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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.

Mathmo
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2 Answers2

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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

$$\int_0^{\infty} \frac{dx}{1+x^n} = \frac{\pi}{n \sin{(\pi/n)}}$$

Ron Gordon
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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.)

mrf
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  • Right, foolish of me to not see that $z^n+1=0$ does not have a solution at $z=0$.... So the only singularity in the sector lies at $e^{\frac{i\pi}{n}}$ (assuming we take the sector you suggested). How do we calculate the residue there? – Mathmo May 16 '13 at 20:20
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    @Mathmo: The easiest way to compute the residue is $$\operatorname{Res}\limits_{z=\exp(i\pi/n)} \frac{1}{1+z^n} = \left.\frac{1}{nz^{n-1}}\right|_{z=\exp(i\pi/n)}$$ – mrf May 16 '13 at 20:23