= Mellin integral of a reciprocal binomial
{c}
{title2=$\int_0^\infty\frac{x^{s-1}}{1+x^p}dx=\frac\pi{p\sin(\pi s/p)}$}
For real $p>0$ and $0<s<p$, the displayed <improper integral> converges: its endpoint powers are $x^{s-1}$ near zero and $x^{s-p-1}$ near infinity. The <change of variables> $t=x^p$ gives
$$
\frac1p\int_0^\infty\frac{t^{s/p-1}}{1+t}dt=\frac1pB(s/p,1-s/p)=\frac{\Gamma(s/p)\Gamma(1-s/p)}p=\frac\pi{p\sin(\pi s/p)}.
$$
Here the <beta function>, the <beta-gamma identity>, and the <Gamma reflection formula> evaluate the integral. The same formula holds for complex $s$ in the <Mellin fundamental strip> $0<\operatorname{Re}s<p$, by absolute convergence and analytic continuation within the strip. For integer $p$, a sector contour of angle $2\pi/p$ gives a direct <residue theorem> proof, with a consistent branch of $z^{s-1}$.
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