Given the following: $$\int_{\varGamma_R} {z\,dz\over e^{2\pi iz^2}-1}, \ \ \ \varGamma_R=\{z\in \Bbb C:|z|=R\},\quad n<R^2<n+1,\,n\in\Bbb N.$$
I want to use the residue for this, but I can't seem to find the poles for the functions
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Did you consider a change of variable which makes the integral very simple ? – Claude Leibovici Jan 28 '14 at 08:01
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well, no. i'm trying to understand where the poles are – user107761 Jan 28 '14 at 08:02
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Because has no poles? $\forall z\in{\Bbb C}\ e^z\ne 0$. And your $\Gamma_R$ isn't a curve! – Martín-Blas Pérez Pinilla Jan 28 '14 at 08:02
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$\Gamma R$ is a curve. and I do have poles.with order of 4n+1. I just don't know why. – user107761 Jan 28 '14 at 08:04
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How did you know the order of the poles or it is given a solution. – Mhenni Benghorbal Jan 28 '14 at 08:17
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Is a curve when $\Gamma_R={z\in{\Bbb C}:|z|=R}$. And ${z\over e^{2\pi iz^2}}$ has zero poles in ${\Bbb C}$. Check again your formulas. – Martín-Blas Pérez Pinilla Jan 28 '14 at 08:19
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yep, I was wrong. edited. – user107761 Jan 28 '14 at 08:22
2 Answers
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Note that the poles occur whenever $z^2=k$, where $k \in \mathbb{N}\cup \{0\}$ such that $k \le n^2$. Thus, the poles are at $z_k = \sqrt{k}$, $k \in \{0,1,2,\ldots,n^2\}$. A typical residue is
$$\operatorname*{Res}_{z=z_k} \frac{z}{e^{i 2 \pi z^2}-1} = \frac{z_k}{i 4 \pi z_k e^{i 2 \pi z_k^2}} = \frac1{i 4 \pi}$$
as $z_k^2$ is an integer. Note that the poles are all simple; the factor of $z$ in the numerator ensures that even the pole at $z=0$ is simple. Thus the integral is, by the residue theorem,
$$i 2 \pi \sum_{k=0}^{n^2} \operatorname*{Res}_{z=z_k} \frac{z}{e^{i 2 \pi z^2}-1} = \frac12 \left ( n^2+1\right )$$
Ron Gordon
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Hint: First, find the poles as
$$e^{2\pi iz^2} -1=0 \implies 2\pi\,iz^2 = 2k\pi i, \quad k \in \mathbb{Z}. $$
To determine the order, see here.
Mhenni Benghorbal
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