A interesting improper integral, $ \int_{0}^1\frac{\ln x}{x^2-x-1}\text{d}x$ $$\displaystyle \int_{0}^1\frac{\ln x}{x^2-x-1}\text{d}x$$
I think there should be a smart way to evaluate this. But I cant see..
 A: 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.
A: $$\begin{align*}
I &= \int_0^1 \frac{\log(x)}{x^2-x-1} \, dx \\[1ex]
&= - \int_0^1 \int_x^1 \frac1{y(x^2-x-1)} \, dy \, dx \tag{1} \\[1ex]
&= - \int_0^1 \int_0^y \frac1{y(x^2-x-1)} \, dx \, dy \tag{2} \\[1ex]
&= - \frac1{\sqrt5} \int_0^1 \log\left(\frac{\phi y+1}{(1-\phi)y+1}\right) \, \frac{dy}y \tag{3} \\[1ex]
&= \frac1{\sqrt5} \int_0^1 \frac{\log((1-\phi)y+1) - \log(\phi y+1)}y \, dy
\end{align*}$$

*

*$(1)$ : definition of $\log(x)$

*$(2)$ : change order of integration

*$(3)$ : integrate w.r.t. $x$, where $\phi=\frac{1+\sqrt 5}2$ is the golden ratio


Now let
$$J(a) = \int_0^1 \frac{\log(ay+1)}y \, dy,$$
noting that $J(0)=0$. Differentiating under the integral sign gives
$$J'(a) = \int_0^1 \frac{dy}{1+ay} = \frac{\log(1+a)}a$$
and integrating yields the dilogarithm $\operatorname{Li_2}$,
$$J(a) = J(0) + \int_0^a \frac{\log(1+b)}b \, db = -\operatorname{Li}_2(-a)$$
Hence
$$I = \frac{J(\phi) - J(1-\phi)}{\sqrt5} = \boxed{\frac{\operatorname{Li}_2(\phi-1)-\operatorname{Li}_2(-\phi)}{\sqrt5}} = \boxed{\frac{\pi^2}{5\sqrt5}}$$
where the last equality follows from known values of $\operatorname{Li}_2$. (Some discussion at MO here)
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{{\displaystyle #1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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.
