 Solution

For a rational function, the stipulated decay implies $R(z)=O(|z{|}^{-2})$ at infinity, unless it vanishes identically. Thus its real-line integral converges absolutely. Close the segment $[-T,T]$ with an upper semicircle, choosing $T$ beyond all finite poles. The arc contribution has modulus at most $\pi T{\sup }_{|z|=T}|R(z)|=O(T^{-1})$, tending to zero. The residue theorem therefore givesThere are no real poles requiring indentations.

Apply this to $R(z)=1/(1+z^{2n})$. Its upper-half-plane poles are ${\zeta }_k=e^{i(2k+1)\pi /(2n)}$, $k=0,\dots ,n-1$, all simple. Since ${\zeta }_k^{2n}=-1$, their residues areFor $\theta =\pi /(2n)$, the finite geometric series givesThe full real-line integral is consequently $\pi /[n\sin (\pi /(2n))]$. Its integrand is even, so the requested half-line integral is