Contour integration: Difference between revisions
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In mathematics, '''contour integration''', invented by [[wikipedia:Augustin-Louis_Cauchy|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) | |||
{{corner|br|[[File:Chudblackboard.png|frameless]]}} | |||
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 : | Consider the following contour in the complex plane : | ||
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where R is a limit going to infinity...or something | where R is a limit going to infinity...[[Amerimutt|or something]] | ||
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if we set : | if we set : | ||
<math> P=\int\limits_{\pi}^{0} \frac{Rie^{iv} \mathrm{d} v}{R^2e^{2iv}+1}</math></blockquote> | <math> P=\int\limits_{\pi}^{0} \frac{Rie^{iv} \mathrm{d} v}{R^2e^{2iv}+1}</math></blockquote> | ||
Now take the absolute value | Now take the absolute value | ||
<math> |P|=|\int\limits_{\pi}^{0} \frac{Rie^{iv} }{R^2e^{2iv}+1}\mathrm{d} v|</math></blockquote> | <math> |P|=|\int\limits_{\pi}^{0} \frac{Rie^{iv} }{R^2e^{2iv}+1}\mathrm{d} v|</math></blockquote> | ||
from the triangle inequality for integrals : | from the triangle inequality for integrals : | ||
<math> |P|\le | |||
<math> |P|\le | <math> |P|\le \int\limits_{\pi}^{0} | \frac{Rie^{iv} }{R^2e^{2iv}+1}|\mathrm{d} v</math></blockquote> | ||
<math> |P|\le \int\limits_{\pi}^{0} \frac{R|e^{iv}| }{|R^2e^{2iv}+1|}\mathrm{d} v</math></blockquote> | |||
It's known that if z is real, then | It's known that if z is real, then | ||
<math>|e^{iz}|=1</math></blockquote> | <math>|e^{iz}|=1</math></blockquote> | ||
thus : | thus : | ||
<math> |P|\le | |||
<math> |P|\le \int\limits_{\pi}^{0} \frac{R }{|R^2e^{2iv}+1|}\mathrm{d} v</math></blockquote> | |||
Using the triangle inequality : | Using the triangle inequality : | ||
<math> |P|\le | |||
<math> |P|\le | <math> |P|\le \int\limits_{\pi}^{0} \frac{R }{|R^2e^{2iv}|+|1|}\mathrm{d} v</math></blockquote> | ||
<math> |P|\le \int\limits_{\pi}^{0} \frac{R }{|R^2| +|1|}\mathrm{d} v</math></blockquote> | |||
Since there is no more v, we can move the thing out of the integral, or so i think | Since there is no more v, we can move the thing out of the integral, or so i think | ||
<math> |P|\le \frac{-R \pi }{R^2 +1}</math></blockquote> | <math> |P|\le \frac{-R \pi }{R^2 +1}</math></blockquote> | ||
If we set a limit with R go to infinity : | If we set a limit with R go to infinity : | ||
<math> |P|\le \lim_{R \to \infty} \frac{-R\pi }{R^2 +1}</math></blockquote> | <math> |P|\le \lim_{R \to \infty} \frac{-R\pi }{R^2 +1}</math></blockquote> | ||
<math> |P|\le \lim_{R \to \infty} \frac{-\pi }{R +1/R}</math></blockquote> | <math> |P|\le \lim_{R \to \infty} \frac{-\pi }{R +1/R}</math></blockquote> | ||
letting the limit go, we get | letting the limit go, we get | ||
<math> |P|\le 0</math></blockquote> | <math> |P|\le 0</math></blockquote> | ||
Since the absolute value of something is always greater than 0 unless its 0, then P=0, this gives us | Since the absolute value of something is always greater than 0 unless its 0, then P=0, this gives us | ||
<math> \oint\limits_{H}\frac{1}{x^2+1} \mathrm{d} x= \int\limits_{-R}^{R}\frac{1}{x^2+1} \mathrm{d} x</math></blockquote> | <math> \oint\limits_{H}\frac{1}{x^2+1} \mathrm{d} x= \int\limits_{-R}^{R}\frac{1}{x^2+1} \mathrm{d} x</math></blockquote> | ||
We can let R go to infinity | We can let R go to infinity | ||
<math> \oint\limits_{H}\frac{1}{x^2+1} \mathrm{d} x= \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x</math></blockquote> | <math> \oint\limits_{H}\frac{1}{x^2+1} \mathrm{d} x= \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x</math></blockquote> | ||
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: | 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: | ||
<math> 2\pi i \underset{x=i}{\ | |||
<math> 2\pi i \underset{x=i}{Res}\frac{1}{x^2+1}= \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x</math></blockquote> | |||
<math> 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</math></blockquote> | |||
x^2+1=(x-i)(x+i), hence : | |||
<math> 2\pi i \lim_{x \to i } \frac{1}{x+i}= \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x</math></blockquote> | |||
Setting the limit | |||
<math> \pi = \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x</math></blockquote> | |||
mathGODS won |
Latest revision as of 18:51, 2 January 2024
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
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)
the other half of the contour, , can be represented as the equation : where v is a variable that goes from to 0
differentating, we have
plugging it back into our original integrals
if we set :
Now take the absolute value
from the triangle inequality for integrals :
It's known that if z is real, then
thus :
Using the triangle inequality :
Since there is no more v, we can move the thing out of the integral, or so i think
If we set a limit with R go to infinity :
letting the limit go, we get
Since the absolute value of something is always greater than 0 unless its 0, then P=0, this gives us
We can let R go to infinity
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:
x^2+1=(x-i)(x+i), hence :
Setting the limit
mathGODS won