# WINSEM2019-20_MAT3003_TH_VL2019205000322_Reference_Material_IV_07-Feb-2020_Module_4.4_Evaluation_of_

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Evaluation of Real Integrals through Residues Complex Variables and Partial Differential Equations (MAT3003) Module 4 Dr. T. Phaneendra Professor of Mathematics (Higher Academic Grade) Vellore Institute of Technology, Vellore - 632014 (TN) February 7, 2020
Contour Integration Module 4 1 Cauchy’s Residue Theorem If f ( z ) is analytic inside and on a simple (positively oriented) closed contour γ , except at a finite number of singularities z 1 , z 2 , ··· , z n within it, then γ f ( z ) d z = 2 π i n k = 1 Res ( f ; z k ) . (1.1) 2 Evaluation of Real Integrals Type 1 2 π ˆ 0 Q ( cos t , sin t ) d t : We reduce it into a contour integral γ f ( z ) d z , where γ is the unit circle | z | = 1 . General Procedure I Let z = e i t be any point on γ : | z | = 1 so that d z = i e i t d t or d t = d z / i z , (2.1) where 0 t 2 π . I Using Euler’s identity: e i t = cos t + i sin t , we get cos t = e i t + e - i t 2 = z 2 + 1 2 z , sin t = e i t - e - i t 2 i = z 2 - 1 2 z i · (2.2) I Therefore, 2 π ˆ 0 Q ( cos t , sin t ) d t = γ 1 i z · Q z 2 + 1 2 z , z 2 - 1 2 z i | {z } = f ( z ) d z . (2.3) I Find the poles z 1 , z 2 , ··· , z n , of f ( z ) lying within γ , and the residues of f ( z ) there at. I Finally, apply (1.1). CVPDE(MAT3003) 1 SJT, 511-A10
Contour Integration Module 4 Example 2.1. Evaluate 2 π ˆ 0 sin t 5 + 4cos t d t . Solution. Consider the unit circle γ : | z | = 1 . Employing (2.1) and (2.2), we get I = 2 π ˆ 0 sin t 5 + 4cos t d t = γ 1 i z · ( z 2 - 1 ) / 2 i z 5 + 4 ( z 2 + 1 ) / 2 z d z = - 1 4 γ z 2 - 1 z ( z + 1 / 2 )( z + 2 ) | {z } = f ( z ) d z Note that 0 , - 1 / 2 and - 2 are simple poles of f , of which 0 and - 1 / 2 lie inside γ . Therefore, Res ( f ; - 1 / 2 ) = lim z →- 1 / 2 ( z + 1 / 2 ) f ( z ) = lim z →- 1 / 2 z 2 - 1 z ( z + 2 ) = 1 , Res ( f ;0 ) = lim z 0 z f ( z ) = lim z 0 z 2 - 1 ( z + 1 / 2 )( z + 2 ) = - 1 . In view of (1.1), I = - 1 4 · 2 π i [ Res ( f ; - 1 / 2 )+ Res ( f ;0 )] = - 1 4 · 2 π i ( - 1 + 1 ) = 0 . Example 2.2. Evaluate 2 π ˆ 0 d t 2 + sin t · Solution. Consider the unit circle γ : | z | = 1 . Employing (2.1) and (2.2), we get I = 2 π ˆ 0 d t 2 + sin t = γ 1 i z · 1 2 +( z 2 - 1 ) / 2 i z d z = 2 γ 1 z 2 + 4 i z - 1 | {z } = f ( z ) d z The poles of f are given by z 2 + 4 i z - 1 = 0 , which are the simple poles - 2 i ± 3 . But a = - 2 i + 3 lies inside γ . Therefore, Res ( f ; a ) = lim z a ( z + a ) f ( z ) = lim z →- 2 i + 3 1 z - ( - 2 i - 3 ) = 1 2 3 i In view of (1.1), I = 2 · 2 π i Res ( f ; a ) = 2 · 2 π i ( 1 / 2 3 i ) = 2 π / 3 = 2 3 π / 3 . CVPDE(MAT3003) 2 SJT, 511-A10
Contour Integration Module 4 Example 2.3. Evaluate 0 2 π ˆ d t sin 2 t + 4cos 2 t · Example 2.4. Compute