Contour integration

This page is a ruby.

In mathematics, contour integration, invented by Augustin-Lous 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.

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-\infty}^{\infty} \frac{1}{x^2+1} \mathrm{d} x}

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)

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \oint\limits_{H}\frac{1}{x^2+1} \mathrm{d} x=\int\limits_{\gamma_R}\frac{1}{x^2+1} \mathrm{d} x+\int\limits_{-R}^{R}\frac{1}{x^2+1} \mathrm{d} x}

the other half of the contour, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma_R} , can be represented as the equation : Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=Re^{iv}} where v is a variable that goes from Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi} to 0

differentating, we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{d}x=Rie^{iv} \mathrm{d} v}

plugging it back into our original integrals

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \oint\limits_{H}\frac{1}{x^2+1} \mathrm{d} x=\int\limits_{\pi}^{0} \frac{Rie^{iv} \mathrm{d} v}{(Re^{iv})^2+1} +\int\limits_{-R}^{R}\frac{1}{x^2+1} \mathrm{d} x}

if we set :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P=\int\limits_{\pi}^{0} \frac{Rie^{iv} \mathrm{d} v}{R^2e^{2iv}+1}}

Now take the absolute value

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |P|=|\int\limits_{\pi}^{0} \frac{Rie^{iv} }{R^2e^{2iv}+1}\mathrm{d} v|}

from the triangle inequality for integrals :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |P|\le \int\limits_{\pi}^{0} | \frac{Rie^{iv} }{R^2e^{2iv}+1}|\mathrm{d} v}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |P|\le \int\limits_{\pi}^{0} \frac{R|e^{iv}| }{|R^2e^{2iv}+1|}\mathrm{d} v}

It's known that if z is real, then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |e^{iz}|=1}

thus :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |P|\le \int\limits_{\pi}^{0} \frac{R }{|R^2e^{2iv}+1|}\mathrm{d} v}

Using the triangle inequality :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |P|\le \int\limits_{\pi}^{0} \frac{R }{|R^2e^{2iv}|+|1|}\mathrm{d} v}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |P|\le \int\limits_{\pi}^{0} \frac{R }{|R^2| +|1|}\mathrm{d} v}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |P|\le \frac{-R \pi }{R^2 +1}}

If we set a limit with R go to infinity :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |P|\le \lim_{R \to \infty} \frac{-R\pi }{R^2 +1}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |P|\le \lim_{R \to \infty} \frac{-\pi }{R +1/R}}

letting the limit go, we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |P|\le 0}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \oint\limits_{H}\frac{1}{x^2+1} \mathrm{d} x= \int\limits_{-R}^{R}\frac{1}{x^2+1} \mathrm{d} x}

We can let R go to infinity

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \oint\limits_{H}\frac{1}{x^2+1} \mathrm{d} x= \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x}

inside 1/(x^2+1), there is only 2 poles, where x=-i, and where x=i, inside our contour, there is only one pole, x=i, hence, using the residue theorem:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\pi i \underset{x=i}{Res}\frac{1}{x^2+1}= \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \pi i \lim_{x \to i } \frac{(x-i)}{x^2+1}= \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x}

x^2+1=(x-i)(x+i), hence :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\pi i \lim_{x \to i } \frac{1}{x+i}= \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x}

Setting the limit

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi = \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x}

mathGODS won