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

Example[edit | edit source]

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

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

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

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

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

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

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

If we set:

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

Then take the absolute value:

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

From the triangle inequality for integrals:

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

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

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

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

Using the triangle inequality :

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

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

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

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

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

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

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

${\displaystyle |P|\le 0}$

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

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

Then, letting ${\displaystyle R\to }$:

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

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

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

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

${\displaystyle x^2+1=(x-i)(x+i)}$, hence:

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

Setting the limit:

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

mathGODS won

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