# 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..

-
Could you please elaborate on why you find this integral interesting? –  Jonas Meyer Dec 29 '12 at 10:15
It converges to $\dfrac{\pi^2}{5 \sqrt{5}}$ according to Mathematica. To just see that it does converge it suffices to note that the denominator is bounded below in absolute value by a positive number, and $\int_0^1\ln(x)dx$ converges, e.g. by comparison (in absolute value) with $\frac{1}{\sqrt x}$ near $x=0$. –  Jonas Meyer Dec 29 '12 at 10:20
This integral is given by my teacher, I run it on Maple, it does converge. –  Ryan Dec 29 '12 at 10:21
Have you tried partial fractions and series expansions? –  Eckhard Dec 29 '12 at 10:49
@Ryan: Yes. I think the integral may be evaluated elementarily. Chris. –  Chris's sis the artist Dec 29 '12 at 11:25

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 $$\int_0^1\frac{\log(x)}{x^2-x-1}d\,x = \frac{2}{\sqrt{5}}\times\sum_{n=0}^\infty\left[-\frac{1}{x_+}\int_0^1{\log(x)\left(\frac{x}{x_+}\right)^nd\,x} + \frac{1}{x_-}\int_0^{-x_-}{\log(x)\left(\frac{x}{x_-}\right)^n d\,x} - \int_{-x_-}^1{\frac{1}{x}\log(x)\left(\frac{x_-}{x}\right)^n d\,x}\right].$$
The final result is $\pi^2/5\sqrt{5}$, as has already been mentioned.