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

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 $$r$$ ?

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. 