(\frac{3+u^{2}}{4}\right)}}{1-u^{2}}\,\mathrm{d}u}$

Since : $ \left(\forall u\in\left[0,1\right]\right),\ \displaystyle\int_{0}^{1}{\displaystyle\frac{1-u^{2}}{\left(1-u^{2}\right)v-4}\,\mathrm{d}v}=\ln{\left(\displaystyle\frac{3+u^{2}}{4}\right)}$, we have :

\begin{aligned}\displaystyle\int_{0}^{1}{\displaystyle\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x}&=-2\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}{\displaystyle\frac{\mathrm{d}v\,\mathrm{d}u}{v u^{2}+4-v}}\\&=-2\displaystyle\int_{0}^{1}{\displaystyle\int_{0}^{1}{\displaystyle\frac{\mathrm{d}u}{v u^{2}+4-v}}\,\mathrm{d}v}\\&=-2\displaystyle\int_{0}^{1}{\displaystyle\frac{1}{\sqrt{v}\sqrt{4-v}}\displaystyle\int_{0}^{1}{\displaystyle\frac{\sqrt{\frac{v}{4-v}}}{1+\left(\sqrt{\frac{v}{4-v}}u\right)^{2}}\,\mathrm{d}u}\,\mathrm{d}v}\\ \displaystyle\int_{0}^{1}{\displaystyle\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x}&=-2\displaystyle\int_{0}^{1}{\displaystyle\frac{1}{\sqrt{v}\sqrt{4-v}}\arctan{\left(\sqrt{\frac{v}{4-v}}\right)}\,\mathrm{d}v}\end{aligned}

With the substitution : $ \left\lbrace\begin{aligned}\alpha &=\frac{\sqrt{v}}{2} \\ \mathrm{d}\alpha &=\displaystyle\frac{\mathrm{d}v}{4\sqrt{v}}\end{aligned}\right. $, and the fact that $ \left(\forall x\in\left]-1,1\right[\right),\ \arctan{\left(\displaystyle\frac{x}{\sqrt{1-x^{2}}}\right)}=\arcsin{x} $, we get :

$ \displaystyle\int_{0}^{1}{\displaystyle\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x}=-4\displaystyle\int_{0}^{\frac{1}{2}}{\displaystyle\frac{\arcsin{\alpha}}{\sqrt{1-\alpha^{2}}}\,\mathrm{d}\alpha}=-2\left[\arcsin^{2}{\alpha}\right]_{0}^{\frac{1}{2}}=-\displaystyle\frac{\pi^{2}}{18} \cdot $

Thus : $$ \int_{0}^{1}{\frac{\ln{\left(1-t+t^{2}\right)}}{t\left(1-t\right)}\,\mathrm{d}x}=-\frac{\pi^{2}}{9} $$

Answer 7

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Seeking $\ds{\color{blue}{\underline{\ alternative\ }}}$ methods for $\ds{\bbox[15px,#ffd]{\int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x\pars{1 - x}}\,\dd x}: \ {\Large ?}}$.

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\begin{align} &\bbox[15px,#ffd]{\int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x\pars{1 - x}}\,\dd x} = 2\int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x}\,\dd x \\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\, & -2\int_{0}^{1}\ln\pars{x}{2x - 1 \over x^{2} - x + 1}\,\dd x = -2\int_{0}^{1}\ln\pars{x}{2x^{2} + x - 1 \over 1 + x^{3}}\,\dd x \\[5mm] = &\ 2\int_{0}^{1}\ln\pars{x}{\sum_{n = 0}^{5}a_{n}x^{n} \over 1 - x^{6}}\,\dd x\quad \mbox{where}\quad \left\{\begin{array}{lcr} \ds{a_{0}} & \ds{=} & \ds{1} \\ \ds{a_{1}} & \ds{=} & \ds{-1} \\ \ds{a_{2}} & \ds{=} & \ds{-2} \\ \ds{a_{3}} & \ds{=} & \ds{-1} \\ \ds{a_{4}} & \ds{=} & \ds{1} \\ \ds{a_{5}} & \ds{=} & \ds{2} \end{array}\right. \end{align} Then, \begin{align} &\bbox[15px,#ffd]{\int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x\pars{1 - x}}\,\dd x} = 2\sum_{n = 0}^{5}a_{n}\int_{0}^{1}\ln\pars{x}{x^{n} \over 1 - x^{6}}\,\dd x \\[5mm] = &\ \left. -2\sum_{n = 0}^{5}a_{n}\, \partiald{}{\mu}\int_{0}^{1}{1 - x^{\mu} \over 1 - x^{6}}\,\dd x\,\right\vert_{\large\ \mu\ =\ n} \\[5mm] = &\ \left. -\,{1 \over 3}\sum_{n = 0}^{5}a_{n}\, \partiald{}{\mu}\int_{0}^{1}{x^{-5/6} - x^{\mu/6 - 5/6} \over 1 - x}\,\dd x\,\right\vert_{\large\ \mu\ =\ n} \\[5mm] = &\ \left. -\,{1 \over 3}\sum_{n = 0}^{5}a_{n}\, \partiald{}{\mu}\int_{0}^{1}{1 - x^{\mu/6 - 5/6} \over 1 - x}\,\dd x\,\right\vert_{\large\ \mu\ =\ n} \\[5mm] &\ \left. -\,{1 \over 3}\sum_{n = 0}^{5}a_{n}\, \partiald{}{\mu}\Psi\pars{{\mu \over 6} + {1 \over 6}}\right\vert_{\large\ \mu\ =\ n} \\[5mm] = &\ -\,{1 \over 18} \sum_{n = 0}^{5}a_{n} \Psi\, '\pars{{n \over 6} + {1 \over 6}} \\[5mm] = &\ -\,{1 \over 18}\left[% \color{#88f}{\Psi\, '\pars{1 \over 6}} - \color{#f88}{\Psi\, '\pars{1 \over 3}} - 2\Psi\, '\pars{1 \over 2} - \color{#f88}{\Psi\, '\pars{2 \over 3}} \right. \\[2mm] &\ \phantom{= -\,{1 \over 18}} \left. +\ \color{#88f}{\Psi\, '\pars{5 \over 6}} + 2\Psi\, '\pars{1}\right]\label{1}\tag{1} \\[5mm] = &\ -\,{1 \over 18}\ \underbrace{\bracks{% \pi^{2}\csc^{2}\pars{\pi \over 6} - \pi^{2}\csc^{2}\pars{\pi \over 3} - \pi^{2} + {\pi^{2} \over 3}}} _{\ds{2\pi^{2}}} \\[5mm] = &\ \bbx{-\,{\pi^{2} \over 9\phantom{^{2}}}} \\ & \end{align} In (\ref{1}), pairs of "same color functions" are coupled via the Euler Reflection Formula $$ \Psi\,'\pars{1 - z} + \Psi\,'\pars{z} = \pi^{2}\csc^{2}\pars{\pi z} $$ In addittion, $\ds{\Psi\,'\pars{1 \over 2} = {\pi^{2} \over 2}}$ and $\ds{\Psi\,'\pars{1} = {\pi^{2} \over 6}}$.

Answer 8

Note $ \frac1{x(1-x)} = \frac1x+\frac1{1-x}$ to obtain

\begin{align} I=\int_0^1\frac{\ln\left(x^2-x+1\right)}{x(1-x)}dx &=2\int_0^1\frac{dx}x \ln\left(1-x+x^2\right) \end{align}

Let $J(a)=\int_0^1\frac{dx}x\ln\left(1-2\cos a x+x^2\right)$

$$J’(a)=\int_0^1 \frac{2\sin a}{(x-\sin a)^2+\sin^2a}dx=\pi-a $$

Note $$J(\frac\pi2)= \int_0^1\frac{\ln(1+x^2)}{x}dx\overset{x^2\to x}=\frac12\int_0^1\frac{\ln(1+x)}{x}dx\\ = \frac12\int_0^1\frac{\ln(1+x^3)-\ln(1-x+x^2)}{x}dx =\frac16\int_0^1\frac{\ln(1+x)}{x}dx-\frac14I =-\frac38 I $$ Then

\begin{align} I=2J(\frac\pi3) = 2\left(J(\frac\pi2)- \int_{\pi/3}^{\pi/2}J’(a)da\right)=-\frac34I -2 \int_{\pi/3}^{\pi/2}(\pi-a)da \end{align} which leads to $I= -\frac{\pi^2}9$.