Contour integration: Difference between revisions
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The residue theorem, invented by Cauchy, is a method of evaluating definite integrals. | [[Shit Nobody Cares About|The residue theorem, invented by Cauchy, is a method of evaluating definite integrals.Consider the following contour in the complex plane :[[File:chrome_VzRESuQOYt.png|400px<nowiki>]]</nowiki>where R is a limit going to infinity...or somethingLet's find <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 Rhence, this can be represented as, (H is the entire contour, because i said so)<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>the other half of the contour, <math>\gamma_R</math></blockquote>, can be represented as the equation :<math>x=Re^{iv}</math></blockquote>where v is a variable that goes from <math>\pi</math></blockquote>to 0differentating, we have<math>\mathrm{d}x=Rie^{iv} \mathrm{d} v</math></blockquote>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} +\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>]] | ||
Consider the following contour in the complex plane : | |||
[[File:chrome_VzRESuQOYt.png|400px]] | |||
where R is a limit going to infinity...or | |||
<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 | |||
<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> | |||
the other half of the contour, <math>\gamma_R</math></blockquote>, can be represented as the equation : | |||
<math>x=Re^{iv}</math></blockquote> | |||
where v is a variable that goes from <math>\pi</math></blockquote>to | |||
<math>\mathrm{d}x=Rie^{iv} \mathrm{d} v</math></blockquote> | |||
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} +\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 14:52, 21 July 2023
[[Shit Nobody Cares About|The residue theorem, invented by Cauchy, is a method of evaluating definite integrals.Consider the following contour in the complex plane :![400px]]where R is a limit going to infinity...or somethingLet's find '"`UNIQ--postMath-00000001-QINU`"' the first half of the contour is a straight line that goes from -R to Rhence, this can be represented as, (H is the entire contour, because i said so)'"`UNIQ--postMath-00000002-QINU`"' the other half of the contour, '"`UNIQ--postMath-00000003-QINU`"' , can be represented as the equation :'"`UNIQ--postMath-00000004-QINU`"' where v is a variable that goes from '"`UNIQ--postMath-00000005-QINU`"' to 0differentating, we have'"`UNIQ--postMath-00000006-QINU`"' plugging it back into our original integrals'"`UNIQ--postMath-00000007-QINU`"' if we set :'"`UNIQ--postMath-00000008-QINU`"' Now take the absolute value'"`UNIQ--postMath-00000009-QINU`"'](/File:Chrome_VzRESuQOYt.png)
