Contour integration: Difference between revisions
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<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}{textup{Res}} \frac{1}{x^2+1}= \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x</math></blockquote> | <math> 2\pi i \underset{x=i}{\textup{Res}} \frac{1}{x^2+1}= \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x</math></blockquote> |
Revision as of 15:01, 21 July 2023
The residue theorem, invented by Cauchy, is a method of evaluating definite integrals.
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:
Failed to parse (unknown function "\textup"): {\displaystyle 2\pi i \underset{x=i}{\textup{Res}} \frac{1}{x^2+1}= \int\limits_{-\infty}^{\infty}\frac{1}{x^2+1} \mathrm{d} x}