INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 04,2025
Journal:
https://www.academicpublishers.org/journals/index.php/ijai
page 1110
ANALYSIS OF THE FUNCTION AT INFINITY
Risolat Esomurodova,Atajanova Ruxiya, Rajapov Xursand
Abstract:
This article describes methods for solving integrals using Function deduction with
respect to a point at infinity. This article can be a great guide for independent learners
Keywords:
Complex area, deduction ,continuous, integral, curved line.
Let the point at infinity
=
z
be an isolated singular point of the function
)
(
z
f
. Let
{
}
R
z
z
U
>
=
:
)
(
be a neighborhood of a point at infinity and let
)
(
z
f
be analytic at
)
(
U
. Denote by
C
the closed contour that lies entirely in
)
(
U
.
Definition.
The residue of the function
)
(
z
f
with respect to the infinitely distant
point is the value of the integral
-
C
dz
z
f
i
)
(
2
1
p
, where the integration along the contour
C
occurs in the negative direction.
The Laurent series expansion in
)
(
U
for the function
)
(
z
f
has the form
+
+
+
+
+
+
=
-
-
2
2
1
2
2
1
0
)
(
z
c
z
c
z
c
z
c
c
z
f
Since this series converges uniformly on the contour
C
, we can integrate it term by term.
Noticing that
0
=
-
C
n
n
dz
z
c
, если
1
n
;
i
c
z
dz
c
C
p
2
1
1
-
-
=
-
we get after integration
1
)
(
2
1
-
-
=
-
c
dz
z
f
i
C
p
thus, the residue of a function with respect to a point at infinity
1
)
(
-
-
=
c
resf
.
Comment.
In the case of a removable singular point lying at a finite distance, the
residue is always zero. This may not be the case for a point at infinity.
INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 04,2025
Journal:
https://www.academicpublishers.org/journals/index.php/ijai
page 1111
For example, the function
z
1
at infinity has a removable singularity, and the corresponding
residue is -1.
Theorem.
If
)
(
z
f
is analytic at any point of the extended complex plane except for a
finite number of singular points, then the sum of the residues with respect to all its singular
points (including the point at infinity) is always equal to zero.
Proof.
Let us describe a circle of finite radius such that all singular points fall into this
circle. According to the main residue theorem, the value of
C
dz
z
f
i
)
(
2
1
p
is equal to the sum
of residues with respect to all singular points lying inside
C
.
C
on the other hand, the value
-
C
dz
z
f
i
)
(
2
1
p
is equal to the residue of the
function
)
(
z
f
with respect to the point at infinity. Therefore, the sum of all calculations is
equal to
=
+
-
C
C
dz
z
f
i
dz
z
f
i
0
)
(
2
1
)
(
2
1
p
p
.
Examples with solutions.
Example 1.
Find the residue with respect to the point at infinity for a function
.
1
)
(
z
z
z
f
+
=
Decision.
1
1
lim
=
+
®
z
z
z
point
=
z
is a removable singular point. Expression
z
z
f
1
1
)
(
+
=
can be considered as its Laurent expansion in the neighborhood of an
infinitely distant point. So
1
1
=
-
c
and therefore
1
)
(
-
=
resf
.
Example 2.
Calculate the integral
=
+
2
4
1
z
z
dz
Decision.
The roots
)
4
,
3
,
2
,1
(
=
k
z
k
of the equation
0
1
4
=
+
z
are the poles (finite)
of the function . Note that all these roots lie inside the circle
2
=
z
.
INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 04,2025
Journal:
https://www.academicpublishers.org/journals/index.php/ijai
page 1112
Function
4
1
1
)
(
z
z
f
+
=
in a neighborhood of a point at infinity has a decomposition
,
1
1
1
1
1
1
1
1
)
(
12
8
4
1
4
4
4
-
+
-
=
+
=
+
=
z
z
z
z
z
z
f
z
that's why
.
0
)
(
1
=
-
=
-
c
resf
according to the above theorem, we have
0
)
(
)
(
4
1
=
+
=
resf
z
resf
k
k
this implies
.
0
)
(
2
)
(
2
1
4
1
2
4
=
-
=
=
+
=
=
iresf
z
resf
i
z
dz
k
k
z
p
p
Example 3.
Calculate the integral
.
)
3
(
)
2
(
3
4
3
3
2
17
dz
z
z
z
z
=
+
+
Decision.
The function
4
3
3
2
17
)
3
(
)
2
(
)
(
+
+
=
z
z
z
z
f
has five singular points
3
=
z
inside the circle
k
z
, which are multiple poles. According to him, it is convenient to use the
above theorem to calculate this integral.
0
)
(
)
(
5
1
=
+
=
resf
z
resf
k
k
hence,
)
(
)
3
(
)
2
(
3
4
3
3
2
17
-
=
+
+
=
resf
dz
z
z
z
z
compute
)
(
resf
:
.
)
1
(
)
1
(
1
1
)
1
(
)
1
(
)
3
(
)
2
(
)
(
4
3
3
2
4
3
12
2
2
6
17
4
3
3
2
17
3
2
3
2
z
z
z
z
z
z
z
z
z
z
z
z
f
+
+
=
+
+
=
+
+
=
INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 04,2025
Journal:
https://www.academicpublishers.org/journals/index.php/ijai
page 1113
This shows that the correct part of the Laurent expansion of this function in the vicinity of the
point at infinity begins with the term
z
1
.
Hence,
1
)
(
-
=
resf
. Thus,
i
dz
z
z
z
z
p
2
)
3
(
)
2
(
3
4
3
3
2
17
=
+
+
=
Conclusion
It is shown to give a complete idea of the deduction of a function to a point in
infinity, which is one of the basic concepts of the theory of complex variable functions. The
deduction of a function with respect to an infinite point is explained, and simple methods of
solving examples are explained.
References:
1.
V.T. Dubrovin. Theory of functions of a complex variable theory of practice, Kazan,
2010.
2. Jumarie G (2010). Cauchy’s integral formula via the modified Riemann–Liouville
derivative for analytic functions of fractional order. Applied Mathematical Letters, 23(12)
1444-1450.
3. Blaya R.A., Reyes J. B., Brackx F, Schepper H.D. and Sommen F. 2012, Cauchy Integral
Formulae in Quaternionic Hermitean Clifford Analysis, Complex Analysis and Operator
Theory, 6(5) 971-985.
4. Estrada R, Vindas J. A general integral. Disertationes Mathematicae (Rozprawy
Mat). 2012;483. 49 pages.
5. Vindas J, Estrada R. A tauberian theorem for distributional point values. Arch Math
(Basel). 2008;91:247–253.
