Question

19 = 1 7. (15 points) Sections 6.4-6.6 Use the residue theorem to calculate the following integrals. a. 1 = 1 d 2 +9 dar b. 12 = 1. <+1) h c. Cos 2x da 2? + 4 2i Vr+e? da x2 + 4 d. 1 = 2 va + che de

Answer 1

solution's We know that Cauchy Residue Residue Theorem say f Z) d Z ani ( Sum of Residue I = dn 2 n49 let tz) then Pole of tez) is 2tg zo 2. 7 z=131 Im(z) let c is a cemi-cirde in above half Plane This only Pole si lies inabove halp plane R Relz) Residue at z=31 lim (2-3i). - ?>si Zxg By Cauchy Residue Theorem fizidz R dz zzg (tinte () dz take lim Ryo dz 29 km du 249 프 6 dn (x+1)2 let f(2)= (2+1) Pole of fez): (271)² = 07 ZzI1', ti

(n-1)! Zazo dzmarliz-zo) fiz)) Take cas semicircle in above half Plane as before Part 9 les in above half Plane. I only Pole Residue at tin ( zi) fiz) Zi z=i ie Pole of order 2 then Res f(2) Zxzo lim d fiz) Res fLZ) d zbi (2-1)! zoi lim a 22 d dz zti .2 lim Zrie -2 M (z+i} (21)3 41 Then by Cauchy Residure Theorem ( tezld 2 = R dz ani ( 4 ) = 7 7 -R take, lim R70 dz (271)2 KIN dz (2?172 9 du (x²4172 4 cosqu du azi tuke tez) = 2 take only n74 274 Polei 2=1 gi zadi in above halt Plane aiz lim (2-2i) e (244) 4 i Residue at z-qi 4 e 2721

By candy Residue Theorem ziz flz)dz -dz ani ( Sum of Residue) 274 R 9 2iZ e d Z 244 anil e N 2 ziz e 274 7 dz رودم 9 e 2 any so eziz e -dz zt 4 Kld 2 212 e dz o z274 4 Cox 2 2 z dzt 9 е я o 815 2 Zdz Z+4 274 Iz ( I =colz + Isonz comporring Real and imaginary Parta Col2Z ₂44 a dz = 4 og corqu 2 xt 4 dn te 4 4

ce tering 212 d 14 In t du let fez)= 21/2 te 249 274 same as above Pole in above half Plane is z=gi Residue at z=gi lim (2-21) f(z) Z721 3 21 3 2 t Zari = lim (2_2i) 2i 3 z te (374) + eziz (2421) (21) Aste 41 11 21 (21) + 21121) 41 using Cauchy Residue Theorem lim fizlez tayz = smil 4*+2) 2700 - R ziz S gi 32 t dz ala 274 (21) Sté) 12 zin 21 3nte dn 244 A l (21) tét