Cauchy's integral theorem, Cauchy's integral formula, Cauchy's integral formula for derivatives, Taylor's Series, Maclaurin’s Series,Laurent's Series,Singularities and zeros, Cauchy's Residue theorem,Evaluation various types of complex integrals.

1. COMPLEX INTEGRATION

2. If f(z) is a single-valued, continuous function in some region
R in the complex plane then we define the integral of along a
path C in R as
Where

3. Connected Region
A connected region is one in which any two points can be
connected by a curve which entirely within the region.

4. Simply Connected Region
A connected region which has no hole in it, is called simply
connected region.
Multiple Connected Region
A region which is not simply connected is called multiple
connected region.

5. MULTIPLE CONNECTED REGION
CONVERTED INTO SIMPLY
CONNECTED REGION

6. Contour :
The shape of the outer surface of one region.
Contour integration:
An integral along a simple closed curve is called contour integration.
Note:
1.Positive direction of a curve C is anticlock wise direction.
Figure 1 Positive direction Figure 2. Negative direction
2. If the curve C is in reverse direction (Negative direction) then
3. If the curve then then

7. CAUCHY’S INTEGRAL
THEOREM
If f(z) is analytic and exists & continuous at all points in
and on a simply closed curve C, then
In general, C contains ‘n’ closed curves , then

8. CAUCHY’S INTEGRAL
FORMULA
If f(z) is analytic inside and on the closed curve C of a simply
connected region R and a is any point within C, then
, C is positive oriented curve

9. CAUCHY’S INTEGRAL
FORMULA FOR
DERIVATIVES
If f(z) is analytic inside and on the closed curve C and a is
any point on it, then

10. TAYLOR’S SERIES
If f(z) is analytic in and on the simply closed curve C and a is
any point in C, then
which is convergent at every point in C.

11. MACLAURIN’S SERIES
In Taylor’s series, take a = 0,
which is Maclaurin’s series.

12. LAURENT’S SERIES
If 𝐶1 𝑎𝑛𝑑 𝐶2 are two concentric circles with centre a and radii
𝑟1 𝑎𝑛𝑑 𝑟2 , (𝑟1 > 𝑟2) and if f(z) is analytic on 𝐶1 𝑎𝑛𝑑 𝐶2 and
throughout the annular region R between them, then at each point z
in the region R,
𝑓 𝑧 = 𝑛=0
∞
𝑎 𝑛 𝑧 − 𝑎 𝑛
+ 𝑛=1
∞
𝑏 𝑛 𝑧 − 𝑎 −𝑛
Where 𝑎 𝑛 =
1
2𝜋𝑖 𝐶1
𝑓 𝑤 𝑑𝑤
𝑤−𝑎 𝑛+1 , 𝑛 = 0,1,2 … and 𝑏 𝑛 =
1
2𝜋𝑖 𝐶2
𝑓 𝑤 𝑑𝑤
𝑤−𝑎 −𝑛+1 , 𝑛 = 1,2 …

13. 𝐶1
𝐶2
Note:
1. 𝑛=0
∞
𝑎 𝑛 𝑧 − 𝑎 𝑛 is called analytic part of Laurent’s series
(positive integral powers)
2. 𝑛=1
∞
𝑏 𝑛 𝑧 − 𝑎 −𝑛
is called principle part of Laurent’s series
(negative integral powers)
3. If f(z) is analytic inside 𝐶2 , then Laurent’s series becomes
Taylor’s series (since negative power terms in Laurent’s series
are zero)
4. Laurent’s series can be written as 𝑓 𝑧 = 𝑛=−∞
∞
𝑎 𝑛 𝑧 − 𝑎 𝑛

14. Results:
1. 𝑒 𝑧
= 1 +
𝑧
1!
+
𝑧2
2!
+
𝑧3
3!
+ ⋯ , (𝑤ℎ𝑒𝑛 |𝑧| < ∞)
2. 𝑠𝑖𝑛𝑧 = 𝑧 −
𝑧3
3!
+ ⋯ +
−1 𝑛+1 𝑧2𝑛−1
2𝑛−1 !
+ ⋯ , (𝑤ℎ𝑒𝑛 |𝑧| < ∞)
3. 𝑐𝑜𝑠𝑧 = 1 −
𝑧2
2!
+
𝑧4
4!
− ⋯
−1 𝑛 𝑧2𝑛
2𝑛 !
+ ⋯ , (𝑤ℎ𝑒𝑛 𝑧 < ∞)
4. 𝑆𝑖𝑛ℎ 𝑧 = 𝑧 +
𝑧3
3!
+ ⋯ +
𝑧2𝑛−1
2𝑛−1 !
+ ⋯ , (𝑤ℎ𝑒𝑛 |𝑧| < ∞)
5.cosh 𝑧 = 1 +
𝑧2
2!
+
𝑧4
4!
+ ⋯
𝑧2𝑛
2𝑛 !
+ ⋯ , (𝑤ℎ𝑒𝑛 𝑧 < ∞)
6.
1
1+𝑧
= 1 + 𝑧 −1
= 1 − 𝑧 + 𝑧2
− 𝑧3
+ ⋯ −1 𝑛
𝑧 𝑛
+ ⋯ , (𝑤ℎ𝑒𝑛 |𝑧| < 1)
7.
1
1−𝑧
= 1 − 𝑧 −1
= 1 + 𝑧 + 𝑧2
+ 𝑧3
+ ⋯ + 𝑧 𝑛
+ ⋯ , (𝑤ℎ𝑒𝑛 |𝑧| < 1)
8.
1
1+𝑧 2 = 1 + 𝑧 −2
= 1 − 2𝑧 + 3𝑧2
− 4𝑧3
+ ⋯ −1 𝑛
(𝑛 + 1)𝑧 𝑛
+ ⋯ , (𝑤ℎ𝑒𝑛 |𝑧| < 1)
9.
1
1−𝑧 2 = 1 − 𝑧 −2
= 1 + 2𝑧 + 3𝑧2
+ 4𝑧3
+ ⋯ + (𝑛 + 1)𝑧 𝑛
+ ⋯ , (𝑤ℎ𝑒𝑛 |𝑧| < 1)
10.
1
1+𝑧 𝑛 = 1 + 𝑧 −𝑛
= 1 − 𝑛𝑧 +
𝑛 𝑛+1
2!
𝑧2
−
𝑛 𝑛+1 𝑛+2
3!
𝑧3
+ ⋯ , (𝑤ℎ𝑒𝑛 |𝑧| < 1)
11.
1
1−𝑧 𝑛 = 1 − 𝑧 −𝑛
= 1 + 𝑛𝑧 +
𝑛 𝑛+1
2!
𝑧2
+
𝑛 𝑛+1 𝑛+2
3!
𝑧3
+ ⋯ , (𝑤ℎ𝑒𝑛 |𝑧| < 1)

15. SINGULARITIES AND
ZEROS
Definition (Singularity)
If a point z = α is called a singular point, or singularity of the complex function f(z) if f is not
analytic at z = α , but every neighborhood DR(α) ={z: |z - α| < R} of α contains at least one point
at which f(z) is analytic.
Example.1
The function 𝑓 𝑧 =
1
1−𝑧
is not analytic at z = 1, but is analytic for all other values of z. Thus the
point z = 1 is a singular point of f (z).
Example.2
The function 𝑓 𝑧 =
1
𝑧
is not analytic at z = 0, but is analytic for all other values of z. Thus the
point z = 0 is a singular point of f (z).
Example.3
Consider g (z) = log z , g (z) is analytic for all z except at the origin and at all points on the
negative real-axis. Thus, the origin and each point on the negative real axis is a singularity of g(z).
Example.4
The function 𝑓 𝑧 =
1
𝑧(𝑧−𝑖)
is not analytic at z = 0 and z = i, but is analytic for all other values of
z. Thus the point z = 0 and z = i are the singular points of f (z).

16. Definition. (Isolated Singularity)
The point α is called an isolated singularity of the complex function f(z) if f is not analytic at z = α, but
there exists a real number R > 0 such that f(z) is analytic everywhere in the punctured disk DR
*(α)={z: 0
< |z - α| < R}.
Example.1
The function 𝑓 𝑧 =
1
1−𝑧
has an isolated singularity at z = 1.
Example.2
The function 𝑓 𝑧 =
1
𝑧
has an isolated singularity at z = 0.
Example.3
Consider the function 𝑓 𝑧 =
(𝑧+1)
𝑧2(𝑧2+1)
=
(𝑧+1)
𝑧2(𝑧−𝑖)(𝑧+𝑖)
has three isolated singularities at z = 0, z = i and z = -i
Example.4
The function g(z) = log z, however, the singularity at z = 0 (or at any point of the negative real axis) that
is not isolated, because any neighborhood of contains points on the negative real axis, and g(z) = log z is
not analytic at those points. Functions with isolated singularities have a Laurent series because the
punctured disk DR
*(α) is the same as the annulus A (α, 0, R). The logarithm function g (z) does not have a
Laurent series at any point z = -a on the negative real-axis. We now look at this special case of Laurent's
theorem in order to classify three types of isolated singularities.

17. REMOVABLE SINGULARITY,
POLE OF ORDER K,
ESSENTIAL SINGULARITY
Let f (z) has an isolated singularity at α with Laurent series
expansion
f (z) = ∑cn(z –α)n, valid for z Є A(α, 0, R) and -∞ ≤ n ≤ ∞.
Then we distinguish the following types of singularities at α.

18. Definition (Removable Singularities)
If cn= 0 for n = -1, -2, -3…, then we say that f (z) has a removable singularity at α. That is no
negative powers terms in the Laurent series expansion of f (z).
If f (z) has a removable singularity at z = α, then it has a Laurent series𝑓 𝑧 = 𝑛=0
∞
𝑐 𝑛( 𝑧 −

19. Definition. (Pole of order k)
If k is a positive integer such that c-k ≠ 0 but cn = 0 for n = -k-1, -k-2, -k-3…, then we say that f (z) has a pole of order k atα. That is in
the Laurent series expansion of f (z) thereare only (k terms) finite number of negative power terms.
If f (z) has a pole of order k at z = α, the Laurent series for f (z) is
𝑓 𝑧 = 𝑛=−𝑘
∞
𝑐 𝑛 𝑧 − 𝛼 𝑛
, valid for z Є A (α, 0, R) ,where c-k ≠ 0.
Example.1
Consider the function𝑓 𝑧 =
𝑠𝑖𝑛𝑧
𝑧3
=
1
z3 z −
z
3
3
! +
z
5
5
! −
𝑧
7
7
! + ⋯
=
1
z2 −
1
3
! +
z
2
5
! −
z
4
7
! + ⋯ ,
Here the function f (z)has a pole of order k = 2 at z = 0.
Example.2
Consider the function𝑓 𝑧 =
5𝑧+1
𝑧−2 3 𝑧+3 𝑧−2
has a pole of order 3 at z = 2 and simple poles at z = -3 and z = 2.
Definition (Simple Pole)
If f (z) has a pole of order 1 at z = α, we say that f (z) has a simple pole at z = α.
Example.1
Consider the function 𝑔 𝑧 =
𝑒 𝑧
𝑧
=
1
𝑧
1 +
𝑧
1!
+
𝑧2
2!
+
𝑧3
3!
+ ⋯
=
1
z
+ 1 +
z
2!
+
z2
3!
+
z3
4!
+ ⋯ ,
Clearly the function g (z) has a simple pole at z = 0.

20. Definition (Essential Singularities)
If cn≠ 0 for infinitely many negative integers n, then we say that f(z) has
anessential singularity at z = α. That is in the Laurent series expansion of f
(z), there are infinite number of negative power terms.
Example.1
Consider the function f (z) = z2 sin (1/z)
= z2 1
z
−
1
z
3
3!
+
1
z
5
5!
−
1
z
7
7!
+ ⋯
= z −
1
3!
z−1
+
1
5!
z−3
−
1
7!
z¯5
+ ⋯ ,
Here the function f (z) has an essential singularity at the origin. Z = 0
Example.2
Consider the function f (z) = 𝑒
1
𝑧
= 1 +
1
𝑧
1!
+
1
𝑧
2
2!
+
1
𝑧
3
3!
+ ⋯ ,
= 1 +
1
1!
z¯1
+
1
2!
z¯2
+
1
3!
z¯3
+ ⋯ ,
Here the function f (z) has an essential singularity at the origin. Z = 0

21. Definition (Zero of order k)
A function f(z) analytic in DR(α) has a zero of order k at the point z = α if and only
if f(n)(α) = 0 for n = 0,1,2,..., k-1, and f(k)(α) ≠ 0 (kth derivative of f(z) )
Example.1
In the following function
f (z) = z sin z2
= z³ −
1
3!
z7
+
1
5!
z11
−
1
7!
z15
+ ⋯ ,
We have f ′ (z) = 2 z2 cos z2 + sin z2
f ′′ (z) = 6 z cos z2 – 4 z3 sin z2
f ′′′(z) = 6 cos z2 - 8 z4 cos z2 – 24 z2 sin z2
Then, f (0) = f′ (0) = f ′′ (0) = 0 , but f ′′′ (0) = 6 ≠ 0.
Hence the function f (z) has a zero of order k = 3 atz = 0.
Definition (Simple Zero)
If the function f (z) has a zero of order one, then we say that f (z) has a simple zero.
Example.1
The function f(z) = z has a simple zero at z = 0
We have f ′ (z) = 1 , then f ′ (0) = 1 ≠ 0 , hence the function f (z) has zero of order one.

22. Results
1. A function f (z) analytic in DR(α) has a zero of order k at the point z = α iff its Taylor series given by
𝑓 𝑧 = 𝑛=0
∞
𝑐 𝑛 𝑧 − 𝛼 𝑛
has c0 = c1 = … = ck-1 = 0 and ck ≠ 0.
2. Suppose f(z) is analytic in DR(α). Then f(z) has a zero of order k at the point z =α if and only if
it can be expressed in the form𝑓 𝑧 = 𝑧 − 𝛼 𝑘
𝑔 𝑧 where g (z) is analytic at z = α and g(α) ≠ 0.
3. If f(z) and g(z) are analytic at z =α, and have zeros of orders m and n, respectively at z =α, then their
product h (z) = f(z)g(z) has a zero of order m + n at z =α.
4. A function f (z) analytic in the punctured disk DR
*(α) has a pole of order k at z = α if and only if
it can be expressed in the form𝑓 𝑧 =
ℎ 𝑧
𝑧− 𝛼 𝑘 where the function h (z) is analytic at the point z = α and h (α) ≠ 0.
5. If f (z) is analytic and has a zero of order k at the point z = α, then 𝑔 𝑧 =
1
𝑓 𝑧
has a pole of order k at z = α.
6. If f (z) has a pole of order k at the point z = α, then 𝑔 𝑧 =
1
𝑓 𝑧
has a zero of order k at z = α.
7. If f(z) and g(z) have poles of orders m and n, respectively at the point z = α, then their product h(z) =
f(z)g(z) has a pole of order m +n at z = α.
8. Let f (z) and g (z) be analytic with zeros of orders m and n, respectively at z = α. Then their
quotient ℎ 𝑧 =
𝑓(𝑧)
𝑔 𝑧
has the following behavior:
(i) If m > n, then h(z) has a zero of order m - n at z = α.
(ii) If m < n , then h(z) has a pole of order n - m at z = α.
(iii) If m = n, then h(z) has a removable singularity at z = α, and can be defined so that h(z) is analytic at z =
α, by ℎ 𝛼 =
lim
𝑧 → 𝛼
ℎ (𝑧).

23. Example .1 Let f (z) = z³sinz. Then f(z) can be factored as the product
of z³ and sinz, which have zeros of orders m = 3 and n = 1, respectively,
at z = 0. Hence z = 0 is a zero of order 4 of f (z).
Let g (z) = z³ and h (z) = sinz and f (z) = g (z) h (z)
Clearly g (z) and h (z) have zeros of orders m = 3 and n =1 respectively at z = 0 and
hence by result 3, f (z) has zero of order m +n = 3 + 1 = 4 at z = 0
Example .2 Locate the zeros and poles of ℎ 𝑧 =
𝑡𝑎𝑛𝑧
𝑧
, and determine their order.
Given ℎ 𝑧 =
𝑡𝑎𝑛𝑧
𝑧
=
𝑠𝑖𝑛𝑧
𝑧𝑐𝑜𝑠𝑧
=
𝑓(𝑧)
𝑔(𝑧)
We know that the zeros of f (z) = sinz occur at the points z = nπ, where n is an
integer. Because f ′ (nπ) = cosnπ ≠ 0, the zeros of f (z) are simple. Similarly, the
function g (z) = zcosz has simple zeros at the points
z = 0and z= 𝑛 +
1
2
𝜋, where n is an integer. From the information given, we
find that h 𝑧 =
𝑓 𝑧
𝑔 𝑧
behaves as follows:
i. h (z) has simple zeros at z = nπ, where n = ±1, ±2…
ii. h (z) has simple poles at z = 𝑛 +
1
2
𝜋, where n is an integer; and
iii. h (z) is analytic at z = 0if we define ℎ 0 =
lim
𝑧 → 0
ℎ 𝑧 = 1.

24. Example.3 Locate the poles of𝑔 𝑧 =
1
5𝑧4+26𝑧2+5
and specify their order.
The roots of the quadratic equation 5z² + 26z + 5 = 0 are z = - 5 and z = −
1
5
.
If we replace z with z² in this equation, the function𝑓 𝑧 = 5𝑧4
+ 26𝑧2
+ 5 has roots z2 = - 5 and z2= −
1
5
, Therefore the
roots of f (z) are z = ±i√5 and z = ±
𝑖
5
.
That is f (z) has simple zeros at the points z = ±i√5 and z = ±
𝑖
5
.
By the above result 5, g(z) has simple poles at z = ±i√5 and z = ±
𝑖
5
.
Example. 4 Locate the zeros and poles of g 𝑧 =
𝜋𝑐𝑜𝑡𝜋𝑧
𝑧2 , and determine their order.
The function f (z) = z² sinπz has a zero of order k = 3 at z = 0 and simple zeros at the points z = ±1, ±2,…, result 6
implies that g(z) has a pole of order 3 at the point z = 0 and simple poles at the points z = ±1, ±2,…,
Example.5 Find the poles of f 𝑧 =
1
𝑠𝑖𝑛𝑧−𝑐𝑜𝑠𝑧
Here the poles of f(z) are the zeros of sinz – cosz
Take sinz – cosz = 0 ⇒sinz = cosz
⇒
𝑠𝑖𝑛𝑧
𝑐𝑜𝑠𝑧
= 1 ⇒tanz = 1 ⇒ z = nπ +
𝜋
4
, n = 0,1,2….
Hence the simple zeros of sinz – coszare z = nπ +
𝜋
4
, n = 0,1,2….
Therefore the simple poles of f (z) are z = nπ +
𝜋
4
, n = 0,1,2….

25. RESIDUE THEORY
Definition (Residue)
Let f(z) has a non-removable isolated singularity at the point a
z0. Then f(z) has the Laurent series representation for all z in some
punctured disk DR
*(z0) given by 𝑓 𝑧 = 𝑛=−∞
∞
𝑎 𝑛 z − z0
𝑛
The coefficient a-1 of
1
𝑧−𝑧0
is called the residue of f(z) at z0 .
It is denoted by Res[f,z0] = a-1

26. Example.1
Consider f (z) = e2/z
Then the Laurent series of f about the point z0 = 0 is given by
= 1 +
2
1!z
+
22
2!z2 +
23
3!z3 + ⋯ ,
The co-efficient of
1
𝑧−𝑧0
=
1
𝑧−0
=
1
𝑧
is 2
Hence by definition of residue, residue of f (z) = e2/z at z0 = 0 is given by Res [f, z0] = 2
Example .2 Find residue of f (z) =
3
2𝑧+𝑧2−𝑧3 at z0 = 0
f(z) =
3
z 2+z−z2 =
3
z z+1 2−z
Now
3
z z+1 2−z
=
A
z
+
B
z+1
+
C
2−z
⇒A(z+1)(2-z) + Bz(2-z) + Cz (z+1) = 3
⇒A(-z2 +2 + z) +B(2z –z2) +C(z2 +z) = 3
⇒-A –B +C = 0---- (a)
A +2B +C = 0 ---- (b)
2A =3 ---- ( c) ⇒A = 3/2
(a) ⇒-B + C = A= 3/2
(b) ⇒2B + C = -A = -3/2
-----------------------------
Adding B + 2C =0⇒ B = - 2C
Put B = - 2C in -B + C = 3/2
⇒3C = 3/2⇒C= 1/2
Put C= 1/2, B = -2C = -1⇒ B = -1
Hence f(z) = =
A
z
+
B
z+1
+
C
2−z
=
3
2z
-
1
z+1
+
1
2(2−z)
=
3
2z
- (1+z) -1 +
1
2
(2-z)-1 =
3
2z
- (1- z+ z2 -….) +
1
2
2-1 =
3
2z
- (1- z+ z2 -….) +
1
4
(1 +
z
2
+
z
2
2
+ ….)
=
3
2z
-
3
4
+
9z
8
- …..
The residue of f at 0 is given by Res [f,0] = coefficient of
1
z
=
3
2
Example.3 Find residue of f (z) =
𝑒 𝑧
𝑧3 at z0 = 0
Laurent expansion of f(z) =
1
𝑧3 {1 + 𝑧 +
𝑧2
2!
+
𝑧3
3!
… } =
1
𝑧3 +
1
𝑧2 +
1
𝑧2!
+
1
3!
…
The residue of f at 0 is given by Res [f,0] = coefficient of
1
z
=
1
2

27. CAUCHY'S RESIDUE
THEOREM
Let D be a simply connected domain, and let C ⊂D be a closed
positively oriented contour within and on the function f(z) is
analytic, except finite number of singular z1,z2,….,zn, then
𝐶
𝑓(𝑧) 𝑑𝑧 = 2πi 𝑘=1
𝑛
Res[f, zk]

28. Residues at Poles
(i) If f(z) has a simple pole at z0, then Res[f,z0] =
lim
𝑧 → 𝑧0
( 𝑧 −

29. Results
1. Let P(z) be a polynomial of degree at most 2. If a ,b and c are distinct complex
numbers,
then f(z) =
𝑃 𝑧
(𝑧−𝑎)(𝑧−𝑏)(𝑧−𝑐)
=
𝐴
(𝑧−𝑎)
+
𝐵
(𝑧−𝑏)
+
𝐶
(𝑧−𝑐)
Where A = Res [f,a] =
𝑃 𝑎
(𝑎−𝑏)(𝑎−𝑐)
B = Res [f,b] =
𝑃(𝑏)
(𝑏−𝑎)(𝑏−𝑐)
C = Res [f,c] =
𝑃(𝑐)
(𝑐−𝑎)(𝑐−𝑏)
2. If a repeated root occurs in partial fraction, and P(z) has degree of at most 2, then
f(z) =
𝑃 𝑧
𝑧−𝑎 2(𝑧−𝑏)
=
𝐴
𝑧−𝑎 2 +
𝐵
(𝑧−𝑎)
+
𝐶
(𝑧−𝑏)
Where A = Res [(z-a)f(z),a]
B = Res [f, a]
C = Res [f, b]

30. EVALUATION OF REAL DEFINITE INTEGRALS

31. CASES OF POLES ARE NOT
ON THE REAL AXIS.
Type I
Evaluation of the integral 𝟎
𝟐𝝅
𝒇 𝒄𝒐𝒔𝜽𝒔𝒊𝒏𝜽 𝒅𝜽 where f(cosθsinθ) is a real rational function of
sinθ,cosθ.
First we use the transformation z = eiθ =cosθ + i sinθ ------ ( a)
And
1
𝑧
=
1
𝑒 𝑖𝜃 = e-iθ= cosθ - i sinθ -------- (b)
From (a) and (b) , we have cosθ =
1
2
(𝑧 +
1
𝑧
) , sinθ =
1
2𝑖
(𝑧 −
1
𝑧
)
Now z = eiθ⇒dz = ieiθdθ⇒dθ =
𝑑𝑧
𝑖𝑧
Hence 0
2𝜋
𝑓 𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 𝑑𝜃 = 𝐶
𝑓[
1
2
𝑧 +
1
𝑧
,
1
2𝑖
(𝑧 −
1
𝑧
)]
𝑑𝑧
𝑖𝑧
Where C, is the positively oriented unit circle |z| = 1
The LHS integral can be evaluated by the residue theorem and
𝐶
𝑓[
1
2
𝑧 +
1
𝑧
,
1
2𝑖
(𝑧 −
1
𝑧
)]
𝑑𝑧
𝑖𝑧
= 2πi ∑ Res(zi) , where zi is any pole in the interior of the
circle |z| =1

32. Type II.
Evaluation of the integral −∞
∞
𝒇 𝒙 𝒅𝒙 where f(x) is a real rational function of the real variable x.
If the rational function f(x) =
𝑔(𝑥)
ℎ(𝑥)
, then degree of h(x) exceeds that of g(x) and g(x) ≠ 0.To find the value of the
integral, by inventing a closed contour in the complex plane which includes the required integral. For this we have
to close the contour by a very large semi-circle in the upper half-plane. Suppose we use the symbol “R” for the
radius. The entire contour integral comprises the integral along the real axis from −R to +R together with the
integral along the semi-circular arc. In the limit as R→∞the contribution from the straight line part approaches
the required integral, while the curved section may in some cases vanish in the limit.
The poles z1,z2,….,zk of
𝑔(𝑥)
ℎ(𝑥)
, that lie in the upper half-plane
−∞
∞
𝑓 𝑥 𝑑𝑥 = −∞
∞ 𝑔(𝑥)
ℎ(𝑥)
𝑑𝑥= 2πi∑ Res[f,zk]

33. Type III.
Evaluation of the integral −∞
∞
𝒇 𝒙 𝒔𝒊𝒏 𝒎𝒙 𝒅𝒙 , −∞
∞
𝒇 𝒙 𝒄𝒐𝒔 𝒎𝒙 𝒅𝒙where m > 0 and f(x) is a real rational function of the
real variable x.
If the rational function f(x) =
𝑔(𝑥)
ℎ(𝑥)
, then degree of h(x) exceeds that of g(x) and g(x) ≠ 0.
Let g(x) and h(x) be polynomials with real coefficients, of degree p and q, respectively, where q ≥ p+1.
If h(x) ≠0 for all real x, and m is a real number satisfying m > 0, then
−∞
∞ 𝑔(𝑥)
ℎ(𝑥)
cos 𝑚𝑥 𝑑𝑥 = lim
𝑅→∞ −𝑅
𝑅 𝑔(𝑥)
ℎ(𝑥)
cos 𝑚𝑥 𝑑𝑥 and −∞
∞ 𝑔(𝑥)
ℎ(𝑥)
𝑠𝑖𝑛𝑚𝑥 𝑑𝑥 = lim
𝑅→∞ −𝑅
𝑅 𝑔(𝑥)
ℎ(𝑥)
sin 𝑚𝑥 𝑑𝑥
We know that Euler’s formula 𝑒 𝑖𝑚𝑥
= cos mx + i sin mx , where cos mx = Re[𝑒 𝑖𝑚𝑥
]
and sin mx = Im[𝑒 𝑖𝑚𝑥
] , m is a positive real.
We have −∞
∞ 𝑔(𝑥)
ℎ(𝑥)
𝑒 𝑖𝑚𝑥
𝑑𝑥 = −∞
∞ 𝑔(𝑥)
ℎ(𝑥)
cos 𝑚𝑥 𝑑𝑥 + i −∞
∞ 𝑔(𝑥)
ℎ(𝑥)
𝑠𝑖𝑛𝑚𝑥 𝑑𝑥
Here we are going to use the complex function f(z) =
𝑔(𝑧)
ℎ(𝑧)
𝑒 𝑖𝑚𝑧
to evaluate the given integral.
−∞
∞ 𝑔(𝑥)
ℎ(𝑥)
cos 𝑚𝑥 𝑑𝑥= Re{2πi∑ Res[f,zk]} and
−∞
∞ 𝑔(𝑥)
ℎ(𝑥)
𝑠𝑖𝑛𝑚𝑥 𝑑𝑥 = Im{2πi∑ Res[f,zk]}, where z1,z2,…..zk are the poles lies on the upper half of the semi-circle.

34. Jordan’s Lemma
If f (z) → 0 uniformly as z →∞, then lim
𝑅→∞ 𝐶1
𝑒 𝑖𝑚𝑧
𝑓 𝑧 𝑑𝑧 = 0, (m > 0) where C1 denotes the
semi-circle |z| = R, I(z) > 0.
Note
z = reiθ⇒r = |z| and θ= arg (z)
logz=Logr+iarg(z)
If z = x+iy , r = (x2+y2)1/2 , θ= arg (z) = arg (x+iy) = tan-1(y/x)
log(x+i) = log (x2+1)1/2 + iarg(x) = log (x2+1)1/2 +0 = log (x2+1)1/2

35. CASE OF POLES ARE ON THE REAL AXIS.
Type IV
If the rational function f(z) =
𝑔(𝑧)
ℎ(𝑧)
, then degree of h(z) exceeds that of g(z) and g(z) ≠ 0.
Suppose h(z) has simple zeros on the real axis ( that is simple poles of f(z) on the real axis) , let it be a1,a2,…ak
and h(z) has zeros inside the upper half of semi-circle ( that is poles of f(z) inside the upper half of semi-
circle), let it be b1,b2,…bs,
then −∞
∞
𝑓 𝑥 𝑑𝑥 = πi∑ Res[f,ak] + 2πi∑ Res[f,bs], where k = 1,2,….k and s = 1,2,…s
Where C1,C2,….Ck are the semi circles and b1,b2,…bs are lie upper half of these semi circles.

36. Type V
If the rational function f(z) =
𝑔(𝑧)
ℎ(𝑧)
, then degree of h(z) exceeds that of g(z) and g(z) ≠ 0.
Suppose h(z) has simple zeros on the real axis ( that is simple poles of f(z) on the real axis) ,
let it be a1,a2,…ak and h(z) has zeros inside the upper half of semi-circle ( that is poles of f(z)
inside the upper half of semi-circle), let it be b1,b2,…bs,
Let m be a positive real number and if f(z) =
𝑒 𝑖𝑚𝑧 𝑔(𝑧)
ℎ(𝑧)
, then
−∞
∞
𝑐𝑜𝑠𝑚𝑥
𝑔 𝑥
ℎ 𝑥
𝑑𝑥 = Re −∞
∞
𝑐𝑜𝑠𝑚𝑥 𝑓 𝑥 𝑑𝑥
= 𝑅𝑒 2πi 𝑖=1
𝑠
Res f, bi + 𝑅𝑒 πi 𝑗=1
𝑘
Res f, aj
And
−∞
∞
𝑠𝑖𝑛𝑚𝑥
𝑔 𝑥
ℎ 𝑥
𝑑𝑥 = Img −∞
∞
𝑠𝑖𝑛𝑚𝑥 𝑓 𝑥 𝑑𝑥
= 𝐼𝑚𝑔 2πi 𝑖=1
𝑠
Res f, bi + 𝐼𝑚𝑔 πi 𝑗=1
𝑘
Res f, aj
Whereb1,b2,…bs,are the poles of f(z) that lie in the upper half of the semi-circles
C1,C2,….Ck.