Definite Integral $\int_0^1\frac{\ln(x^2-x+1)}{x^2-x}\,\mathrm{d}x$

$$\int_0^1\frac{\ln(x^2-x+1)}{x^2-x}\,\mathrm{d}x$$

WA gives $\pi^2/9$

• Use the fact that $\dfrac1{x^2-x}=\dfrac1{x(x-1)}=\dfrac1{x-1}-\dfrac1x$ to break it up into two convergent integrals. Then notice that $x^2-x+1=\dfrac{x^3+1}{x+1}$ , and use the fact that $\ln\dfrac ab=\ln a-\ln b$. Then employ the well-known Taylor series for the natural logarithm. – Lucian Feb 26 '14 at 9:54

4 Answers

Mhenni has struck first with the approach I have taken, but I would like to elaborate. Again, the integrand may be Taylor expanded:

\begin{align}-\int_0^1 dx \frac{\log{[1-(x-x^2)]}}{x-x^2} &= \sum_{n=0}^{\infty} \frac1{n+1} \int_0^1 dx \, x^n (1-x)^n\\ &= \sum_{n=0}^{\infty} \frac1{n+1} \frac{n!^2}{(2 n+1)!}\\ &=2 \sum_{n=0}^{\infty} \frac1{(2 n+2) (2 n+1) \binom{2 n}{n}} \end{align}

It turns out that

$$\frac{\arcsin{x}}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{2^{2 n} x^{2 n+1}}{(2 n+1) \binom{2 n}{n}}$$

So the sum in question is simply

$$4 \int_0^1 dx \frac{\arcsin{(x/2)}}{\sqrt{1-x^2/4}} = 8 \int_0^{\pi/6} d\theta \, \theta = \frac{\pi^2}{9}$$

as was to be shown.

• While you have every right to downvote this answer, as well as Mhenni's, it would be nice to know how either of these answers constitutes something not useful. – Ron Gordon Feb 26 '14 at 19:19

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x^{2} - x}\,\dd x:\ {\large ?}}$

Roots of $\ds{x^{2} - x + 1 = 0}$ are given by $\ds{\quad x_{\rm r} \equiv {1 + \root{3}\ic \over 2}\quad}$ and $\ds{\quad{1 \over x_{\rm r}} = x_{\rm r}^{*}}$.

\begin{align}&\color{#c00000}{% \int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x^{2} - x}\,\dd x} =-\int_{0}^{1}\ln\pars{x^{2} - x + 1}\pars{{1 \over x} + {1 \over 1 - x}}\,\dd x \\[3mm]&=-2\int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x}\,\dd x =-2\bracks{\int_{0}^{1}{\ln\pars{1 - x/x_{\rm r}} \over x}\,\dd x+ \int_{0}^{1}{\ln\pars{1 - x/x_{\rm r}^{*}} \over x}\,\dd x} \\[3mm]&=-2\bracks{\int_{0}^{1/x_{\rm r}}{\ln\pars{1 - x} \over x}\,\dd x+ \int_{0}^{1/x_{\rm r}*}{\ln\pars{1 - x} \over x}\,\dd x} =2\bracks{{\rm Li}_{2}\pars{1 \over x_{\rm r}} + {\rm Li}_{2}\pars{x_{\rm r}}} \end{align} where $\ds{{\rm Li}_{2}\pars{z}}$ is the Dilogarithm Function.

Note that $\ds{{\rm Li}_{1}\pars{z} = -\ln\pars{1 - z}}$ and $\ds{{\rm Li}_{2}\pars{z} = \int_{0}^{z}{{\rm Li}_{1}\pars{t} \over t}\,\dd t}$.

By using the Dilogarithm Inversion Formula $\ds{{\rm Li}_{2}\pars{z} + {\rm Li}_{2}\pars{1 \over z} =-\,{\pi^{2} \over 6} - \half\,\ln^{2}\pars{-z}}$ where $\ds{z \not\in \left[\vphantom{\Large A}0, 1\right)}$: \begin{align}&\color{#c00000}{% \int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x^{2} - x}\,\dd x} =2\bracks{-\,{\pi^{2} \over 6} - \half\,\ln^{2}\pars{-1 - \root{3}\ic \over 2}} \\[3mm]&=2\bracks{-\,{\pi^{2} \over 6} - \half\,\pars{-\,{2\pi \over 3}}^{2}} \end{align}

$$\color{#66f}{\large% \int_{0}^{1}{\ln\pars{x^{2} - x + 1} \over x^{2} - x}\,\dd x = {\pi^{2} \over 9}}$$

Here is an approach which is based on the Taylor series and the beta function

$$I = \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} \int_{0}^{1}\frac{(x^2-x)^k}{(x^2-x)}dx = \sum_{k=0}^{\infty}\frac{1}{k} \int_{0}^{1}{x^{k-1}(1-x)^{k-1}}dx$$

$$= \sum_{k=0}^{\infty}\frac{1}{k} \beta(k,k) = \sum_{k=0}^{\infty}\frac{1}{k} \frac{\Gamma(k)\Gamma(k)}{\Gamma(2k)} = 4(\sin^{-1}(1/2))^2 \sim 1.096622711.$$

• How do you go from $\sum_{k=0}^{\infty}\frac{1}{k} \frac{\Gamma(k)\Gamma(k)}{\Gamma(2k)}$ to $4(\sin^{-1}(1/2))^2$? – user85798 Feb 26 '14 at 16:20

Let me present you the simplest approach I could even think about. At start, I have to say I am using a method suggested earlier in comments by @Lucian . So it begins by computation of this kind of integrals, by expanding the integral in power series :

$$I_n = \int_0^1 \frac{\ln(1-x^n)}{x} \;\mathrm{d}x= -\int_0^1 \sum_{k=1}^{\infty} \frac{x^{nk-1}}{k} \;\mathrm{d}x = -\sum_{k=1}^{\infty} \frac{1}{nk^2}=-\frac{\zeta{(2)}}{n}$$

Next, using the fact that,

$$\frac{1}{x^2-x}=-\frac{1}{1-x}-\frac{1}{x}$$

And $x^2-x+1$ is symmetric inder transformation $x \to 1-x$ we get for our original integral :

$$I=-2\int_0^1\frac{\ln\left(x^2-x+1\right)}{x}\;\mathrm{d}x=-2\int_0^1\frac{\ln\left(\frac{1-x^6}{1-x^3}\cdot\frac{1-x}{1-x^2}\right)}{x}\;\mathrm{d}x = -2\left(I_6-I_3-I_2+I_1\right)=2\zeta(2)\left(\frac16-\frac13-\frac12+1\right)=\frac23\zeta(2)=\frac{\pi^2}{9}$$