Difference between revisions of "Jordan's lemma"

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(Created page with " We consider integrals of the form <math> \int_C f(z) e^{i a z} \; dz </math>")
 
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[[File:contour_Jordan_1.png|thumb|right|350px|"Figure 1: the contour C for a Fourier-like integration (upward closure only)"]]
  
We consider integrals of the form  
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We consider integrals of the form <math> \int_C f(z) e^{i a z} \; dz </math>, where <math> C = C_1 + C_2 </math> is a closed simple contour as shown in Figure 1.
<math> \int_C f(z) e^{i a z} \; dz </math>
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The function <math> f(z) </math> has a finite number of poles located inside of the contour. As <math> |z| \rightarrow \infty</math> <math> |f(z)| </math> is assumed to decay at least as fast as <math> \frac{1}{|z|}</math>. By the [[Residue Theorem]] the value of this integral may be written as
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<math> <math> 2 \pi i \sum \mbox{Res} f(z) e^{i a z} = \int_{C_1} f(z) e^{i a z} \; dz + \int_{C_2} f(z) e^{i a z} \; dz . </math> </center>
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Typically, the application of such a contour integral is to find the value of the integral along the real axis, which is the integral along <math> C_1 </math>.
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The next step is to evaluate the integral over the contour <math> C_2 </math> and show that this vanishes. We consider the following estimate
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<center> <math> |\int_{C_2} f(z) e^{i a z} \; dz | \le \int_{C_2}  | f(z) e^{i a z}| \;| dz | </math>. </center>
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By [[Cauchy's theorem]] we may assume that <math> C_2 </math> is a semi-circle, and may consider the polar form of <math> z = |z|e^{i\phi} </math>,
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<math> dz = i|z|e^{i \phi} d\phi </math>, and we may also consider <math> z = |z| (\cos \phi + i \sin \phi)</math> in the exponential. Substituting
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these into the  <math> C_2 </math> integral yields
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<center> <math> \int_{C_2}  | f(z) e^{i a z}| \;| dz | = \int_0^{\pi} |f(z) e^{i a |z|(\cos \phi  - \sin \phi)}| | i|z|e^{i \phi} d\phi| \le \lim_{R\rightarrow \infty} \frac{M}{R} \int_0^{\pi} e^{-aR \sin(\phi)} \; R d\phi,  </math> </center>
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where <math> R </math> is the radius of the semi-circle <math> C_2 .</math> for this integral to decay the <math> \mbox{Re} a > 0 .</math> Here, the [[Maximum Modulus theorem]] has been applied, and we have
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We may further proceed with the estimate by recognizing that integral is twice the same integral, with upper integration
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limit of <math> \frac{\pi}{2}</math> and the extra factor of 2 is absorbed into the <math> M </math> yielding
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<center> <math> \int_{C_2}  | f(z) e^{i a z}| \;| dz | \le \lim_{R\rightarrow \infty} M \int_0^{\pi/2} e^{-aR \sin(\phi)} \; d\phi \le \lim_{R\rightarrow \infty} M \int_0^{\pi/2} e^{-aR \frac{2}{\pi}(\phi)} \; d\phi </math> </center>
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<center> <math>  \qquad \qquad \qquad \qquad \qquad = \lim_{R\rightarrow \infty}
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\left. \frac{ - 2 M}{\pi a R}  e^{-aR \frac{2}{\pi}(\phi)} \right|_0^{\pi/2} = \frac{2 M}{\pi a R} ( 1 - e^{-aR} ) \rightarrow 0 . </math> </center>
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Here we have used the fact that on the interval <math> [0, \pi/2 ]</math>  we have <math> -\sin(\phi) \le - \phi (\pi/2) </math>.

Revision as of 19:53, 8 October 2015

"Figure 1: the contour C for a Fourier-like integration (upward closure only)"

We consider integrals of the form , where is a closed simple contour as shown in Figure 1. The function has a finite number of poles located inside of the contour. As is assumed to decay at least as fast as . By the Residue Theorem the value of this integral may be written as

Typically, the application of such a contour integral is to find the value of the integral along the real axis, which is the integral along .

The next step is to evaluate the integral over the contour and show that this vanishes. We consider the following estimate

.

By Cauchy's theorem we may assume that is a semi-circle, and may consider the polar form of , , and we may also consider in the exponential. Substituting these into the integral yields


where is the radius of the semi-circle for this integral to decay the Here, the Maximum Modulus theorem has been applied, and we have


We may further proceed with the estimate by recognizing that integral is twice the same integral, with upper integration limit of and the extra factor of 2 is absorbed into the yielding



Here we have used the fact that on the interval we have .