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
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F061769ab-7580-4e62-9a7f-e1405e983ea2%2Fb475937d-563a-4f3c-a224-6b9ed9417828%2Frg9hq9c_processed.jpeg&w=3840&q=75)
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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