It seems that the following formulas hold :
$$\int_{0}^{\infty} \frac{1}{\sqrt{x^{2n}+1}} dx = \frac{\Gamma(\frac{n-1}{2n})\Gamma(\frac{2n+1}{2n})}{\sqrt\pi}$$ for any integer $n > 1$
and
$$\int_{0}^{\infty} \frac{1}{\sqrt{x^n+1}} dx = \frac{\Gamma(\frac{n-2}{2n})\Gamma(\frac{n+1}{n})}{\sqrt\pi}$$ for any odd integer $n > 1$
Has anyone an idea how to proof this ?
Perhaps, some substitution removes the square-root, but I do not know which.
Putting $x^p=t$, the integral becomes $$ \int_0^{\infty}\frac{1}{\sqrt{x^p+1}}\textrm{d}x=\frac{1}{p}\int_0^{\infty}\frac{t^{\frac{1}{p}-1}}{(1+t)^{\frac{1}{2}}}\textrm{d}t=\frac{1}{p}B\left(\frac{1}{p},\frac{1}{2}-\frac{1}{p}\right)=\frac{1}{p\sqrt{\pi}}\Gamma\left(\frac{1}{p}\right)\Gamma\left(\frac{1}{2}-\frac{1}{p}\right) $$ where we use the beta function as $B\left(r,s\right)=\int_0^{\infty} \frac{t^{r-1}}{(1+t)^{r+s}} dt$ (see http://dlmf.nist.gov/5.12#E3).