SNCA:Contour integration

	This is actually... LE HELPFUL? You can read this for more info about the topic

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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

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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