INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 09,2025
Journal:
https://www.academicpublishers.org/journals/index.php/ijai
page 3
METHODOLOGY OF SOLVING OLYMPIAD PROBLEMS USING INTEGRAL
TECHNIQUES
Qobiljonov Muhriddin Murodjon ugli
Andijan State University
Faculty of Physics, Mathematics and IT
2nd-year student of Mathematics
Abstract:
This article presents a detailed analysis of two advanced problems located at the
intersection of mathematical analysis and linear algebra. The first problem,demonstrates the
integration of a logarithmic function with a rational expression, solved using series expansion
techniques. The second problem involves the integration of a matrix-valued cosine function
under the Gaussian kernel, offering an approach to operator-valued functions and matrix
analysis. Both problems are designed to enhance students’ theoretical knowledge, logical
reasoning, and familiarity with competition-level mathematical problem-solving. The article
serves as a methodological guide for gifted learners aiming to deepen their understanding of
advanced mathematical concepts.
Аннотация:
В данной статье проводится подробный анализ двух сложных задач,
находящихся на пересечении математического анализа и линейной алгебры. Первая
задачапредставляет собой интеграл логарифмической функции, делённой на
рациональное выражение, и решается методом разложения в ряд. Вторая задача
рассматривает интеграл от матричной косинус-функции с аргументом, зависящим от
переменной, под ядром Гаусса. Обе задачи направлены на развитие у студентов
теоретических знаний, аналитического мышления и навыков решения олимпиадных
задач. Статья ориентирована на одарённых учащихся, стремящихся к глубокому
изучению математики.
Keywords:
Mathematical analysis, definite integral, logarithmic function, matrix-valued
function, linear algebra, operator functions, olympiad problems, methodological approach,
series expansion, analytical solution.
Ключевые слова:
Математический анализ, определённый интеграл, логарифмическая
функция, матричная функция, линейная алгебра, операторные функции, олимпиадные
задачи, методический подход, разложение в ряд, аналитическое решение.
Introduction:
Mathematical analysis and linear algebra are core branches of mathematics that encompass a
wide range of complex yet fundamental problems. In particular, the theory of real-valued
functions, definite integration, and matrix operators are central to many modern theoretical
studies. This article focuses on two challenging problems that lie at the intersection of these
fields, both selected from high-level mathematical olympiads. Through detailed analysis and
exploration of solution strategies, the work aims to develop students’ abilities in independent
reasoning, applying formulas, and approaching unconventional mathematical problems with a
scientific mindset.
INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 09,2025
Journal:
https://www.academicpublishers.org/journals/index.php/ijai
page 4
Введение:
Математический анализ и линейная алгебра являются важнейшими разделами
математики, включающими множество сложных, но фундаментальных задач. Особенно
актуальны в современной науке такие направления, как теория функций действительного
переменного, определённые интегралы и матричные операторы. В данной статье
рассматриваются две сложные задачи олимпиадного уровня, объединяющие указанные
направления. Путём их подробного анализа и пошагового решения автор стремится
сформировать у студентов навыки самостоятельного мышления, грамотного применения
формул и научного подхода к нестандартным математическим задачам.
Calculate
:
I=
0
∞
e
−
x
2
cos
0
x x
0 0
x
0 0 0
dx
Solution
:
To solve the given example, we use the Maclaurin series expansion of the cosx function:
f x ~f x +
f
'
x
1! x+
f
''
x
2! x
2
+
f
'''
x
3! x
3
+…+
f
n
x
n! x
n
cosx=1−
x
2
2! +
x
4
4! −
x
6
6! +
x
8
8! −
x
10
10! +…+
−1
n
x
2
n
2
n ! +…
cos
x =
n
=0
∞
−1
n
x
2
n
2
n !
If we take
A=
0
x x
0 0
x
0 0 0
and replace x with A, it is not difficult to see that
A
0
=
I
3
.(I-unit
matrix)
A
2
=
AA=
0
x x
0 0
x
0 0 0
0
x x
0 0
x
0 0 0
=
0 0
x
2
0 0 0
0 0 0
A
4
=
A
2
A
2
=
0 0
x
2
0 0 0
0 0 0
0 0
x
2
0 0 0
0 0 0
=
0 0 0
0 0 0
0 0 0
A
4
=
A
6
=
A
8
=…=
A
2
n
=
O
3
O
3
−
zero matrix .
So,
cos
A =I
3
−
1
2 A
2
=
1 0 0
0 1 0
0 0 1
−
1
2
0 0
x
2
0 0 0
0 0 0
=
1 0 −
x
2
2
0 1
0
0 0
1
INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 09,2025
Journal:
https://www.academicpublishers.org/journals/index.php/ijai
page 5
it follows that.
� =
0
∞
�
−�
2
���
0 � �
0 0 �
0 0 0
�� =
0
∞
�
−�
2
1 0 −
�
2
2
0 1
0
0 0
1
�� =
=
0
∞
�
−�
2
0
−
�
2
�
−�
2
2
0
�
−�
2
0
0
0
�
−�
2
�� =
0
∞
�
−�
2
0
0
∞
−
�
2
�
−�
2
2
0
0
∞
�
−�
2
0
0
0
0
∞
�
−�
2
1.
The
last
obtained
integrals
are
calculated:
0
∞
e
−
x
2
dx=
x
2
=
t
x=t
1
2
dx=
1
2
t
−
1
2
dt
x=0 in t=0
x=∞ in t=∞
=
0
∞ 1
2
t
−
1
2
e
−
t
dt=
Г
1
2
2
=
π
2
2.
0
∞
− �
2
�
−�
2
�� =
�
2
= �
� = �
1
2
�� =
1
2
�
−
1
2
��
� = 0 �� � = 0
� = ∞ �� � = ∞
=−
0
∞ 1
2
�
−
1
2
�
−�
�� =−
Г
3
2
2
=
−
1
2
Г
1
2
+ 1 =−
1
2
∙
1
2
Г
1
2
=−
�
4
If we rely on the last obtained results,
I=
0
∞
e
−
x
2
cos
0
x x
0 0
x
0 0 0
dx =
π
2
0 −
π
8
0
π
2
0
0
0
π
2
=
π
2
1 0 −
1
4
0 1
0
0 0
1
it follows that.
Answer:
I=
0
∞
e
−
x
2
cos
0
x x
0 0
x
0 0 0
dx =
π
2
1 0 −
1
4
0 1
0
0 0
1
Calculate:
INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 09,2025
Journal:
https://www.academicpublishers.org/journals/index.php/ijai
page 6
0
1
ln
x+1
x
2
+1
Solution:
x= tan α dx=
1
cos
2
α
we introduce the notation.
0
1
ln
x+1
x
2
+1
=
0
π
4
ln tan
α +1
tan
2
α+1
∙
1
cos
2
α
dα=
0
π
4
ln tan
α +1 dα
α=
π
4
−
α dx=dα
0
�
4
ln tan � + 1 �� =
�
4
0
ln tan
�
4 − � + 1 −�� =
0
�
4
ln
tan
�
4 − tan �
1 + tan
�
4 tan �
+ 1 ��
=
0
�
4
ln
1 − tan �
1 + tan � + 1 �� =
0
�
4
ln
2
1 + tan � �� =
0
�
4
ln 2�� −
0
�
4
ln tan � + 1 ��
2
0
π
4
ln tan
α +1 dα =
0
π
4
ln 2
dα
0
π
4
ln tan
α +1 dα =
1
2
0
π
4
ln 2
dα
0
π
4
ln 2
dα = ln 2 α
π
4
0
=
π
4
ln 2
From this,
0
π
4
ln tan
α +1 dα =
π
8
ln 2
it follows that.
Answer:
0
1
ln
x+1
x
2
+1
=
0
π
4
ln tan
α +1 dα =
π
8
ln 2
Conclusion:
INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 09,2025
Journal:
https://www.academicpublishers.org/journals/index.php/ijai
page 7
In this article, two complex problems of integral calculus are examined, requiring profound
knowledge of mathematical analysis and linear algebra. The first problem demonstrates the
analysis of a logarithmic expression through series integration, while the second provides new
approaches to working with matrix functions. Both problems serve to develop skills in
Olympiad-level thinking, formula manipulation, and generalization. Furthermore, the article is
structured on a methodological approach, which is of great importance in fostering students’
abilities for deep analysis, independent thinking, and preparation for scientific communication.
References:
1.
Gradshteyn, I. S., Ryzhik, I. M.
Table of Integrals, Series, and Products
. Academic
Press, 2014.
2.
Higham, N. J.
Functions of Matrices: Theory and Computation
. SIAM, 2008.
3.
Hardy, G. H.
A Course of Pure Mathematics
. Cambridge University Press, 1908.
4.
Arfken, G., Weber, H.
Mathematical Methods for Physicists
. Elsevier, 2005.
5.
Putnam Competition Archive – https://kskedlaya.org/putnam-archive/
6.
Nielsen, M. A., Chuang, I. L.
Quantum Computation and Quantum Information
.
Cambridge University Press, 2010.
