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4 Contour integration

Figure: Integral contour $ \mathcal{C}$ enclosing the upper half plane.
\includegraphics[width=0.3\textwidth]{mp/contour.1}

As a brief review, some definite integrals from $ -\infty$ to $ \infty$ can be evaluated by integrating along the contour $ \mathcal{C}$ shown in Figure 1.

A sufficient condition on the function $ f(z)$ to be integrated, is that $ \lim_{\vert z\vert\rightarrow\infty}\vert f(z)\vert$ falls off at least as fast as $ \frac{1}{z^2}$. When this is the case, the integral around the outer semicircle of $ \mathcal{C}$ is 0, so the $ \ensuremath{\int_{\ensuremath{\mathcal{C}}}^{} {f(z)} \dd{z}} = \ensuremath{\int_{-\infty}^{\infty} {f(z)} \dd{z}}$.

We can evaluate the integral using residue theorem

$\displaystyle \ensuremath{\int_{\ensuremath{\mathcal{C}}}^{} {f(x)} \dd{z}} = \...
...ensuremath{\mathcal{C}}}} 2\pi i \operatorname{Res}\left({z=z_p},{f(z)}\right),$ (63)

where for simple poles (single roots)

$\displaystyle \operatorname{Res}\left({z=z_p},{f(z)}\right) = \lim_{z \rightarrow {z_p}}(z-z_p) f(z),$ (64)

and in general for a pole of order $ n$

$\displaystyle \operatorname{Res}\left({z=z_p},{f(z)}\right) = \frac{1}{(n-1)!} ...
...\lim_{z \rightarrow {z_p}} \nderiv{n-1}{z}{}\left[ (z-z_p)^n \cdot f(z) \right]$ (65)


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4 Contour integration
Copyright © 2009-10-12, W. Trevor King (contact)
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Drexel Physics