Help in proving that $\int_{0}^{\infty} \frac{dx}{1+x^n}=\int_{0}^{1} \frac{dx}{(1-x^n)^{1/n}},n>1$

Question

I wanted to prove that $$\int_{0}^{\infty} \frac{dx}{1+x^n}=\int_{0}^{1} \frac{dx}{(1-x^n)^{1/n}}, n>1.$$ I converted them to Gamma functions but I could not prove it. Please help me.

Answer 1

To do it in one substitution, use $y=(1+x^n)^{-1/n}$, or equivalently $x=(y^{-n}-1)^{1/n}$.

Answer 2

$$I=\int_{0}^{\infty} \frac{1}{1+x^n} dx$$ Let $x=1/y\implies dx=-dy/y^2$, then $$I=\int_{0}^{\infty} \frac{y^{n-2} dy}{1+y^n}$$ Next, use $1+y^n=z^n \implies y^{n-1}dy=z^{n-1} dz$ to get $$I=\int_{1}^{\infty} \frac{dz}{z (z^n-1)^{1/n}}.$$ Lastly, let us take $z=1/u \implies dz=-du/u^2$, to get $$I=\int_{0}^{1} \frac{du}{(1-u^n)^{1/n}}.$$

Answer 3

A chance is to bring both integrals into the standard Beta form $$\int_{0}^{1}z^\alpha(1-z)^\beta\,dz =\frac{\Gamma(\alpha+1)\Gamma(\beta+2)}{\Gamma(\alpha+\beta+2)}$$ by letting $\frac{1}{1+x^n}=z$ in the first integral and $x^n=z$ in the second integral.