11
$\begingroup$

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

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

  • 0
    Could you please elaborate on why you find this integral interesting?2012-12-29
  • 2
    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 $x=0$.2012-12-29
  • 0
    This integral is given by my teacher, I run it on Maple, it does converge.2012-12-29
  • 2
    Have you tried partial fractions and series expansions?2012-12-29
  • 0
    @Ryan: Yes. I think the integral may be evaluated elementarily. Chris.2012-12-29
  • 0
    This integral is a special case of the [more general problem](http://math.stackexchange.com/questions/188732/methods-to-evaluate-int-a-b-frac-ln-left-tx-u-right-mx/188828#188828).2013-02-15

1 Answers 1

6

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

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.

  • 0
    Wow! very sharp! Thank you! : )2012-12-29