Application of Residue Theorem on Some Different Types of Integrals
Yida Wu
a
Jinan Foreign Language School, Jinan, China
Keywords: Residue Theorem, Singularities, Complex Number, Definite Integral.
Abstract: Complex analysis in the complex plane is an important branch of integration in the field of mathematics, and
it is also an efficient mathematical tool to study and analyse the behaviour of complex variables. Complex
functions are used in many scientific fields such as physics, computer science, and engineering. Complex
analysis can solve many problems that are difficult to be solved by integrals of real variables alone or the
solutions are very complex. Many problems in physics, chemistry, and engineering can be efficiently solved
using complex analysis. This paper mainly introduces how to solve some specific types of integrals by using
the residue theorem skilfully, and how to simplify the complexity of calculation and integration by using the
residue theorem. Moreover, this paper illustrates the basic application of the residue theorem in detail through
several examples. The discussion in this paper is helpful to popularize the idea of calculating and solving
these types of integrals, and promote the application of these types of integrals in solving practical problems.
1 INTRODUCTION
Integral calculus serves as a foundational pillar of
advanced mathematics and plays an indispensable
role in interdisciplinary domains grounded in
mathematical frameworks (Bak & Newman, 2010).
When addressing practical problems in applied
disciplines, scholars often encounter scenarios
requiring holistic solutions. While elementary real
integrals can be resolved through conventional
techniques, such as computing integrals and applying
the Newton-Leibniz formula, many specialized forms
of integrals prove intractable via these classical
approaches. This limitation obstructs the application
of the Newton-Leibniz framework, creating
significant challenges for research in affected fields.
To overcome this, mathematicians turn to the residue
theorem, a cornerstone of complex analysis, as a
transformative tool for evaluating such integrals.
Central to this methodology is the concept
of residues, defined as coefficients of the minus-
power term in the Laurent series expansion. Residues
enable the computation of integrals involving isolated
singularities in which the functions exhibit undefined
or divergent behavior. The residue theorem simplifies
these calculations by reducing contour integrals to a
summation of residues enclosed within a specified
path. This innovation not only circumvents the need
a
https://orcid.org/0009-0007-6783-1879
for indefinite integrals but also streamlines the
evaluation of previously unsolvable integrals,
marking a paradigm shift in definite integral
computation (Zhu et al, 2022). Beyond theoretical
mathematics, the residue theorem holds profound
implications for mathematical physics, underpinning
advancements in electromagnetism, quantum
mechanics, fluid dynamics, and other fields reliant on
complex variable functions.
The basic idea of calculating the integral by using
the residual theorem is as follows: First, the
transformation function transforms the real variable
along the closed loop curve into the integral of the
complex variable; Then, the problem is transformed
to solve the residual values at isolated singularities in
the closed loop. Finally, the solution of the product
function is obtained by using the residual theorem.
The purpose is to summarize the rest theorem
systematically and understand its application, and to
calculate the integral of this important theorem.
2 METHOD AND THEOREMS
2.1 Cauchy-Goursat Theorem
Supposed that f(z) is a complex function, and let
curve C be a simple, closed positively oriented
114
Wu, Y.
Application of Residue Theorem on Some Different Types of Integrals.
DOI: 10.5220/0013815000004708
Paper published under CC license (CC BY-NC-ND 4.0)
In Proceedings of the 2nd International Conference on Innovations in Applied Mathematics, Physics, and Astronomy (IAMPA 2025), pages 114-118
ISBN: 978-989-758-774-0
Proceedings Copyright © 2025 by SCITEPRESS – Science and Technology Publications, Lda.

contour curve. If the expression is all analytic inside
the curve, then the integral of this function along this
curve is 0 (Fan, 2022). Namely,
𝑓
(
𝑧
)

𝑑𝑧=0
(
1
)
Let 𝑓(𝑧) denote a complex function defined
within the annular region bounded by two concentric
circles centered at 𝑧

. The radius of the two
concentric circles is 𝑅

and 𝑅

( 𝑅

> 𝑅

). If the
function is all analytic within the area ( 𝑅

< |z −
z

|<𝑅

), then the function at the point z could express
into Loran series uniform. Namely,
𝑓
(
𝑧
)
=𝑎



(
𝑧−𝑧

)

+𝑏



(
𝑧−𝑧

)

(
2
)
Laurent expansion is a generic form of Taylor
expansion. If the function is fully resolved in that
region, then the second part vanishes. That is, 𝑏

=
0.
2.2 Definition About Residues and
Residue Theorem
Suppose that f(z) is a complex function defined in a
area containing finitely many singularities and C is a
contour curve enclosing all these singularities within
the region (Qiu, 2020). In such cases, the integral of
𝑓(𝑧) along C can be solved by residue theorems. The

𝑓
(
𝑧
)
𝑑𝑧

equals to the adduct of all residues of f(z)
(Res


f(z)) by 2πi.
𝑓
(
𝑧
)
𝑑𝑧

=2𝜋𝑖𝑅es




𝑓
(
𝑧
)(
3
)
in which 𝑐

are calculated by
𝑐

=
1
2𝜋𝑖

𝑓
(
𝜉
)
(𝜉− 𝑧

)

𝑑𝜉

(
4
)
Here, 𝐶 represents arbitrary sealing contour lying
totally within the domain of integration and traversing
counterclockwise around 𝑧

. This contour integral is
evaluated by parameterizing the path 𝐶 and
integrating the resulting expression with respect to the
parameter (Zhou et al, 2022). Additionally, the
residue at 𝑧

corresponds to the coefficient of the
(𝑧− 𝑧

)

term in the Laurent series expansion
off(z), i.e., Res


𝑓(𝑧) = 𝑐

.
There are several different Classification of
singularities (Zhang et al, 2023). The first is
removable singularity. In this type of singularities,
though there is no definition at this point, the value
exists at the area near the point. In other terms, the
there are no negative power terms in the Laurent
expansion. The second is pole. In this type of
singularities, when a point approaches a singularity,
the value of the function is infinite. In addition, there
are limited power terms in the Laurent expansion. The
third is essential singularity. In this type, the value of
this point is oscillating, unstable and tends to any
value in any complex number.
3 APPLICATIONS
3.1 𝟏
(
𝟏+𝒙
𝒏
)⁄
-type Integral
In this type, n is an integer and 𝑛 > 2.
First, the author will assume 𝑛=3 (Zhou &
Huang, 2022). By using the residue theorem, the
integral could be expressed to closed loop integral in
the complex plane. The function has three single
poles, 𝑧

=𝑒
/
, 𝑧

=𝑒

, 𝑧

=𝑒
/
. Then, the
author will construct an anticlockwise curve 𝐶


with
argument is 2π/3 and radius 𝑟

is 1/3. This curve
only cover pole 𝑧

. According to the residue theorem:
𝐼=
𝑑𝑥
1+𝑥



+
𝑑𝑧
1+𝑧



+
𝑑𝜁
1+𝜁


=2𝜋𝑖Res𝑓
(
𝑧

)(
5
)
As r→∞ the first term of the integral corresponds to
the target integral. Meanwhile, the second term
vanishes (approaches zero), and the third term is
directly related to the target integral through a
symmetry or transformation.
For the second term, substituting z=Re

, (with
r→∞) and observing that |f(z)z|<ε→0 ,people
conclude that this term becomes negligible in the
limit. Thus, the integral could express into:
 𝑓
(
𝑧
)
𝑑𝑧


=𝑧𝑓
(
𝑧
)
𝑑𝑧/𝑧


≤
|
𝑧𝑓
(
𝑧
)|


|
𝑑𝑧
|
|
𝑧
|
< 𝜀
𝑟𝑑𝜃
𝑟
/

=
2𝜋
3
𝜀→0
(
6
)
In the third term of the integral, let 𝜁=𝜌𝑒
/
,
then the integral could express into:

𝑑𝜁
1+𝜁


=
𝑒
/
𝑑𝜌
1+𝜌



=−𝑒
/

𝑑𝑥
1+𝑥



=−𝑒
/
𝐹
(
𝑥
)(
7
)
Here, 𝜌=𝑥,𝐹
(
𝑥
)
=

𝑓
(
𝑥
)
𝑑𝑥. The singular residue
on the right can be obtained by L'Hospital's rule:
−


𝑒
/
. Substituting this result one can obtain that
𝐹
(
𝑥
)
=
𝑑𝑥
1+𝑥



=
2𝜋𝑖
1−𝑒
/
=
𝜋
3
2𝑖
𝑒
/
−𝑒
/
=
𝜋
3
𝑐𝑠𝑐
𝜋
3
(
8
)
Application of Residue Theorem on Some Different Types of Integrals
115

Next, the author will assume 𝑛=4. When 𝑛=4,
the function has four singularities in the complex
plane, they are z

=e
()/
( 𝑘 = 1,2,3,4).
Select a 1/4 great circular loop with a positive real
axis, and the loop surrounds only 𝑧

. According to the
residue theorem, the integration satisfies the function.
Let ϑ=ρe
/
, the third term of the function is

𝑑𝜗
1+𝜗


=
𝑒
/
1+𝜌



=−𝑒
/

𝑑𝑥
1+𝑥



=−𝑒
/
𝐹
(
𝑥
)(
9
)
The integral is solved by substituting all the above
results into formula:
𝐹
(
𝑥
)
=
𝑑𝑥
1+𝑥



=
2𝜋𝑖
1−𝑒
/
Res𝑓
(
𝑧

)
=
𝜋
4
2𝑖
𝑒
/
−𝑒
/
=
𝜋
4
𝑐𝑠𝑐
𝜋
4
(
10
)
In the same way, one could choose a 1/2 large
semi-circular loop containing the positive real axis.
The loop surrounds two singularities z

and z

, and
they can also be selected by positive real axis, 3/4
great arc C

and a closed loop consisting of a ray 𝑙

with an argument principal value of 3𝜋/2. Clearly,
the loop surrounds the three singularity points z

, z

and z

. The final value of these integrals of different
curve is same, according to the residue theorem.
3.2 𝑹(𝐜𝐨𝐬𝜽,𝐬𝐢𝐧𝜽) -type Integral
The function is characterized as a rational form in real
variables. By using the Euler’s formula, the integral
of the function could be transformed into z=e

,
cosθ=



,sinθ=



,dθ=


(Loney, 2001).
Then the integral of the function along the curve
could be transformed into this form:
𝐹= 𝑅(
𝑧+𝑧

2
|

|

,
𝑧−𝑧

2𝑖
)
𝑑𝑧
𝑧𝑖
(
11
)
The integral can then be significantly streamlined
through an application of the residue
theorem.

𝑅(𝑐𝑜𝑠𝜃,𝑠𝑖𝑛𝜃)𝑑𝜃= 2𝜋𝑖
∑
𝑅es
[
𝑓(𝑧)
]



.
There is an example which is helpful to this paper
to introduce the method for solving this type of
integration (Shen, 2017), i.e.,
𝐹=
𝑐𝑜𝑠2𝜃
5−4𝑐𝑜𝑠𝜃


𝑑𝜃
(
12
)
The point with in the unit circle 𝐶 could be
defined as z=e

(0≤θ≤2π). After applying the
Euler’s formula, it can be translated into 𝑧

=
−𝑖𝑠𝑖𝑛2𝜃. 𝑐𝑜𝑠2𝜃 =





,𝑠𝑖𝑛2𝜃=





,𝑑𝜃=


.
The integral could be converted into this form,
𝑓(𝑧) =




[(

)]
=
(

)


()()
. The function
has the singularities that is 𝑧

=0 and two simple
poles, 𝑧

=1/2 and 𝑧

=2 (not inside). So, it only
to calculate the residues 𝑧

and 𝑧

. This is because
𝑅es 
1
2
,𝑓
(
𝑧
)
= lim
→/
𝑧 −
1
2
𝑓
(
𝑧
)
=−
17𝑖
24
(
13
)
It then turns out to be
𝐹=2𝜋𝑖
5𝑖
8
−
17𝑖
24
=
𝜋
6
(
14
)
There is another example:
F= cos

θdθ


(
15
)
in which 𝑛∈ 𝑁. Since cos𝜃=
(
𝑧+𝑧

)
/2, thus
the mentioned integral can be recast into
F=
z+z

2



dz
zi
=
−i
2


(z

+1)

2

dz

(
16
)
Clearly, this integral has a
(
2𝑛+ 1
)
-order singularity
at 𝑧=0. Thus, it is calculated that
F=2πi
−i
2

Res
[
0,f
(
z
)
]
=
π
2

lim
→

1
(
2𝑛
)
!
d

dz

[(z

+1)]


=
π
2
n-
C


3.3 𝑷(𝒙) 𝑸(𝒙)
⁄
-type Integral
In this subsection, the author will consider the integral
of the form
𝐹=
𝑃
(
𝑥
)
𝑄
(
𝑥
)
𝑐𝑜𝑠

𝜃𝑑𝜃


(
17
)
This is an example to introduce the solution of this
type integral
𝐼=
𝑥𝑒

𝑥

−1


𝑑𝑥
(
18
)
After applying the partial fraction (Wang & Li,
2016), the singularities of the function would be
obvious
𝐼= 𝑙𝑖𝑚
→


,

→


𝑥𝑒

𝑥

−1
𝑑𝑥



+
𝑥𝑒

𝑥

−1




+
𝑥𝑒

𝑥

−1



(
19
)
Let 𝐼

=









𝑑𝑧,𝐼

=









𝑑𝑧,𝐼

=








𝑑𝑧.Then, 𝑓(𝑧)=





is all holomorphic
inside the curve. After using the Cauchy integral
theorem,









𝑑𝑥+










𝑑𝑥 +










𝑑𝑥+ 𝐼

+𝐼

+𝐼

=0. By virtue of the
IAMPA 2025 - The International Conference on Innovations in Applied Mathematics, Physics, and Astronomy
116

Jordan Lemma, it works that when z→0,z/z

−
1→0. Thus 𝑙𝑖𝑚
→
𝐼

=0.Because 𝑓 has the simple
poles on 𝑧=±1, lim


→

𝐼

=−𝑖𝜋𝑅es(𝑓,−1)=
−𝑖𝜋 𝑙𝑖𝑚
→
(𝑧 + 1)𝑓(𝑧)=



. Then, lim


→

𝐼

=
−𝑖𝜋𝑅es(𝑓,1)=



. Therefore,
𝑃.𝑉.
𝑥𝑒

𝑥

−1


𝑑𝑥=
𝑖𝜋𝑒

2
+
𝑖𝜋𝑒

2
=𝑖𝜋𝑐𝑜𝑠(2)
There is another example:
𝐼=
𝑠𝑖𝑛2𝑥
𝑥

+𝑥+1


𝑑𝑥
(
20
)
Firstly, the partial fraction is applied on this
function to simplify the structure of the function
(Zhou & Wu, 2018). Then, it is clearly to notice that
the function 𝑓=




does not have a singularity.
Also, it is found that 𝑙𝑖𝑚
→




=0 but the integral
has simple poles which are 𝑧

=𝑒


and 𝑧

=𝑒



.
According to the virtue of the Jordan Lemma, the
function could be transformed into that:
𝐼=𝐼𝑚
𝑒

𝑥

+𝑥+1


𝑑𝑥=𝐼𝑚 
𝑒

𝑧

+𝑧+1
𝑑𝑧

The third example of the function in this type is
that
𝐼=
𝑥

(𝑥

+ 9)(𝑥

+4)



𝑑𝑥
(
21
)
At first, this is because the integral is an even
function, 𝐼=





(

)(

)



𝑑𝑥 . It possesses
singularities at z=±3i (second-order poles) and z=
±3i (simple poles), while remaining holomorphic
everywhere else on the complex plane. To apply the
residue theorem, one can construct a semi-circular
contourC

in the higher half-plane with radius 𝑟>3.
This contour encloses all singularities z=
±3i and z=±3i within the region bounded by: the
real axis part [−𝑟,𝑟], the upper semicircle C

defined
by |𝑟|=𝑟. By integrating 𝑓(𝑧) counter-clockwise
around this boundary, the residue theorem transforms
the integral into the following form:
𝐼=𝑓
(
𝑥
)
𝑑𝑥


+𝑓
(
𝑧
)
𝑑𝑧


=2𝜋𝑖𝑅es2𝑖,𝑓
(
𝑧
)
+𝑅es3𝑖,𝑓
(
𝑧
)

(
22
)
The residues at the points are 𝑅es[2𝑖,𝑓(𝑧)]=
𝑙𝑖𝑚
→




(

)()

=


, and 𝑅es[3𝑖,𝑓(𝑧)]=
𝑙𝑖𝑚
→
(
𝑧−3𝑖
)


(

)(

)(



)
=


, thus , the total
integral

𝑓(𝑥)𝑑𝑥=




−

𝑓(𝑧)𝑑𝑧


.
As r→∞, this contour integral over 𝐶

vanishes,
i.e.,

𝑓(𝑧)𝑑𝑧| = 0


. For any point z on the semi-
circular contour 𝐶

, it is observed that |𝑧

|=|𝑧|

.
Applying the triangle inequality |z + w|≥||z| −
|w||, people can derive the following estimate, so that
by this equation, |

𝑓(𝑧)𝑑𝑧| = |



(



)(



)




|≤


(



)(



)

𝐿(𝐶

), where 𝐿(𝐶

)=𝜋𝑟 is the length
of the semicircle 𝐶

. Thus, it is derived that
|𝑓(𝑧)𝑑𝑧|


=
𝑧

(
𝑧

−9
)(
𝑧

+4
)



≤
𝑟

(
𝑟

−9
)(
𝑟

+4
)

(
23
)
As 𝑟→∞, the right-hand side of the inequality
approaches zero, implying that the contour
integral

𝑓
(
𝑧
)
𝑑𝑧


vanishes. Consequently, the
Cauchy principal value of the integral over the real
line is:
𝑃.𝑉. 𝑓
(
𝑥
)
𝑑𝑥=
𝜋
100


(
24
)
Since the integrand 𝑓(𝑥) =


(



)(



)
is an even
function, the principal value simplifies to twice the
integral from 0 to ∞, i.e.,

𝑥

(
𝑥

+9
)(
𝑥

+4
)


𝑑𝑥=
𝜋
200
(
25
)
When 𝑟→∞, the right-hand goes to 0, rendering

𝑓(𝑧)𝑑𝑧= 0


. The principal part is therefore
𝑃.𝑉.

𝑓(𝑥)𝑑𝑥=




. As the integral is even, one
can find that



(



)(



)


𝑑𝑥=


.
4 CONCLUSION
The residue theorem plays a very important role in
dealing with complex function problems. In this
paper, the definition and application range of the
residue theorem are given in detail, and the basic
reserve knowledge about the residue theorem is
introduced, such as the classification of singularities,
Laurent expansion, and the definition of residue. In
this paper, the author focuses on the application of the
residue theorem to some specific integrals. This paper
describes in detail how to convert the object integral
into a complex function form which can be used by
the residue theorem, and then greatly reduces the
difficulty of real integration by using the residue
theorem. The residue theorem is a powerful tool for
dealing with the integration of complex functions.
Types of integrals that may be difficult to solve with
Application of Residue Theorem on Some Different Types of Integrals
117

only real variables can be solved by clever application
of the residue theorem. The residue theorem offers a
systematic approach to evaluating complex integrals
by leveraging the singularities of the integrand. This
method involves identifying the singular points of the
function within the contour and computing their
associated residues to determine the integral’s value.
This method generally simplifies the computational
difficulty of the original method and provides a novel
and concise approach to processing integrals. This
paper discusses the application of residue theorems
for certain kinds of integrals, which helps to extend
the application of residue theorems, and promotes the
application of residue theorems in solving practical
problems.
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