P: Cómo uno demostrar que $\int_{0}^{1}{1-x^{\sqrt5}\over (1-x^{\phi})^{\phi}}\mathrm dx={\phi^2}?$

Question

Tener en cuenta

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

¿Cómo podemos mostrar que $(1)$ converge a $\color{red}{\phi^2}$?

Un intento:

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

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

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

Después de simplificar, esto es donde llegué a

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

Yo no puedo proceder alguno.

Answer 1

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

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