Contour integration: Difference between revisions
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plugging it back into our original integrals | plugging it back into our original integrals | ||
<math> \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} \mathrm{d} x | <math> \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</math></blockquote> | ||
if we set : | |||
<math> P=\int\limits_{\pi}^{0} \frac{Rie^{iv} \mathrm{d} v}{(Re^{iv})^2+1}</math></blockquote> | |||
Now take the absolute value | |||
<math> |P|=|\int\limits_{\pi}^{0} \frac{Rie^{iv} \mathrm{d} v}{(Re^{iv})^2+1}|</math></blockquote> |
Revision as of 14:47, 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