SNCA:Contour integration

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In mathematics and science, contour integration, invented by Augustin-Louis Cauchy, is a method in complex analysis that can:

• evaluate definite integrals (residue theorem)
• find location of poles and zeroes of a function (argument principle )
• give upper and lower bounds for integrals (ML inequality)

It is a mandatory course in TND school. It can also be used to demonstrate the usage of thrembo, as well as proving the Holocaust death toll was 271,000, not 6 million.

Note: I half assed this page and it may contain some errors

Use

Consider a function $g\left(z\right)$ which is holomorphic in domain $D$, and $\gamma$ is a closed curve in the complex plane. Then let $z:D\to \mathbb{ℂ}$ parameterize $\gamma$. Thus:

${\oint }_{\gamma }g\left(z\right)dz=\underset{D}{\int }g\left(z\left(t\right)\right)z^{\prime }\left(t\right)dt=0$

A common parameterization of $\gamma$ is $z\left(t\right)=R{e}^{it}$. We can see clearly why the integral of a holomorphic function is zero, due to the fundamental theorem of calculus:

${\oint }_{\gamma }g\left(z\right)dz=\underset{0}{\overset{2\pi }{\int }}iR{e}^{it}g\left(R{e}^{it}\right)dt=G\left(R\right)-G\left(R\right)=0$

But things aren't so simple when it comes to meromorphic functions, that is, functions which have countably many poles. Examples of some meromorphic functions in the complex plane are $\frac{1}{z^2+1}$, $\frac{1}{z}$ and $\tan \left(z\right)$; functions with essential singularities are not meromorphic (example(s): $e^{\frac{1}{z}}$, $\sin \left(\frac{1}{z}\right)$). An integral of a meromorphic function over a closed curve is given in terms of the 'residues' of the function (this has to do with homology, or something). The manner which we calculate a residue for a meromorphic function $f\left(z\right)$ is extracting the coefficient $a_{-1}$ of the Laurent series representation of $f\left(z\right)$, given by:

$f\left(z\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n{(z-z_k)}^n$ $=\dots +a_{-2}{(z-z_k)}^{-2}+a_{-1}{(z-z_k)}^{-1}+a_0+a_1(z-z_k)+a_2{(z-z_k)}^2+\dots$

Assuming (for simplicity) that the poles of $f\left(z\right)$ are of order one, we can easily compute the residues for the set of poles $P=\{z_1,z_2,\dots ,z_k\}$:

${\oint }_{\gamma }f\left(z\right)dz=2\pi i\times \sum _{\alpha \in P}\left(\underset{z\to \alpha }{\lim }(z-\alpha )f\left(z\right)\right)$

Given a pole of order one, $f\left(z\right)$ decomposes into $f\left(z\right)=\frac{A\left(z\right)}{B\left(z\right)}$, then:

${\oint }_{\gamma }f\left(z\right)dz=2\pi i\times \sum _{\alpha \in P}\frac{A\left(\alpha \right)}{B^{\prime }\left(\alpha \right)}$

Example

Consider the following contour in the complex plane, where $R$ is a limit going to infinity, or something.

Let's find $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2+1}$.

The first half of the contour is a straight line that goes from $-R$ to $R$, hence, this can be represented as, ($H$ is the entire contour, because I said so)

$\underset{H}{\oint }\frac{dx}{x^2+1}=\underset{{\gamma }_R}{\int }\frac{dx}{x^2+1}+\underset{-R}{\overset{R}{\int }}\frac{dx}{x^2+1}$

The other half of the contour, ${\gamma }_R$, can be represented as the equation $x=R{e}^{iv}$ where $v$ is a variable that goes from $\pi$to $0$/

Differentiating, we have $dx=Ri{e}^{iv}dv$. Plugging it back into our original integrals:

$\underset{H}{\oint }\frac{dx}{x^2+1}=\underset{\pi }{\overset{0}{\int }}\frac{Ri{e}^{iv}dv}{(R{e}^{iv}{)}^2+1}+\underset{-R}{\overset{R}{\int }}\frac{dx}{x^2+1}$

If we set:

$P=\underset{\pi }{\overset{0}{\int }}\frac{Ri{e}^{iv}dv}{R^2{e}^{2iv}+1}$

Then take the absolute value:

$|P|=\left|\underset{\pi }{\overset{0}{\int }}\frac{Ri{e}^{iv}}{R^2{e}^{2iv}+1}dv\right|$

From the triangle inequality for integrals:

$|P|\le \underset{\pi }{\overset{0}{\int }}\left|\frac{Ri{e}^{iv}}{R^2{e}^{2iv}+1}\right|dv$

$|P|\le \underset{\pi }{\overset{0}{\int }}\frac{R|e^{iv}|}{|R^2{e}^{2iv}+1|}dv$

It's known that if $z$ is real, then $|e^{iz}|=1$. Thus:

$|P|\le \underset{\pi }{\overset{0}{\int }}\frac{R}{|R^2{e}^{2iv}+1|}dv$

Using the triangle inequality :

$|P|\le \underset{\pi }{\overset{0}{\int }}\frac{R}{|R^2{e}^{2iv}|+|1|}dv$

$|P|\le \underset{\pi }{\overset{0}{\int }}\frac{R}{|R^2|+|1|}dv$

Since there is no more $v$, we can move the thing out of the integral, or so I think

$|P|\le \frac{-R\pi }{R^2+1}$

If we allow $R\to \mathrm{\infty }$:

$|P|\le \underset{R\to \mathrm{\infty }}{\lim }\frac{-R\pi }{R^2+1}$

$|P|\le \underset{R\to \mathrm{\infty }}{\lim }\frac{-\pi }{R+1/R}$

$|P|\le 0$

Since the absolute value of something is always greater than $0$ unless it is $0$, then $P=0$, this gives us:

$\underset{H}{\oint }\frac{dx}{x^2+1}=\underset{-R}{\overset{R}{\int }}\frac{dx}{x^2+1}$

Then, letting $R\to$:

$\underset{H}{\oint }\frac{dx}{x^2+1}=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2+1}$

inside $\frac{1}{x^2+1}$, there are only 2 poles, where $x=\pm i$, inside our contour, there is only one pole, $x=i$, hence, using the residue theorem:

$2\pi i\underset{x=i}{Res}\frac{1}{x^2+1}=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2+1}$

$2\pi i\underset{x\to i}{\lim }\frac{(x-i)}{x^2+1}=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2+1}$

$x^2+1=(x-i)(x+i)$, hence:

$2\pi i\underset{x\to i}{\lim }\frac{1}{x+i}=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2+1}$

Setting the limit:

$\pi =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2+1}$

mathGODS won

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