This document discusses contour integration, which is calculating the value of a contour integral around a given contour in the complex plane. It describes four types of contour integration based on the form of the integral. Type I involves integrals from 0 to 2π of a rational function of sinθ and cosθ. Type II involves integrals from -∞ to ∞ or 0 to ∞. Type III involves integrals with polynomials in the numerator and denominator. Type IV involves integrals with poles on the real axis or contour identation. Examples are provided for each type to demonstrate the working rules and solutions.

2. Presented to: Ma’am Hunza
Presentedby: AmenahGondal
Class: BS.ED VI
Roll no: EDU(s)-2017-F-11

3. OBJECTIVES
WE WILL DISCUSS:
i. CONOTUR INTEGRATION AND ITS TYPES AND
EXAMPLES
ii. MITTAG LEFFLER THEOREM AND ITS EXAMPLE

5. CONTOUR INTEGRATION
“Contour integration is the process of calculating the values
of a contour integral around a given contour in the complex
plane.”

6. TYPES OF CONTOUR INTEGRATION
Form of the integral is 0
2𝜋
𝑓 𝑠𝑖𝑛𝜃, 𝑐𝑜𝑠𝜃 𝑑𝜃 where 𝑓 is a rational function of 𝑠𝑖𝑛𝜃, 𝑐𝑜𝑠𝜃.
WORKING RULE:
STEP I: Make the substitutions
𝑍 = 𝑒 𝑖𝜃
.
𝑐𝑜𝑠𝜃 =
1
2
(𝑍 +
1
𝑍
)
𝑠𝑖𝑛𝜃 =
1
2𝑖
(𝑍 −
1
𝑍
)
𝑑𝜃 =
𝑑𝑍
𝑖𝑍
𝑎𝑛𝑑 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙
0
2𝜋
𝑓 𝜃 𝑑𝜃 =
𝑐
𝑓 𝑍 𝑑𝑍
Where C is positively oriented unit circle 𝑍 = 1.
TYPE I

7. STEP II: Calculate the poles of 𝑓 𝑍 . Say at 𝑍0, 𝑍1, 𝑍2, … . , select those poles which lie in the unit circle
𝑍 = 1.then find the residues at the selected poles
𝑅1 𝑓, 𝑍0 , 𝑅2 𝑓, 𝑍1 𝑒𝑡𝑐.
STEP III: 0
2𝜋
𝑓 𝑠𝑖𝑛𝜃, 𝑐𝑜𝑠𝜃 𝑑𝜃 = 𝑐
𝑓 𝑍 𝑑𝑍 = 2𝜋𝑖 𝑗=1
𝑛
𝑅𝑗 (𝑢𝑠𝑖𝑛𝑔 𝐶𝑎𝑢𝑐ℎ𝑦𝑠′ 𝑅𝑒𝑠𝑖𝑑𝑢𝑒 𝑇ℎ𝑒𝑜𝑟𝑒𝑚)
EXAMPLE:
Prove that 𝟎
𝟐𝝅 𝒅𝝑
(𝒂+𝒃𝒄𝒐𝒔𝜽) 𝟐 =
𝟐𝝅𝒂
(𝒂 𝟐−𝒃 𝟐) 𝟑 𝟐.
SOLUTION:
As it is prove that 𝟎
𝟐𝝅 𝒅𝝑
𝒂+𝒃𝒄𝒐𝒔𝜽
=
𝟐𝝅
(𝒂 𝟐−𝒃 𝟐) 𝟏 𝟐 (A)
Differentiating A w.r.t a
0
2𝜋
−1 𝑑𝜗
𝑎 + 𝑏𝑐𝑜𝑠𝜃 2
= 2𝜋(−1 2)(𝑎2
− 𝑏2
)3 2
. 2𝑎
Cancelling negative signs from both sides,
0
2𝜋
𝑑𝜗
𝑎 + 𝑏𝑐𝑜𝑠𝜃 2
= =
2𝜋𝑎
(𝑎2 − 𝑏2)3 2

8. TYPE II
Form of the integral will be either −∞
∞
𝑓 𝑥 𝑑𝑥 𝑜𝑟 0
∞
𝑓 𝑥 𝑑𝑥.
WORKING RULE:
STEP I:
Replace x by Z in the integral and test whether 𝑍 𝑓 𝑍 → 0𝑎𝑠 𝑍 → ∞.
STEP II:
Find the poles of 𝑓(𝑍), locate those poles which lie in the upper half plane. Find the
residues at the locates poles.
STEP III:
Formula to be used is −∞
∞
𝑓 𝑥 𝑑𝑥 = 2𝜋𝑖 𝑅+ 𝑜𝑟 0
∞
𝑓 𝑥 𝑑𝑥 = 𝜋𝑖 𝑅+ 𝑤ℎ𝑒𝑟𝑒
𝑅+ 𝑑𝑒𝑛𝑜𝑡𝑒𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑠𝑖𝑑𝑢𝑒𝑠 𝑎𝑡 𝑝𝑜𝑙𝑒𝑠 𝑙𝑦𝑖𝑛𝑔 𝑖𝑛 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 ℎ𝑎𝑙𝑓 𝑝𝑙𝑎𝑛𝑒.

9. EXAMPLE:
Prove that −∞
∞ 𝒅𝒙
(𝒙 𝟐+𝟏) 𝟑 =
𝟑𝝅
𝟖
.
SOLUTION:
Given 𝑓 𝑥 =
1
(𝒙 𝟐+𝟏) 𝟑
𝑓 𝑍 =
1
(𝒁 𝟐 + 𝟏) 𝟑
𝑍𝑓 𝑍 → 0 𝑎𝑠 𝑍 → ∞
The poles of f(Z) are at Z=±𝑖 of order 3 the only pole which lies in the upper half plane is Z=i of order 3.
The residue at Z=i is
𝑅 𝑓, 𝑖 =
1
2!
𝑑2
𝑑𝑍2
[(𝑍 − 𝑖)3
1
𝑍 − 𝑖 3 𝑍 + 𝑖 3
] 𝑍=𝑖
=
1
2
𝑑2
𝑑𝑍2
1
𝑍 + 𝑖 3
=
1
2
−3 −4
1
(2𝑖)5
=
6
32𝑖
−∞
∞
𝑑𝑥
(𝑥2 + 1)3
= 2𝜋𝑖 ×
6
32𝑖
=
3𝜋
8

10. TYPE III:
Form of the integral is either
−∞
∞ 𝑃(𝑥)
𝑄(𝑥)
sin 𝑚𝑥 𝑑𝑥 𝑜𝑟 −∞
∞ 𝑃(𝑥)
𝑄(𝑥)
𝑐𝑜𝑠𝑚𝑥 𝑑𝑥 𝑜𝑟 𝑙𝑖𝑚𝑖𝑡 𝑖𝑠 𝑓𝑟𝑜𝑚 0 𝑡𝑜 ∞ , 𝑚 > 0 𝑤ℎ𝑒𝑟𝑒,
• 𝑃 𝑥 𝑎𝑛𝑑 𝑄(𝑥) are polynomials of x.
• 𝐷𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑄 𝑥 exceeds the degree of 𝑃 𝑥 .
• 𝑇ℎ𝑒 equation 𝑄 𝑥 = 0 has no real roots.
WORKING RULE:
STEP I:
Replace x by Z and sinmx or cosmx by 𝑒 𝑖𝑚𝑧
. Find the poles of 𝑓(𝑍)𝑒 𝑖𝑚𝑧
and choose those poles which
lie in the upper half plane. Find the residue at the chosen plane.
STEP II:
Then by Cauchy Residue theorem
 −∞
∞
𝑓 𝑥 𝑠𝑖𝑛𝑚𝑥𝑑𝑥 = 𝐼𝑚 (2𝜋𝑖 𝑅) or −∞
∞
𝑓 𝑥 𝑐𝑜𝑠𝑚𝑥𝑑𝑥 = 𝑅𝑒 (2𝜋𝑖 𝑅)
 0
∞
𝑓 𝑥 𝑠𝑖𝑛𝑚𝑥𝑑𝑥 = 𝐼𝑚 (𝜋𝑖 𝑅) or 0
∞
𝑓 𝑥 𝑐𝑜𝑠𝑚𝑥𝑑𝑥 = 𝑅𝑒 (𝜋𝑖 𝑅)

11. EXAMPLE:
Prove that −∞
∞ 𝐜𝐨𝐬 𝒙 𝒅𝒙
(𝒙 𝟐+𝒂 𝟐)(𝒙 𝟐+𝒃 𝟐)
=
𝝅
𝒂 𝟐−𝒃 𝟐 (
𝒆−𝒃
𝒃
𝒆−𝒂
𝒂
) (a>b>0)
SOLUTION:
Given integral can be written as
−∞
∞
𝑒 𝑖𝑧
𝑑𝑍
(𝑍2 + 𝑎2)(𝑍2+𝑏2)
The poles are at 𝑍 = ±𝑎𝑖 𝑎𝑛𝑑 𝑍 = ±𝑏𝑖.The only poles which lie in the upper half plane are at 𝑍 = 𝑎𝑖 𝑎𝑛𝑑 𝑍𝑏𝑖.
𝑅1 𝑓, 𝑎𝑖 = lim
𝑍→𝑎𝑖
𝑒 𝑖𝑧
𝑍 + 𝑎𝑖 𝑍2 + 𝑏2
=
𝑒−𝑎
2𝑎𝑖(𝑏2 − 𝑎2)
𝑅2 𝑓, 𝑏𝑖 = lim
𝑍→𝑏𝑖
𝑒 𝑖𝑧
𝑍 + 𝑎2 𝑍2 + 𝑏𝑖
=
𝑒−𝑏
2𝑏𝑖(𝑎2 − 𝑏2)
−∞
∞
cos 𝑥 𝑑𝑥
(𝑥2 + 𝑎2)(𝑥2+𝑏2)
= 𝑅𝑒[2𝜋𝑖 ×
1
2𝑖
(
𝑒−𝑏
𝑏 𝑎2 − 𝑏2
−
𝑒−𝑎
𝑎 𝑎2 − 𝑏2
]
= 𝜋
𝑎𝑒−𝑏
− 𝑏𝑒−𝑎
𝑎𝑏 𝑎2 − 𝑏2
=
𝜋
𝑎2 − 𝑏2
[
𝑒−𝑏
𝑏
−
𝑒−𝑎
𝑎
]

12. Form of the integral is either −∞
∞
𝑓 𝑥 [sin 𝑚𝑥 𝑜𝑟 cos 𝑚𝑥] 𝑑𝑥 𝑜𝑟 0
∞
𝑓 𝑥 [𝑠𝑖𝑛𝑚𝑥 𝑜𝑟 𝑐𝑜𝑠𝑚𝑥]𝑑𝑥.
WORKING RULE:
STEP I: Replace x by Z and cos mx or sin mx by 𝑒 𝑖𝑚𝑧
. Find poles of the integrand. Locate those poles
which lie in the upper half plane and on the real axis.
STEP II: Find the residues at poles in the upper half plane say 𝑅 𝑝 and residues at poles on the x-axis
say 𝑅 𝑥.
STEP III: −∞
∞
𝑓 𝑥 𝑠𝑖𝑛 𝑚𝑥 𝑑𝑥 = 𝐼𝑚[2𝜋𝑖𝑅 𝑝 + 𝜋𝑖𝑅 𝑥] 𝑜𝑟 −∞
∞
𝑓 𝑥 𝑐𝑜𝑠 𝑚𝑥 𝑑𝑥 = 𝑅𝑒 [2𝜋𝑖𝑅 𝑝 + 𝜋𝑖𝑅 𝑥] for
integral 0
∞
𝑑𝑖𝑣𝑖𝑑𝑒 𝑡ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡 𝑏𝑦 2.
TYPE IV: (CASE OF POLES ON THE REAL AXIS OR IDENTATION
OF CONTOURS)

13. EXAMPLE:
Prove that 𝟎
∞ 𝒄𝒐𝒔𝒂𝒙−𝒄𝒐𝒔𝒃𝒙
𝒙 𝟐 𝒅𝒙 =
𝝅
𝟐
𝒃 − 𝒂 𝒘𝒉𝒆𝒓𝒆 𝒃 > 𝒂 > 𝟎.
SOLUTION:
The given integral
0
∞
𝑐𝑜𝑠𝑎𝑥 − 𝑐𝑜𝑠𝑏𝑥
𝑥2
𝑑𝑥 𝑐𝑎𝑛 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠
𝑐
𝑒 𝑎𝑖𝑧 − 𝑒 𝑏𝑖𝑧
𝑍2
𝑑𝑍
The pole is at Z=0 of order 2 lies on the real axis, no pole lies in upper half plane.
𝑅 𝑥 𝑓, 0 =
𝑑
𝑑𝑍
[
𝑒 𝑎𝑖𝑧 − 𝑒 𝑏𝑖𝑧
𝑍2
𝑍2] 𝑍=0 = [𝑎𝑖𝑒 𝑎𝑖𝑧 − 𝑏𝑖𝑒 𝑏𝑖𝑧] 𝑍=0 = 𝑖(𝑎 − 𝑏)
0
∞
𝑐𝑜𝑠𝑎𝑥 − 𝑐𝑜𝑠𝑏𝑥
𝑥2
𝑑𝑥 =
1
2
𝑅𝑒 𝜋𝑖 × 𝑖 𝑎 − 𝑏 =
𝜋
2
𝑏 − 𝑎

14. In this type, we shall consider cases where the contour integration involves multiple valued function.
CASE I: When the pole lies on the negative part of the real axis
Form of the integral is either 0
∞
𝑥 𝛼−1 𝑓 𝑥 𝑑𝑥 𝑜𝑟 0
∞
𝑥 𝛼 𝑓 𝑥 𝑑𝑥 𝑤ℎ𝑒𝑟𝑒 𝛼 𝑖𝑠 𝑎 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛(𝑛𝑜𝑛 − 𝑖𝑛𝑡𝑒𝑔𝑒𝑟)
WORKING RULE:
STEP I: Replace x by Z and 𝑥 𝛼−1 = 𝑍 𝛼−1 = 𝑒 𝛼−1 𝑙𝑜𝑔𝑧 = 𝑒 𝛼−1 log 𝑧 +𝑖𝜋
(∴ 𝑝𝑜𝑙𝑒 𝑙𝑖𝑒𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑎𝑙 𝑎𝑥𝑖𝑠 𝑙𝑜𝑔 𝑍 = 𝑙𝑜𝑔 −𝑍 𝑎𝑛𝑑 𝜃 = 𝜋)
∴ 𝜑 𝑍 = 𝑒 𝛼−1 log −𝑧 +𝑖𝜋 𝑓(𝑍)
STEP II: Calculate the poles of f(Z) and consider all the poles which lie in the whole complex plane.
Consider residues at these poles.
STEP III: 0
∞
𝑥 𝛼−1
𝑓 𝑥 𝑑𝑥 =
−𝜋
sin 𝜋𝑎
𝑒−𝑎𝜋𝑖
𝑅
TYPE V: (WHEN THE INTEGRAND INVOLVES MULTIPLE
VALUED FUNCTION)

15. EXAMPLE:
Prove that 𝟎
∞ 𝒙 𝒂−𝟏
𝟏+𝒙
𝒅𝒙 =
𝝅
𝒔𝒊𝒏𝒂𝝅
(𝟎 < 𝒂 < 𝟏).
SOLUTION:
The given integral is
0
∞
𝑥 𝑎−1
1 + 𝑥
𝑑𝑥 =
𝑐
𝑍 𝑎−1
1 + 𝑍
𝑑𝑍
𝜑 𝑍 = 𝑒 𝑎−1 [log −𝑧 +𝜋𝑖].
1
1 + 𝑍
∴ 𝑓 𝑍 =
1
1 + 𝑍
The pole of 𝑓 𝑍 =
1
1+𝑍
𝑖𝑠 𝑎𝑡 𝑧 = −1
𝑅 𝜑, −1 = [𝑒 𝑎−1 log −𝑧 +𝜋𝑖 .
1 + 𝑍
1 + 𝑍
] 𝑍=−1 ∴ log 1 = 0
𝑅 𝜑, −1 = 𝑒(𝑎−1)𝜋𝑖 = 𝑒 𝑎𝜋𝑖. 𝑒−𝜋𝑖 = −𝑒 𝑎𝜋𝑖
0
∞
𝑥 𝑎−1
1 + 𝑥
𝑑𝑥 =
−𝜋
𝑠𝑖𝑛𝛼𝜋
𝑒−𝛼𝜋𝑖
−𝑒 𝑎𝜋𝑖
=
𝜋
𝑠𝑖𝑛𝛼𝜋
= 𝜋𝑐𝑜𝑠𝑒𝑐𝛼𝜋

16. CASE II: When the pole lies on the positive part of the real axis
Form of the integral is either
0
∞
𝑥 𝛼−1 𝑓 𝑥 𝑑𝑥 𝑜𝑟 0
∞ 𝑥 𝛼 𝑓 𝑥 𝑑𝑥 𝑤ℎ𝑒𝑟𝑒 𝛼 𝑖𝑠 𝑎 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛(𝑛𝑜𝑛 − 𝑖𝑛𝑡𝑒𝑔𝑒𝑟)
WORKING RULE:
STEP I: Replace x by Z and 𝑥 𝛼−1 = 𝑍 𝛼−1 = 𝑒 𝛼−1 𝑙𝑜𝑔𝑧 = 𝑒 𝛼−1 log 𝑧 +𝑖𝜃
(𝑤ℎ𝑒𝑟𝑒 𝑍 𝑖𝑠 + 𝑣𝑒 𝑎𝑛𝑑 𝜃 = 0)
∴ 𝜑 𝑍 = 𝑒 𝛼−1 logz 𝑓(𝑍)
STEP II: Find the poles of f(Z) and calculate residues at those poles.
STEP III: 0
∞
𝑥 𝛼−1
𝑓 𝑥 𝑑𝑥 = −𝜋𝑐𝑜𝑡𝛼𝜋 𝑅

17. EXAMPLE:
Prove that 𝟎
∞ 𝒙 𝒂−𝟏
𝟏−𝒙
𝒅𝒙 = 𝝅𝒄𝒐𝒕𝜶𝝅.
SOLUTION:
Replacing x by Z the given integral becomes
0
∞
𝑍 𝑎−1
1 − 𝑍
𝑑𝑍 𝑛𝑜𝑤 𝜑 𝑍 = 𝑒 𝛼−1 𝑙𝑜𝑔𝑧 . 𝑓(𝑍)
𝑓 𝑍 =
1
1 − 𝑍
The pole of 𝑓 𝑍 =
1
1−𝑍
𝑖𝑠 𝑎𝑡 𝑍 = 1
𝑅 𝜑, 1 = [𝑒 𝑎−1 [𝑙𝑜𝑔𝑧]
𝑍 − 1
1 − 𝑍
] 𝑍=1
= −𝑒 𝑎−1 log 1
∴ log 1 = 0 = −𝑒0
= −1
0
∞
𝑥 𝑎−1
1 − 𝑥
𝑑𝑥 = −𝜋𝑐𝑜𝑡𝛼𝜋 −1 = 𝜋𝑐𝑜𝑡𝛼𝜋

19. STATEMENT:
Let f(Z) be a meromorphic function whose only singularities in the
finite part of the plane are simple poles at 𝒂 𝟏, 𝒂 𝟐, … . . , 𝒂 𝒏 which can
be arranged as 𝟎 < 𝒂 𝟏 < 𝒂 𝟐 < ⋯ < 𝒂 𝒏 which residues
𝒃 𝟏, 𝒃 𝟐, … . . , 𝒃 𝒏 respectively.
Consider a sequence of closed contours 𝑪 𝟏, 𝑪 𝟐, … . . , 𝑪 𝒏
𝑪 𝒏 encloses 𝒂 𝟏, 𝒂 𝟐, … . . , 𝒂 𝒏 and no other poles.
The minimum distances 𝑹 𝒏 of 𝑪 𝒏 from the origin tends to infinity
when 𝒏 → ∞. then for all values of Z except poles,
𝒇 𝒁 = 𝒇 𝟎 +
𝒏=−∞
∞
𝒃 𝒏[
𝟏
𝒁 − 𝒂 𝒏
+
𝟏
𝒂 𝒏
]

20. PROOF:
Suppose that we have an integral
𝐼 =
1
2𝜋𝑖 𝑐
𝑓( 𝑡 𝑑𝑡
𝑡 𝑡 − 𝑍
𝑤ℎ𝑒𝑟𝑒 𝑍 𝑖𝑠 𝑎 𝑝𝑜𝑖𝑛𝑡
𝑤𝑖𝑡ℎ𝑖𝑛 𝐶 𝑛. the poles of the integrand are at:
𝑖. 𝑡 = 𝑎 𝑚 where m=1,2,3,…. (given) simple poles
𝑖𝑖. 𝑡 = 0 𝑜𝑓 𝑜𝑟𝑑𝑒𝑟 1
𝑖𝑖𝑖. 𝑡 = 𝑍 𝑆𝑖𝑚𝑝𝑙𝑒 𝑝𝑜𝑙𝑒
The residues at the poles are
𝑅1 𝑓, 𝑎 𝑚 = lim
𝑡→𝑎 𝑚
[
𝑡−𝑎 𝑚 𝑓(𝑡
𝑡(𝑡−𝑍)
] =
𝑏 𝑚
𝑎 𝑚(𝑎 𝑚−𝑍)
[ 𝑏 𝑚 given]
𝑅2 𝑓, 0 = lim
𝑡→0
[
𝑓(𝑡)
𝑡 − 𝑍
] =
𝑓(0)
−𝑍
𝑅3 𝑓, 𝑍 = lim
𝑡→𝑍
𝑓 𝑡
𝑡
=
𝑓(𝑍)
𝑍
By Cauchy’s residue theorem,
Sum of all residues= 𝑚=1
𝑛 𝑏 𝑚
𝑎 𝑚(𝑎 𝑚−𝑍)
−
𝑓 0
𝑍
+
𝑓(𝑍)
𝑍
𝐼 = 2𝜋𝑖 ×
𝑏 𝑚
𝑎 𝑚(𝑎 𝑚 − 𝑍)
−
𝑓 0
𝑍
+
𝑓 𝑍
𝑍
(𝐴)

21. Now consider
𝐼 =
1
2𝜋𝑖 𝑐 𝑛
𝑓 𝑡 𝑑𝑡
𝑡(𝑡 − 𝑍)
Taking modulus of both sides
𝐼 =
1
2𝜋 𝑐 𝑛
𝑓 𝑡 𝑑𝑡
𝑡(𝑡 − 𝑍)
∴ 𝑡 − 𝑍 > 𝑡 − 𝑍
1
𝑡 − 𝑍
<
1
𝑡 − 𝑍
∴ 𝐼 ≤
1
2𝜋 𝑐 𝑛
𝑑𝑡 𝑓 𝑡
𝑡 (𝑡 − 𝑍)
(∴ 𝑓 𝑡 ≤ 𝑀 𝑎𝑛𝑑
𝑐 𝑛
𝑑𝑡 = 2𝜋𝑅 𝑛)
≤
2𝜋𝑅 𝑛 𝑀
2𝜋𝑅 𝑛 𝑅 𝑛 − 𝑍
→ 0 𝑎𝑠 𝑅 𝑛 → ∞ ∴ 𝐴 𝑡𝑎𝑘𝑒𝑠 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚.
𝑓(𝑍)
𝑍
=
𝑓(0)
𝑍
+
𝑚=1
𝑛
𝑏 𝑚
𝑎 𝑚(𝑍 − 𝑎 𝑚)
In general
𝒇 𝒁 = 𝒇 𝟎 +
𝒎=−∞
∞
𝒃 𝒎
𝟏
𝒁 − 𝒂 𝒎
+
𝟏
𝒂 𝒎
𝒂𝒔 𝒎 → ∞ (𝑩)
(B) is the Mittag-Lefflers’ Theorem

22. WORKING RULE:
STEP I: Given a function𝑓(𝑍). Find 𝑓(0) then
find the poles of integrand and corresponding
residues.
STEP II: 𝑓 𝑍 = 𝑓 0 + 𝑛=1
∞ 1
𝑍−𝑎 𝑛
+
1
𝑎 𝑛

23. EXAMPLE:
Prove that 𝒄𝒐𝒕𝒁 =
𝟏
𝒁
+ 𝟐𝒁 𝒏=𝟏
∞ 𝟏
𝒁 𝟐−𝒏 𝟐 𝝅 𝟐
Solution:
Consider 𝑓 𝑍 = 𝑐𝑜𝑡𝑍 − 1
𝑍 =
𝑍𝑐𝑜𝑠𝑍−𝑠𝑖𝑛𝑍
𝑍𝑠𝑖𝑛𝑍
lim
𝑧→0
𝑓 𝑍 = lim
𝑍→0
[
𝑍𝑐𝑜𝑠𝑍 − 𝑠𝑖𝑛𝑍
𝑍𝑠𝑖𝑛𝑍
](
0
0
𝑓𝑜𝑟𝑚)
= lim[
𝑍→0
𝑍 −𝑠𝑖𝑛𝑍 + 𝑐𝑜𝑠𝑍 − 𝑠𝑖𝑛𝑍
𝑍𝑐𝑜𝑠𝑍 + 𝑠𝑖𝑛𝑍
](
0
0
𝑓𝑜𝑟𝑚)
= lim
𝑍→0
𝑍(−𝑐𝑜𝑠𝑍) − 𝑠𝑖𝑛𝑍
𝑐𝑜𝑠𝑍 + 𝑍 −𝑠𝑖𝑛𝑍 + 𝑐𝑜𝑠𝑍
=
0
2
= 0
Since f(0)=0, therefore there is no singularity at Z=0 the poles of f(Z) are at 𝑍 = 𝑛𝜋, 𝑛 = ±1, ±2, … .
𝑅 𝑓, 𝑛𝜋 = [ 𝑍 − 𝑛𝜋
𝑍𝑐𝑜𝑠𝑍 − 𝑠𝑖𝑛𝑍
𝑍𝑠𝑖𝑛𝑍
] 𝑍=𝑛𝜋(
0
0
𝑓𝑜𝑟𝑚)
= [ 𝑍 − 𝑛𝜋 {
𝑍 −𝑠𝑖𝑛𝑍) + 𝑐𝑜𝑠𝑍 − 𝑐𝑜𝑠𝑍
𝑍𝑐𝑜𝑠𝑍 + 𝑠𝑖𝑛𝑍
] 𝑍=𝑛𝜋
= [
𝑍𝑐𝑜𝑠𝑍 − 𝑠𝑖𝑛𝑍
𝑍𝑐𝑜𝑠𝑍 + 𝑠𝑖𝑛𝑍
] 𝑍=𝑛𝜋 =
𝑛𝜋(−1) 𝑛
𝑛𝜋(−1) 𝑛
= 1

24. Now using formula,
𝑓 𝑍 = 𝑓 0 +
−∞
∞
𝑏 𝑛[
1
𝑍 − 𝑎 𝑛
+
1
𝑎 𝑛
]
= 0 +
−∞
∞
1[
1
𝑍 − 𝑛𝜋
+
1
𝑛𝜋
]
=
𝑛=1
∞
[
1
𝑍 − 𝑛𝜋
+
1
𝑛𝜋
+
1
𝑍 + 𝑛𝜋
−
1
𝑛𝜋
]
= 2𝑍
𝑛=1
∞
1
𝑍2 − 𝑛2 𝜋2