Contour Integral: $\int^{\infty}_{0}(1+z^n)^{-1}dz$ I'm working through Priestley's Complex Analysis (really good book by the way) and this Ex 20.2:
Evaluate $\int^{\infty}_{0}(1+z^n)^{-1}dz$ round a suitable sector of angle $\frac{2\pi}{n}$ for $n=1,2,3,...$
Can someone advise what the contour may be? If we use a sector that includes zero, surely we'll have to indent it to avoid the singularity (since it isn't a simple pole on account of the negative power).
Thanks.
 A: The contour $C$ is a wedge-shaped contour of angle $2 \pi/n$, as you state, with respect to the positive real axis.  This contour may be broken into 3 pieces:
$$\oint_C \frac{dz}{1+z^n} = \int_0^R \frac{dx}{1+x^n} + i R \int_0^{2 \pi/n} d\phi \, e^{i \phi} \frac{1}{1+R^n e^{i n \phi}} - e^{i 2 \pi/n} \int_0^R \frac{dx}{1+x^n} $$
The second integral vanishes in the limit as $R \to \infty$; in fact, it vanishes as $1/R^{n-1}$ in this limit.  The rest is equal to $i 2 \pi$ times the residue at the pole at $z=e^{i \pi/n}$; note that this pole is interior to $C$ and therefore no further deformation of $C$ is necessary.
The residue theorem then implies
$$\left ( 1-e^{i 2 \pi/n}\right) \int_0^{\infty} \frac{dx}{1+x^n} = \frac{i 2 \pi}{n e^{i \pi (n-1)/n}}$$
The final result is
$$\int_0^{\infty} \frac{dx}{1+x^n} = \frac{\pi}{n \sin{(\pi/n)}}$$
A: The suitable contour is already given: a circle sector, with center at the origin, with one "edge" along the positive real axis and an opening angle of $2\pi/n$. 
The function has simple poles at points where $z^n = -1$, i.e. at $z = \exp(i\pi/n + 2\pi i k/n)$, and only one of these lie inside the suggested contour. (None on the boundary.)
