The golden ratio always appears at the right moment

Synopsis : We start with the transformation $x\longmapsto \dfrac1x$ , add these two integrals together to get a simplified integrand. Finish if off by rewriting the integral in terms of the Beta function .

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The given value of $\phi = \dfrac{1+\sqrt5}2$ is just a red herring, in fact, $\displaystyle f(A) := \int_0^\infty \dfrac{dx}{\sqrt{1+x^4} (1+x^A)}$ is equal to $\dfrac{\left [\Gamma\left( \frac14\right)\right]^2}{8\sqrt\pi}$ for all $A$ .

We start with the transformation $y = \dfrac1x\Rightarrow \dfrac{dy}{dx} = -\dfrac{1}{x^2}\Rightarrow dx =-\dfrac{dy}{y^2}$ , then

$f(A) = \int_{\infty}^0 \dfrac{1}{\sqrt{\left(\frac1{y^4}\right)^4 + 1} \left(1 + \frac1{y^A}\right)} \cdot -\dfrac{dy}{y^2} \; = \; \int_0^\infty \dfrac{dy}{\sqrt{y^4+1}} \cdot \dfrac{y^A}{y^A + 1} \; .$

Adding the equation above with the original form of $f(A)$ gives

$\begin{aligned} f(A) + f(A) &= &\int_0^\infty \dfrac{dy}{\sqrt{y^4+1}} \cdot \cancelto1{ \left( \dfrac1{y^A + 1} + \dfrac{y^A}{y^A + 1} \right)} \\ 2 f(A)& =& \int_0^\infty \dfrac{dy}{\sqrt{y^4+1}} \qquad ,\qquad \text{ let }y = \sqrt{\tan z} \Rightarrow \dfrac{dy}{dz} =\dfrac12(\tan z)^{-1/2} \sec^2 z \\ 2 f(A)& =& \dfrac12\int_0^{\pi /2} \dfrac{(\tan z)^{-1/2} \sec^2 z}{\sec z}\, dz \\ 4 f(A)& =& \int_0^{\pi /2} (\sin z)^{-1/2} (\cos z)^{-1/2} \, dz \\ 8 f(A)& =& 2\int_0^{\pi /2} (\sin z)^{2m-1} (\cos z)^{2n-1} \, dz \qquad , \qquad \text{ where } 2m-1 = 2n-1 = -\dfrac12 \Leftrightarrow m=n=\dfrac14 \\ 8 f(A)& =& \dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} = \dfrac{\left[\Gamma \left( \frac14\right) \right]^2}{\Gamma\left( \frac12\right)} \\ f(\phi) = f(A) & =&\dfrac{\left[\Gamma \left( \frac14\right) \right]^2}{8\sqrt\pi} \approx \boxed{0.9270} \; . \end{aligned}$

Note : The penultimate step follows from the properties of the trigonometric representation of the Beta function , $\displaystyle B(x,y)=\dfrac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} = \int_0^{\pi/2} 2\sin^{2x-1}t\cos^{2y-1}t \, dt$ , where $\Gamma( \cdot )$ denotes the Gamma function with $\Gamma \left( \dfrac12\right) = \sqrt\pi$ .