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 : | 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) | *evaluate definite integrals (residue theorem) | ||
*find location of poles and zeroes of a function (argument principle ) | *find location of poles and zeroes of a function (argument principle ) | ||
*give upper and lower bounds for integrals (ML inequality) | *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 | '''note''' : I half assed this page and it may contain some errors | ||
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