4
$\begingroup$

Consider

$$\int_{0}^{1}{1-x^{\sqrt5}\over (1-x^{\phi})^{\phi}}\mathrm dx=\color{red}{\phi^2}\tag1$$ $\phi$;Golden ratio

How can we show that $(1)$ converges to $\color{red}{\phi^2}$?

An attempt:

$u=1-x^{\phi}$ then $du=-\phi x^{\phi-1} dx$

$(1-u)^{\sqrt5/\phi}=x^{\sqrt5}$

$(1-u)^{1/\phi^2}=x^{1/\phi}$

After simplifying, this is where I got to

$${1\over \phi}\int_{0}^{1}{1\over u^{\phi}(1-u)^{1/\phi^2}\mathrm du}-{1\over \phi}\int_{0}^{1}{1-u\over u^{\phi}}\mathrm du\tag2$$

I can't proceed any further.

  • 0
    You've got an expression containing two _divergent_ integrals, which is not good. – 2017-02-24
  • 0
    Ideally substitution is not good for integral $(1)$, what other approach can I use? – 2017-02-24

1 Answers 1

5

$\displaystyle \int\limits_0^1\frac{1-x^{2\phi-1}}{(1-x^{\phi})^{\phi}}=\frac{1}{\phi}\int\limits_0^1 \frac{(1-u)^{\phi-2} -(1-u)}{u^{\phi}}du$

$\hspace{2.2cm}\displaystyle =\frac{1}{ \phi }\left( \frac{u^{2-\phi}}{2-\phi} + \frac{1-(1-u)^{ \phi -1}}{u^{\phi -1}(\phi-1)} \right)|_0^1 =\frac{1}{ \phi }\left( \frac{1}{2-\phi} + \frac{1}{\phi-1} -0-0\right)= \phi^2$