How does one show that?
$$\int_\phi^{\phi^2}{((\ln{x}+{1\over2})^e+\gamma)^\pi \over ((\ln{(\phi^3-x)}+{1\over2})^e + \gamma)^{\pi}+((\ln{x}+{1\over2})^e+\gamma)^\pi} \, dx={1\over 2} \tag1$$
$\gamma$;Euler's constant
$\phi$; Golden ratio
Integral $(1)$ seems to be a trivial but I can't see it (I think so).
It takes the form of
$$\int_{\phi}^{\phi^2}{f(x)\over g(x)+f(x)} \, dx = {1\over 2}\tag2$$
Any help?
More specifically, since $\phi^2 + \phi = \phi^3$, the integral takes the form
$$\int_{\phi}^{\phi^2} \frac{f(x)}{f(\phi^2+\phi - x) + f(x)}\,dx.$$
Whenever we have an integral
$$\int_a^b \frac{f(x)}{f(a+b-x) + f(x)}\,dx,$$
the substitution $y = a+b-x$ shows it is equal to
$$\int_a^b \frac{f(a+b-x)}{f(x) + f(a+b-x)}\,dx = \int_a^b 1 - \frac{f(x)}{f(a+b-x) + f(x)}\,dx.$$
Adding the two forms and dividing by $2$ then yields
$$\int_a^b \frac{f(x)}{f(a+b-x) + f(x)}\,dx = \frac{1}{2}\int_a^b 1\,dx = \frac{b-a}{2}.$$
Since $\phi^2 - \phi = 1$, the result here follows.
