# Residue theorem

• February 10th 2009, 07:04 AM
asi123
Residue theorem
Hey guys.
So I got this integral I need to solve, of curse using the residue theorem.
The thing is, that I don't understand the curve.
I know that whenever Z^2 = integer, this function has a singularity point because e^(2*pi*i*n) = 1.
But again, I'm not sure what this curve has enclosed in.

Thanks.
• February 10th 2009, 10:15 AM
ThePerfectHacker
Quote:

Originally Posted by asi123
Hey guys.
So I got this integral I need to solve, of curse using the residue theorem.
The thing is, that I don't understand the curve.
I know that whenever Z^2 = integer, this function has a singularity point because e^(2*pi*i*n) = 1.
But again, I'm not sure what this curve has enclosed in.

Thanks.

The poles occur when $e^{2\pi iz^2} - 1 = 0 \implies 2\pi i z^2 = 2\pi i k, k\in \mathbb{Z}$.
Therefore, the poles of this function are at $z^2 = k$.

We are also told that $n< R^2 < n+1$ for some integer $n\geq 0$.
For the point of illustration say $n=3$ then $3.
By definition $\Gamma_R$ is the circle $|z| = R$.
The poles need to occur inside this circle, so $|z| < R \implies |z^2| < R^2$.
Thus, the poles are all $k\in \mathbb{Z}$ with $|k| < R^2$.
In this case $k=0,\pm 1, \pm 2, \pm 3$.

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