Contour integration with its types and examples and Mittag Leffler theorem with example

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