Transcription of Definite Integrals by Contour Integration
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Definite Integrals by Contour Integration
UNIVERSITY OF BRISTOLD epartment of PhysicsLevel 3 Mathematical Methods 33010 (2009/2010)Definite Integrals by Contour IntegrationMany kinds of (real) definite Integrals can be found using the results we have foundfor Contour Integrals in the complex plane. This is because the values of contourintegrals can usually be written down with very little difficulty. We simply haveto locate the poles inside the Contour , find the residues at these poles, and thenapply the residue theorem . The more subtle part of the job is to choose a suitablecontour integral one whose evaluation involves the definite integral required. Weillustrate these steps for a set of five types of definite 1 IntegralsIntegrals of trigonometric functions from 0 to 2 :I= 2 0(trig function)d By trig function we mean a function of cos and sin .The obvious way to turn this into a Contour integral is to choose the unit circle asthe Contour , in other words to writez=expi , and integrate with respect to .Onthe unit circle, both cos and sin can be written as simple algebraic functions ofz:cos =12(z+1/z)sin =12i(z 1/z)and making this replacement turns the trigonometric function into an algebraicfunction ofzwhose poles can be easily :I= 2 0d 1+acos where 1<a<+1I= 11+a2(z+1z)dziz=2i dz2z+az2+aThe poles of the function being integrated lie at the roots of the equationaz2+2z+a= 0 at the pointsz =1a( 1 1 a2)1CZ+Z-Of the poles, onlyz+lies inside the unit circle, soI=2 iR+whereR+is theresidue atz+To find the res
Nov 26, 2006 · (x2 +1)2 can be found by these methods, because the integral around the arc, which can be written as π 0 iReiθdθ (R2e2iθ +1)2 R−→→∞ π 0 iReiθdθ R4e4iθ = i R3 π 0 e−3iθdθ clearly vanishes as R →∞. Once again the way is clear for us to use the residue theorem, and inspection of the function 1 (z2 +1)2
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