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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?

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    Try a substitution $y = \phi^3 - x$. – 2017-02-11
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    $\int{g(x)\over f(x)+g(x)}dx$ – 2017-02-11
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    [The same spirit, but different details](http://math.stackexchange.com/questions/167409/evaluating-displaystyle-int-0-frac-pi2-frac-sqrt-sin-x-sqrt-s). Not sure whether that counts as a duplicate by using the same principle, or not because of the details. – 2017-02-11
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    Your integral looks like a surrealist painting: you have just changed the melting watch, the glove, the cross, etc. into $e,\pi,\gamma,\Phi...$ :) – 2017-02-11

1 Answers 1

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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.