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| <math>\int\limits_{-\infty}^{\infty} \frac{1}{x^2+1} \mathrm{d} x</math></blockquote> | | <math>\int\limits_{-\infty}^{\infty} \frac{1}{x^2+1} \mathrm{d} x</math></blockquote> |
| the first half of the contour is a straight line that goes from -R to R | | the first half of the contour is a straight line that goes from -R to R |
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| hence, this can be represented as, (H is the entire contour, because i said so) | | hence, this can be represented as, (H is the entire contour, because i said so) |
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| <math> \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</math></blockquote> | | <math> \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</math></blockquote> |
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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: