How can we show that?

$$\int_{0}^{1}{(1-x)(1-2x^{\phi})+\phi(x-x^{\phi})\over (1-x)^2}\cdot{\left(1-x^{\phi}\over 1-x\right)^{1\over \phi}}\mathrm dx=\phi^{\phi}\tag1$$ Where $\phi$;Golden ratio

This integral look too complicated. I try and make a $u=1-x$ still not simplified

$$\int_{0}^{1}{u[1-2(1-u)^{\phi}]+\phi[1-u-(1-u)^{\phi}]\over u^2}\cdot\left[1-(1-u)^{\phi}\over u\right]^{1\over \phi}\mathrm du$$

$$\int_{0}^{1}{\phi-{u\over \phi}-(1-u)^{\phi}(2u+\phi)\over u^2}\cdot\left[1-(1-u)^{\phi}\over u\right]^{1\over \phi}\mathrm du$$

Simplifed $(1)$:

$$\int_{0}^{1}{1+{x\over \phi}+x^{\phi}(2x-\phi{\sqrt{5}})\over (1-x)^2}\cdot\left({1- x^\phi}\over 1-x\right)^{1\over \phi}\mathrm dx=\phi^{\phi} \tag2$$

I have no idea where to go from here.

  • $\begingroup$ without starting any serious attemempts this really looks like that the solution depends heavily on the algebraic properties of the golden ratio so i would try to exploit them. $\endgroup$ – tired Jan 24 '17 at 10:28
  • $\begingroup$ I am trying and trying can't simplify. Can you direct me to this properties of golden ratio @tired. I know. $\phi^{n+2}=\phi^{n+1}+\phi^n$ $\endgroup$ – gymbvghjkgkjkhgfkl Jan 24 '17 at 11:45
  • $\begingroup$ Also we know $F_m\phi^n-F_n\phi^m=(-1)^n$ $\endgroup$ – gymbvghjkgkjkhgfkl Jan 24 '17 at 11:53
  • $\begingroup$ The key property here is $\frac{1}{\phi} = \phi-1$ $\endgroup$ – C. Dubussy Jan 24 '17 at 12:27

Using the relation $\phi - 1 = 1/\phi$, we find that

\begin{align*} &\int_{0}^{1} \frac{(1-x)(1-2x^{\phi})+\phi(x-x^{\phi})}{(1-x)^2}\cdot\left(\frac{1-x^{\phi}}{1-x}\right)^{1/\phi} \, \mathrm{d}x \\ &= \int_{0}^{1} \left( 2 - \frac{\phi^2}{1-x^{\phi}} + \frac{\phi}{1-x} \right) \left(\frac{1-x^{\phi}}{1-x}\right)^{\phi} \, \mathrm{d}x \\ &= \bigg[ x \left(\frac{1-x^{\phi}}{1-x}\right)^{\phi} \bigg]_{0}^{1} \\ &= \phi^\phi. \end{align*}

As a corollary, we have

$$ \int \frac{(1-x)(1-2x^{\phi})+\phi(x-x^{\phi})}{(1-x)^2}\cdot\left(\frac{1-x^{\phi}}{1-x}\right)^{1/\phi} \, \mathrm{d}x = x \left(\frac{1-x^{\phi}}{1-x}\right)^{\phi} + C. $$

Here is my line of reasoning that led to this solution:

  1. I tried to simplify the integrand so that it minimizes the amount of cancellation as well as mimics partial fraction decomposition.

  2. Now, the integrand in the second line looks similar to what we obtain when we apply the logarithmic differentiation $$ \frac{d}{dx}\left(\frac{1-x^{\phi}}{1-x}\right)^{\phi} = \left( \frac{\phi}{1-x} - \frac{\phi^2 x^{\phi-1}}{1-x^\phi} \right) \left(\frac{1-x^{\phi}}{1-x}\right)^{\phi}. \tag{1} $$ Although this is not exactly the same as what we want, it hints that we might actually compute the antiderivative.

  3. Playing a little bit, we find that $$ \frac{d}{dx}\frac{(1-x^{\phi})^{\phi}}{(1-x)^{\phi-1}} = \left( -2 -\frac{\phi^2 x^{\phi-1}}{1-x^{\phi}} + \frac{\phi^2}{1-x^{\phi}} \right) \left(\frac{1-x^{\phi}}{1-x}\right)^{\phi}. \tag{2}$$ Bingo! $\text{(1)} - \text{(2)}$ gives exactly what we want and we are done.


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