= Logarithmic beta integral from a sector contour
{title2=$\int_0^\infty\frac{\log x}{1+x^\beta}\,dx=-\frac{\pi^2\cos(\pi/\beta)}{\beta^2\sin^2(\pi/\beta)}$}
For real $\beta>1$, choose a <branch of the complex logarithm> whose argument interval contains $[0,2\pi/\beta]$ strictly, and integrate $\operatorname{Log}z/(1+z^\beta)$ 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 $O(\epsilon|\log\epsilon|)$ 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> give
$$
(1-e^{i\alpha})I-i\alpha e^{i\alpha}J=\frac{2\pi^2}{\beta^2}e^{i\pi/\beta}.
$$
Divide 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 $\beta>1$.
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