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ze Sol. Consider the integral f(z) dz where flz) = ,2 taken round a closed contour z* + a C, consisting of a semi-circle CR which is upper half of a large circle | z | = R and the part of real axis from R to R. For poles, 22 + a? = 0 z = t ai z = ai is the only pole which lie inside C. Residue of flz) at (z ai) = lim (z ai). zek ai e 2- ai (z - ai)(z + ai) 2ai 2

ze Sol. Consider the integral f(z) dz where flz) = ,2 taken round a closed contour z* + a C, consisting of a semi-circle CR which is upper half of a large circle | z | = R and the part of real axis from R to R. For poles, 22 + a? = 0 z = t ai z = ai is the only pole which lie inside C. Residue of flz) at (z ai) = lim (z ai). zek ai e 2- ai (z - ai)(z + ai) 2ai 2

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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x +a Sol. Consider the integral f(z) dz where fz) = taken round a closed contour C, consisting of a semi-circle C, which is upper half of a large circle | z | = R and the part of real axis from R to R. For poles, z2 + a2 = 0 z = t ai z = ai is the only pole which lie inside C. -a zek aie-a Residue of flz) at (z = ai) = lim (z ai). 2- ai (z - ai)(z + ai) 2ai 2 :: By Cauchy Residue theorem, R xex е dz 2ni 2 dx + = Tiea -Rx2 +a? CR 2 + a Taking limit as R o, R xe Lt R J-R x2 + a? dx + Lt dz = nie-a 2 R- * JCR 2“ + a“ 2 xe ze" dx + Lt dz = niea ..(1) %3D x+ a* 2 R-o JCR 2 2 + a' Since 0 as | z |→ 0o, therefore by Jordan's Lemma, 2* +a' zez Lt R JCR 22 + a? dz = 0 2 xe dx From (1), = Ti e-a :: +a Comparing imaginary parts, x sin x dx = nea x sin x dx = or Jo x +a
Transcribed Image Text:x +a Sol. Consider the integral f(z) dz where fz) = taken round a closed contour C, consisting of a semi-circle C, which is upper half of a large circle | z | = R and the part of real axis from R to R. For poles, z2 + a2 = 0 z = t ai z = ai is the only pole which lie inside C. -a zek aie-a Residue of flz) at (z = ai) = lim (z ai). 2- ai (z - ai)(z + ai) 2ai 2 :: By Cauchy Residue theorem, R xex е dz 2ni 2 dx + = Tiea -Rx2 +a? CR 2 + a Taking limit as R o, R xe Lt R J-R x2 + a? dx + Lt dz = nie-a 2 R- * JCR 2“ + a“ 2 xe ze" dx + Lt dz = niea ..(1) %3D x+ a* 2 R-o JCR 2 2 + a' Since 0 as | z |→ 0o, therefore by Jordan's Lemma, 2* +a' zez Lt R JCR 22 + a? dz = 0 2 xe dx From (1), = Ti e-a :: +a Comparing imaginary parts, x sin x dx = nea x sin x dx = or Jo x +a
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