Solving $\int_0^\infty\dfrac{dx}{(1+x^n)^n}=1$ A friend has challenged me to find the value of $n$ such that $\displaystyle\int_0^\infty\dfrac{dx}{(1+x^n)^n}=1.$ I know that the value of $n$ is equal to $\phi,$ but I do not know how to prove this.
 A: I don't see how do we do that unless we bring the whole thing to Beta function with some clever (or maybe trivial) substitution.
Say, $t={1\over1+x^n}$. Then $x=\sqrt[n]{{1\over t}-1}$. Then $dx=-{dt\over n t^2({1/t}-1)^{1-1/n}}$, and 
$$\int_0^\infty\dfrac{dx}{(1+x^n)^n}=-\int_1^0t^n\left({1\over t}-1\right)^{{1\over n}-1}{dt\over n\cdot t^2}={1\over n}\int_0^1t^{n-1-{1\over n}}(1-t)^{{1\over n}-1}dt={1\over n}B\left(n-{1\over n},{1\over n}\right)={1\over n}\cdot{\Gamma\left(n-{1\over n}\right)\Gamma\left({1\over n}\right)\over \Gamma(n)}={\Gamma\left(n-{1\over n}\right)\Gamma\left(1+{1\over n}\right)\over \Gamma(n)}$$
At this point, it is pretty easy to verify directly that $n=\varphi$ fits. (Note that $\varphi-{1\over\varphi}=1$ and $1+{1\over\varphi}=\varphi$).
What could we do without that afterknowledge, I wonder...
A: HINT:
Enforce the substitution $x=t^{1/n}$.  Then, verify a resulting equation.
SPOILER ALERT:  Scroll over the highlighted area to reveal the solution

Let $I(n)$ be the integral given by $$I(n)=\int_0^\infty \frac{1}{(1+x^n)^n}\,dx \tag 1$$First note that the integral converges for $n>1$ and diverges otherwise.  Next, enforcing the substitution $x= t^{1/n}$ in $(1)$ reveals $$\begin{align}I(n)&=\frac1n \int_0^\infty \frac{t^{1/n-1}}{(1+t)^n}\,dt\\\\&=\frac1n B(1/n,n-1/n)\\\\&=\frac{\Gamma(1/n)\Gamma(n-1/n)}{n\Gamma(n)}\end{align}$$Therefore, if $I(n)=1$, then $$\frac1n \Gamma(1/n)\Gamma(n-1/n)=\Gamma(n) \tag 2$$If $n=\phi>1$ in $(2)$, then $n-1/n=1$ and $(2)$ states that $\Gamma(\phi)=\frac1\phi \Gamma(1/\phi)=\Gamma(1+1/\phi)$.  Since $1+1/\phi=\phi$, we are done! 

A: The easy part. One may observe that, by the change of variable $x \to \dfrac1x$, one gets
$$
\begin{align}
\int_0^\infty\frac{dx}{(1+x^\phi)^\phi}&=\int_0^\infty\frac{x^{\phi^2-2}}{(1+x^\phi)^\phi}\:dx
\\\\&=\int_0^\infty\frac{x^{\phi-1}}{(1+x^\phi)^\phi}\:dx
\\\\&=\frac1{\phi}\int_0^\infty\frac{(1+x^{\phi})'}{(1+x^\phi)^\phi}\:dx
\\\\&=\frac1{\phi}\int_1^\infty u^{-\phi}\:du
\\\\&=\frac{1}{\phi(\phi-1)}
\\\\&=1
\end{align}
$$  if $\phi>1$ and $\phi^2=\phi+1$.
