Find all values of $r$ if $\int_{0}^{\infty} \frac{dx}{(1+x^r)^r} =1. $

Question

Find all values of $r$ if $$\int_{0}^{\infty} \frac{dx}{(1+x^r)^r} =1. $$

I have found one value of $r$ by a brute force method. I use the substitution $x^r=\tan^2 t$ to convert the required integral as: $$J=\int_{0}^{\infty} \frac{dx}{(1+x^r)^r}= \frac{2}{r} \int_{0}^{\pi/2} \sin^{(2/r-1)} t~~ \cos^ {(-2/r+2r-1)}t ~ dt~~~~~(*)$$ and force $-2/r+2r-1=1$ in (*). I get two values of $r$ as $r_1=\frac{1+\sqrt{5}}{2}$ and $r_2=\frac{1-\sqrt{5}}{2}$. Noting that $J$ diverges for $r^2<1$, I reject $r_2$, then for $r=r_1$, I check that $J=1$. Can there be a better approach to solve this question? Are there other values of $r$ ?

Answer 1

The next step could be to convert your Eq. (*) by using beta-integral in terms of gamma fumctions as $$ J(r)=\frac{1}{r}~ \frac {\Gamma(r-1/r) ~\Gamma(1/r)}{\Gamma(r)}, ~~r>1, ~~~ \Gamma(z+1)= z \Gamma(z).~~~~(1)$$ By setting $(r-1/r)=1$ in (1), we get two roots $r=\frac{1+\sqrt{5}}{2}=r_1$ and $\frac{1-\sqrt{5}}{2}=r_2$. Neglecting $r_2$, from (1), we get $J(r_1)=1.$ It is easy to notiice that $J(\infty)=1.$ Further, it is required to prove the uniqueness of $r$. For this the following graph could help.