Question

5D%7Bx%7D%20%20%5Ctan%5E%7B-%201%7D%20%28x%29%7D%7B%281%20%2B%20%20%7Bx%7D%5E%7B%20%5Cphi%7D%20%7B%29%7D%5E%7B2%7D%20%20%7D%20%7B%7D%5E%7B%7D%20%20%20%7B%7D%20%20%20%5C%3A%20dx%5C%5C%20" id="TexFormula1" title=" \rm \int_{0}^ \infty \frac{ \sqrt[ \scriptsize\phi]{x} \tan^{- 1} (x)}{(1 + {x}^{ \phi} {)}^{2} } {}^{} {} \: dx\\ " alt=" \rm \int_{0}^ \infty \frac{ \sqrt[ \scriptsize\phi]{x} \tan^{- 1} (x)}{(1 + {x}^{ \phi} {)}^{2} } {}^{} {} \: dx\\ " align="absmiddle" class="latex-formula">​

Answer 1

With ϕ ≈ 1.61803 the golden ratio, we have 1/ϕ = ϕ - 1, so that

$I = \displaystyle \int_0^\infty \frac{\sqrt[\phi]{x} \tan^{-1}(x)}{(1+x^\phi)^2} \, dx = \int_0^\infty \frac{x^{\phi-1} \tan^{-1}(x)}{x (1+x^\phi)^2} \, dx$

Replace $x \to x^{\frac1\phi} = x^{\phi-1}$ :

$I = \displaystyle \frac1\phi \int_0^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx$

Split the integral at x = 1. For the integral over [1, ∞), substitute $x \to \frac1x$ :

$\displaystyle \int_1^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx = \int_0^1 \frac{\tan^{-1}(x^{1-\phi})}{\left(1+\frac1x\right)^2} \frac{dx}{x^2} = \int_0^1 \frac{\pi2 - \tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx$

The integrals involving tan⁻¹ disappear, and we're left with

$I = \displaystyle \frac\pi{2\phi} \int_0^1 \frac{dx}{(1+x)^2} = \boxed{\frac\pi{4\phi}}$