Contour integration: Difference between revisions
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Etg bastien (talk | contribs) (Created page with "The residue theorem, invented by Cauchy, is a method of evaluating definite integrals. Consider the following contour in the complex plane : File:ContourIntegral_1000.svg Let's find <math> \int\limits_{-\infty}^{\infty} \frac{1}{x^2+1} \mathrm{d} x <math>") |
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Consider the following contour in the complex plane : | Consider the following contour in the complex plane : | ||
[[ | [[File:chrome_VzRESuQOYt.png|400px]] | ||
where R is a limit going to infinity...or something | |||
Let'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 R | |||
hence, 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, γR, can be represented as the equation : | |||
<math> | |||
<math>x=Re^{iv}</math></blockquote> | |||
where v is a vriable that goes from pi to 0 | |||
the other half of the contour is just a straight line going from -R to R, hence, we have | |||
<math>\oint\limits_x=Re^{iv}</math></blockquote> |
Revision as of 14:40, 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, γR, can be represented as the equation :
where v is a vriable that goes from pi to 0 the other half of the contour is just a straight line going from -R to R, hence, we have