A interesting improper integral, $ \int_{0}^1\frac{\ln x}{x^2-x-1}\text{d}x$

Question

$$\displaystyle \int_{0}^1\frac{\ln x}{x^2-x-1}\text{d}x$$

I think there should be a smart way to evaluate this. But I cant see..

Answer 1

Using a partial fraction decomposition, one can write $$ \frac{1}{x^2-x-1}=\frac{2}{\sqrt{5}}\left[\frac{1}{x-x_+}-\frac{1}{x-x_-}\right], $$ where $x_{\pm}=\frac{1}{2}\left(1\pm\sqrt{5}\right)$ are the roots of the polynomial $x^2-x-1$.

We now observe that we have the uniformly convergent series expansions $$ \frac{1}{x-x_+} = -\frac{1}{x_+}\sum_{n=0}^\infty{\left(\frac{x}{x_+}\right)^n},\quad 0<x<1; $$ $$ \frac{1}{x-x_-} = -\frac{1}{x_-}\sum_{n=0}^\infty{\left(\frac{x}{x_-}\right)^n},\quad 0<x<-x_-; $$ and $$ \frac{1}{x-x_-} = \frac{1}{x}\sum_{n=0}^\infty{\left(\frac{x_-}{x}\right)^n},\quad -x_-<x<1. $$ Putting this together, one obtains $$ \int_0^1\frac{\log(x)}{x^2-x-1}d\,x = \frac{2}{\sqrt{5}}\times\sum_{n=0}^\infty\left[-\frac{1}{x_+}\int_0^1{\log(x)\left(\frac{x}{x_+}\right)^nd\,x} + \frac{1}{x_-}\int_0^{-x_-}{\log(x)\left(\frac{x}{x_-}\right)^n d\,x} - \int_{-x_-}^1{\frac{1}{x}\log(x)\left(\frac{x_-}{x}\right)^n d\,x}\right]. $$

I'll leave the rest of the computations to the OP, but can expand if necessary.

The final result is $\pi^2/5\sqrt{5}$, as has already been mentioned.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\int_{0}^{1}{\ln\pars{x} \over x^{2} - x - 1}\dd x} = \int_{0}^{1}{\ln\pars{x} \over \pars{x - \phi}\pars{x + \phi^{-1}}}\dd x \end{align} where $\ds{\phi \equiv \pars{1 + \root{5}}/2}$ is the Golden Ratio.

Then, \begin{align} & \color{#44f}{\int_{0}^{1}{\ln\pars{x} \over x^{2} - x - 1}\dd x} = -\,{\root{5} \over 5}\bracks{{\tt I}\pars{\phi} - {\tt I}\pars{-\,{1 \over \phi}}} \end{align} where \begin{align} \left.\rule{0pt}{5mm}{\tt I}\pars{a}\right\vert_{a\ \not\in\ \bracks{0,1}}\,\,\, & \equiv \int_{0}^{1}{\ln\pars{x} \over a - x}\dd x \sr{x/a\ \mapsto\ x}{=} \int_{0}^{1/a}{\ln\pars{ax} \over 1 - x}\dd x \\[5mm] & \sr{\rm IBP}{=} \int_{0}^{1/a}\ \overbrace{{\ln\pars{1 - x} \over x}}^{\ds{-\on{Li}_{2}'\pars{x}}}\ \dd x = -\on{Li}_{2}\pars{1 \over a} \end{align} $\ds{\on{Li}_{s}}$ is the order-$\ds{s}$ Polylogarithm Function. Therefore, \begin{align} & \color{#44f}{\int_{0}^{1}{\ln\pars{x} \over x^{2} - x - 1}\dd x} = -\,{\root{5} \over 5}\bracks{-\on{Li}_{2}\pars{1 \over \phi} + \on{Li}_{2}\pars{-\phi}} \\[5mm] = & \ \bbx{\color{#44f}{{\root{5} \over 25}\pi^{2}}} \approx 0.8828 \\ & \end{align} $\ds{\on{Li}_{2}\pars{1/\phi}}$ and $\ds{\on{Li}_{2}\pars{-\phi}}$ are given, for example, in this site.