Logarithmic beta integral from a sector contour (${\int }_0^{\infty }\frac{\log x}{1+x^{\beta }}dx=-\frac{{\pi }^2\cos (\pi /\beta )}{{\beta }^2{\sin }^2(\pi /\beta )}$)

For real $\beta >1$, choose a branch of the complex logarithm whose argument interval contains $[0,2\pi /\beta ]$ strictly, and integrate $\mathrm{Log}z/(1+z^{\beta })$ over an annular sector of angle $\alpha =2\pi /\beta$. The one enclosed simple pole is $e^{i\pi /\beta }$, with residue $-i\pi e^{i\pi /\beta }/{\beta }^2$. The circular arcs tend to zero, with bounds $O(ϵ|\log ϵ|)$ and $O(R^{1-\beta }\log R)$.

Writing $I$ for the title integral and $J={\int }_0^{\infty }(1+x^{\beta }{)}^{-1}dx$, the two radial segments and the residue theorem giveDivide by $e^{i\pi /\beta }$ and equate real and imaginary parts. This gives $J=\pi /[\beta \sin (\pi /\beta )]$ and the title formula. When $1<\beta <2$, the sector extends below the negative real axis; the principal complex logarithm therefore cannot be used throughout that sector. A cut along a ray just below the positive axis avoids this issue for every $\beta >1$.