For real \(n\), determine when \(\int_0^\infty (1+x^n)^{-1}\,dx\) converges, and use residue calculus to evaluate it when finite.
Solution
The integral diverges to \(+\infty\) for \(n\leq1\). For \(n>1\), \[ \boxed{\int_0^\infty\frac{dx}{1+x^n} =\frac{\pi}{n\sin(\pi/n)}}. \]
1. The exact convergence range is \(n>1\).
For \(n>1\), the integrand is at most one on \((0,1)\) and at most \(x^{-n}\) on \([1,\infty)\), so both ends are integrable. If \(n\leq0\), it is at least \(1/2\) for \(x\geq1\). If \(0<n\leq1\), it is at least \(1/(2x)\) there because \(1+x^n\leq2x\). These comparisons prove divergence in all the remaining real cases.
2. An annular sector contains exactly one simple pole.
Fix \(n>1\) and set \(\theta=2\pi/n<2\pi\). Choose a logarithm branch on an open sector slightly wider than \(0\leq\arg z\leq\theta\), and define \(z^n=\exp(n\operatorname{Log}z)\) there. This is possible since the sector can have angular width less than \(2\pi\). For \(0<\varepsilon<1<R\), integrate \(F(z)=(1+z^n)^{-1}\) over the positive boundary of \(\{\varepsilon<|z|<R,\ 0<\arg z<\theta\}\).
In this sector, \(z^n=-1\) requires \(|z|=1\) and \(n\arg z=\pi\), so the only pole is \(\zeta=e^{i\pi/n}\). It is simple, with residue \[ \operatorname{Res}_{\zeta}F =\frac1{n\zeta^{n-1}}=-\frac\zeta n, \] because \(\zeta^n=-1\). On the second radial edge, writing \(z=xe^{i\theta}\) gives \(z^n=x^n\) and \(dz=e^{i\theta}dx\), with \(x\) traversed from \(R\) to \(\varepsilon\). Thus the two radial integrals sum to \[ (1-e^{i\theta})\int_\varepsilon^R\frac{dx}{1+x^n}. \] The residue theorem gives the total contour integral \(-2\pi i\zeta/n\) [@SS03].
3. The arcs vanish and the phase factors simplify.
The outer and inner arc integrals have moduli bounded by \[ \frac{\theta R}{R^n-1}\longrightarrow0, \qquad \frac{\theta\varepsilon}{1-\varepsilon^n}\longrightarrow0, \] respectively, using the arc lengths and \(|z^n|=|z|^n\). Letting \(R\to\infty\) and \(\varepsilon\downarrow0\), step 1 and the residue identity therefore give \[ (1-e^{2\pi i/n})I=-\frac{2\pi i e^{i\pi/n}}n. \] Finally \(1-e^{2\pi i/n}=-2i e^{i\pi/n}\sin(\pi/n)\), whose sine factor is positive for \(n>1\). Dividing gives \(I=\pi/(n\sin(\pi/n))\) as claimed.