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

Solution 1:

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.