Difference between revisions of "Residue calculus"
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The idea is to now take <math>r\rightarrow 0^+</math> and <math>R\rightarrow\infty</math>. Some easy calculations (left to the reader) show that the integrals along the semicircular arcs vanish, leaving us with only the integrals along the [[real axes]]. For <math>(0,\infty)</math> we get back the integral we are trying to calculate <math>I</math>, and on <math>(-\infty,0)</math> we get (since arg here is equal to <math>\pi i</math>): <math>I+\pi i\int_{-\infty}^0 dz/(1+z^2)=I+\pi^2 i/2</math>. Thus, plugging everything back in, <math>2I+\pi^2 i/2=\pi^2 i/2</math>, so <math>I=0</math>. | The idea is to now take <math>r\rightarrow 0^+</math> and <math>R\rightarrow\infty</math>. Some easy calculations (left to the reader) show that the integrals along the semicircular arcs vanish, leaving us with only the integrals along the [[real axes]]. For <math>(0,\infty)</math> we get back the integral we are trying to calculate <math>I</math>, and on <math>(-\infty,0)</math> we get (since arg here is equal to <math>\pi i</math>): <math>I+\pi i\int_{-\infty}^0 dz/(1+z^2)=I+\pi^2 i/2</math>. Thus, plugging everything back in, <math>2I+\pi^2 i/2=\pi^2 i/2</math>, so <math>I=0</math>. | ||
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[[Category:Calculus]] | [[Category:Calculus]] | ||
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Revision as of 14:50, December 30, 2008
The residue calculus is a method of definite integration which relies heavily on Cauchy's residue theorem. The idea is to rewrite a definite integral on the real line as limit of integrals in the complex plane which are, in some sense, easier to compute.
Some mathematicians reject the method of residue calculus, as it does not rely on elementary proofs.
Example
Here is a simple example. Suppose we want to find the integral of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \log(x)/(1+x^2)} from 0 to . We begin by choosing an analytic branch of the logarithm, defined everywhere in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{C}} except for the line consisting of negative purely imaginary numbers (that is, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \log(z)=\log|z|+i\arg(z)} where is specified to take values in . Now, choose real numbers where and . Then, let be the positively oriented contour consisting of the clockwise upper semicircular arc from to , the directed line segment from to , the anticlockwise semicircular arc from to , and the directed line segment from to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -r} (this looks like a rainbow in the complex plane).
The reason we chose this contour is that it necessarily avoids the inevitable singularity of the logarithm at 0, and it includes inside of it, the pole of the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(z)=log(z)/(1+z^2)} . By Cauchy's residue theorem, we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I:=\int_{\Gamma(r,R)}f(z)\,dz} = Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\pi i} times the residue of the pole at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z=i} . This is easily calculated, and it is equal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi/4} . Thus, the integral evaluates to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi^2 i/2} .
The idea is to now take Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\rightarrow 0^+} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R\rightarrow\infty} . Some easy calculations (left to the reader) show that the integrals along the semicircular arcs vanish, leaving us with only the integrals along the real axes. For Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0,\infty)} we get back the integral we are trying to calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} , and on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-\infty,0)} we get (since arg here is equal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi i} ): Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I+\pi i\int_{-\infty}^0 dz/(1+z^2)=I+\pi^2 i/2} . Thus, plugging everything back in, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2I+\pi^2 i/2=\pi^2 i/2} , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I=0} .