Volume 03 Issue 10-2023
316
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
03
ISSUE
10
Pages:
316-323
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
A
BSTRACT
This article discusses a method for calculating functions belonging to the class of functions that are not
intgerable in a standard way using the Feynman method, which allows one to obtain an exact analytical
solution. This article shows a method for calculating some rather complex integrals that cannot be
integrated in the standard way.
K
EYWORDS
Definite integral, improper integrals, Feynman’s trick, Dirichlet integral, differentiation of an integral with
respect to a parameter.
I
NTRODUCTION
Some integrals, which belong to the class of
functions that are not intgerable by standard
methods, can be calculated using a method
created by Nobel Prize winner (1965) Richard
Feynman. Richard developed an integration
method called the Feynman trick. He has achieved
achievements in the field of theoretical physics,
the development of a method of integration along
trajectories from quantum mechanics, and the
reformation of teaching methods in higher
educational institutions.
Journal
Website:
http://sciencebring.co
m/index.php/ijasr
Copyright:
Original
content from this work
may be used under the
terms of the creative
commons
attributes
4.0 licence.
Research Article
METHODOLOGY FOR STUDYING SOME NON
–
INTGERABLE
FUNCTIONS IN AN UNCONVENTIONAL WAY
Submission Date:
October 20, 2023,
Accepted Date:
October 25, 2023,
Published Date:
October 30, 2023
Crossref doi:
https://doi.org/10.37547/ijasr-03-10-49
Saipnazarov Shaylovbek Aktamovich
Associate Professor, Candidate of Pedagogical Sciences, Tashkent State University of Economics, Uzbekistan
Ortikova Malika Turaboyevna
Senior lecturer of Tashkent State University of Economics, Uzbekistan
Volume 03 Issue 10-2023
317
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
03
ISSUE
10
Pages:
316-323
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
Integration is associated with important methods
of analysis and study of numerical functions
–
averages, limits, infinitesimals, differentials,
derivatives, and so on, and therefore without
understanding and studying these concepts, the
study of functions is impossible. To find the value
of the integral, scientists such as Richard
Feynman found unconventional methods.
Integrating functions is a mathematical art. It is
interesting to calculate them, especially when
non-standard methods are used in the solution.
Consider the integral
( )
( )
( )
( )
=
p
b
p
a
dx
p
x
f
p
I
,
where
p
–
is the integral parameter,
x
–
is the integration variable.
( )
( )
=
b
a
dx
p
x
f
p
p
I
,
Consider the improper integral (Dirichlet integral)
−
dx
x
ax
sin
(1)
0
a
The integrand is even and therefore
−
=
0
sin
2
sin
dx
x
ax
dx
x
ax
(2)
To calculate the right side of equality (2), we find the function
x
1
in the form
−
=
0
1
x
dt
e
xt
where
const
a
a
=
,
0
. Then form equality (2) we obtain
−
−
−
=
=
=
0
0
0
0
0
0
0
sin
2
sin
2
sin
2
sin
2
dx
axe
dt
dt
e
axdx
dx
dt
e
ax
dx
x
ax
tx
tx
tx
To calculate
−
0
,
sin
dx
axe
tx
we use Euler’s formula
i
e
e
ax
iax
iax
2
sin
−
−
=
(3)
Then
−
−
−
=
+
=
+
=
−
=
0
0
0
0
2
2
2
2
0
2
2
1
2
2
sin
a
t
dt
a
dt
a
t
ia
i
dx
e
i
e
e
dt
dx
axe
tx
iax
iax
xt
=
=
=
2
2
1
2
0
a
t
arctg
a
a
Volume 03 Issue 10-2023
318
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
03
ISSUE
10
Pages:
316-323
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
=
=
0
0
2
sin
;
sin
2
dx
x
ax
dx
x
ax
Now let’s calculate the same integral (1) using Feynman’s method. Thus, we will demonstrate the power of
Feynman’s method.
0
sin
dx
x
ax
Solution.
Let function
( )
p
I
be defined by the formula
( )
0
,
sin
=
−
−
p
dx
x
ax
e
p
I
px
, where
0
a
Differentiating with respect to the parameter
p
, where
x
is a fictitious integration variable, we obtain
−
−
−
−
−
−
−
=
−
=
=
axdx
e
dx
x
ax
xe
dx
x
ax
e
p
dp
dI
px
px
px
sin
sin
sin
By integrating the last integral twice by parts, it is not difficult to show that
2
2
p
a
a
dp
dI
+
−
=
.
After integration we get
( )
C
a
p
arctg
p
I
+
−
=
, where
C
is an arbitrary integration constant. We can
calculate
C
by noting that
( )
0
=
I
in the original integral definition of
( )
p
g
, because the factor
xp
e
−
in the
integral tends to zero everywhere at
→
p
(because
0
x
throughout the integration interval). So
( )
−
=
arctg
C
0
, where we use a plus sign if
0
a
, and a minus sign if
0
a
. So,
2
=
C
and we have
( )
a
p
arctg
p
I
−
=
2
For
0
=
p
(and therefore
0
=
a
p
arctg
) gives us the following remarkable result, called the
discontinuous Dirichlet integral. Ultimately we see that in both cases the result is the same
−
=
=
0
0
,
2
0
,
0
0
,
2
sin
a
if
a
if
a
if
dx
x
ax
Example
–
2.
Calculate the integral
Volume 03 Issue 10-2023
319
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
03
ISSUE
10
Pages:
316-323
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
+
0
cos
2
cos
3
2
ln
x
dx
x
(4)
Solution.
The integral function is continuous at all points of the segment
;
0
, except for point
2
=
x
. At this point we have uncertainty of the form
0
0
, that is, when
=
=
+
=
=
0
1
ln
2
2
cos
3
2
ln
0
2
cos
2
x
Let us transform the integrand expression as follows:
+
=
+
0
0
cos
cos
2
3
1
ln
cos
2
cos
3
2
ln
x
dx
x
x
dx
x
(5)
In the last integral we denote coefficient
2
3
as parameter
p
. For different values of the parameter
p
,
different values of the integral
will be obtained. Let’s write the integral in general form.
( )
(
)
+
=
0
cos
cos
1
ln
x
dx
x
p
p
I
(6)
Thus, we can say that this integral defines a function with respect to the variable
p
. in equality (6), it makes
sense to consider
p
only at
1
1
−
p
. Only for such values of
p
is the argument of the natural logarithm
greater than zero at all points of the integration segment. This means that the original integral (5) will be
a special case of the integral (6) at
2
3
=
p
. Now we take the derivative with respect to the parameter
p
:
( )
(
)
+
=
0
cos
cos
1
ln
dx
x
x
p
p
p
I
Here the limits of integration do not depend on the variable
p.
The partial derivative of the integrand with
respect to the variable
p
will be a continuous function. Therefore, we can enter the differential under the
integral sign and calculate it. When integrating.
( )
(
)
(
)
+
=
+
=
+
=
0
0
0
cos
1
cos
cos
1
cos
cos
cos
1
ln
x
p
dx
x
x
p
xdx
dx
x
x
p
p
p
I
Volume 03 Issue 10-2023
320
International Journal of Advance Scientific Research
(ISSN
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VOLUME
03
ISSUE
10
Pages:
316-323
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
We use the universal integration method. Let
2
x
tg
t
=
, then
2
sin
2
cos
cos
2
2
x
x
x
−
=
. When replacing, you
need to change the limits integration in the integral. If
0
→
x
to
0
→
t
, if
→
x
, to
→
t
.
Then
( )
(
)
(
)
=
+
−
+
−
=
−
+
+
=
−
−
+
=
0
0
2
2
2
0
2
2
1
1
2
1
1
1
1
2
2
1
1
2
cos
t
p
p
dt
p
t
p
p
dt
x
tg
p
p
x
dx
p
I
2
2
0
1
2
1
2
1
1
1
1
1
2
p
p
p
p
t
arctg
p
p
p
−
=
−
=
−
+
+
−
−
. Thus
( )
2
1
p
p
I
−
=
( )
C
p
p
I
+
=
arcsin
(7)
From here we find
C.
( )
0
0
=
I
means
0
=
C
. The problem statement was that it was necessary to
calculate a special case of integral (6) for
2
3
=
p
. If
( )
p
p
I
arcsin
=
. Then
=
=
=
+
=
0
2
3
3
2
3
arcsin
cos
cos
2
3
1
ln
2
3
x
dx
x
I
Example
–
3.
Calculate the integral
(
)
−
−
1
0
2
2
1
1
dx
x
x
arctg
(8)
Solution.
Let us simplify the integrand expressions by making the following substitution:
t
x
sin
=
and the boundaries will have the following form
2
1
,
0
0
=
→
=
=
→
=
t
x
t
x
and we see that in the first quarter
0
cos
,
0
sin
t
t
. Then
(
)
(
)
=
−
−
=
1
0
2
0
2
2
cos
cos
1
1
t
dt
t
arctg
dx
x
x
arctg
I
(9)
Now let’s use Feynman’s method as follows
( )
(
)
=
2
0
cos
cos
t
dt
t
p
arctg
p
I
(10)
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VOLUME
03
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Pages:
316-323
SJIF
I
MPACT
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(2021:
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)
(2022:
5.636
)
(2023:
6.741
)
OCLC
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From (9) it is clear that the parameter
1
=
p
. In this case, our is to calculate the integral
1
=
p
, if we
know that
( )
0
0
=
I
.
Differentiating with respect to the parameter
p
we get:
( )
(
)
(
)
(
)
=
+
=
+
=
=
=
2
0
2
0
2
2
2
2
2
0
2
0
cos
1
cos
cos
1
cos
cos
cos
cos
cos
t
p
dt
t
t
p
tdt
t
t
p
arctg
p
t
dt
t
p
arctg
dp
d
p
I
(
)
(
)
=
+
+
=
+
+
=
+
+
=
2
0
2
2
2
2
0
2
2
2
2
0
2
2
2
2
1
cos
cos
1
sin
cos
cos
sin
t
tg
p
t
dt
t
p
t
dt
t
p
t
t
dt
( )
( )
(
)
+
=
+
=
+
+
=
+
+
=
+
+
=
2
0
2
2
2
0
2
2
2
2
2
2
0
2
2
1
2
1
1
2
1
1
1
1
1
p
p
p
tgt
arctg
p
t
tg
p
tgt
d
t
tg
p
tgt
d
Means
( )
( )
(
)
C
p
p
p
I
p
p
I
+
+
+
=
+
=
2
2
1
ln
2
,
1
2
( )
( )
(
)
2
1
ln
2
0
0
0
p
p
p
I
C
I
+
+
=
=
=
( )
(
)
+
=
−
−
=
=
1
0
2
2
2
1
ln
2
1
1
1
dx
x
x
arctg
I
I
Example
–
4.
Calculate the integral
( )
−
+
dx
x
x
4
3
cos
2
Solution.
( )
( )
−
+
=
+
0
2
2
4
3
cos
2
4
3
cos
dx
x
x
x
dx
x
( )
( )
+
=
0
2
4
cos
2
dx
x
px
p
I
( )
2
0
=
I
px
u
=
p
u
x
=
p
du
dx
=
Volume 03 Issue 10-2023
322
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
03
ISSUE
10
Pages:
316-323
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
( )
( )
( )
( )
(
)
=
+
−
=
+
−
=
+
=
0
2
2
0
2
2
0
4
sin
2
4
sin
2
4
cos
2
x
x
dx
px
x
dx
x
px
x
dx
x
px
p
p
I
(
)
( )
(
)
( )
( )
(
)
=
+
+
−
=
+
−
+
−
=
0
2
0
0
2
2
4
sin
8
sin
2
4
sin
4
4
2
dx
x
x
px
dx
x
px
dx
x
x
px
x
( )
(
)
( )
(
)
( )
(
)
+
+
−
=
+
+
−
=
+
+
−
=
0
2
0
2
0
2
0
4
sin
8
4
sin
8
2
2
4
sin
8
sin
2
dx
x
x
px
dx
x
x
px
dx
x
x
px
du
u
u
Means
( )
( )
(
)
+
+
−
=
0
2
4
sin
8
dx
x
x
px
p
I
( )
( )
(
)
( )
(
)
( )
( )
=
+
+
=
+
=
+
+
=
0
2
0
2
2
0
4
4
cos
8
4
cos
8
4
sin
8
0
p
I
dx
x
px
dx
x
x
px
x
dx
x
x
px
p
p
I
We have obtained a homogeneous differential equation of the second order.
( )
( )
( )
( )
0
4
4
=
−
=
p
I
p
I
p
I
p
I
Make up a characteristic equation
2
,
2
0
4
2
1
2
−
=
=
=
−
k
k
k
General solution equation is equal
( )
p
p
e
c
e
c
p
I
2
2
2
1
−
+
=
( )
2
0
2
1
=
+
=
c
c
I
( )
p
p
e
c
e
c
p
I
2
2
2
1
2
2
−
−
=
( )
−
=
−
=
2
1
2
2
0
c
c
I
2
,
0
2
2
2
2
1
2
1
2
1
=
=
−
=
−
=
+
c
c
c
c
c
c
( )
p
e
p
I
2
2
−
=
( )
( )
6
6
2
2
2
4
3
cos
3
e
e
dx
x
x
I
=
=
+
=
−
−
Volume 03 Issue 10-2023
323
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
03
ISSUE
10
Pages:
316-323
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
C
ONCLUSIONS
Integrating functions is a mathematical art. It is
interesting to calculate them, especially when
non-standard methods are used in the solution.
T
he article uses Feynman’s method to calculate
some rather complex integrals that cannot be
integrated in the standard way.
R
EFERENCES
1.
R. Feynman “Surely You’re Joking, Mr.
Feynman!”, Bantam edition, 1985, c. 400.
2.
S. Frederick Woods, Advanced calculus:
Acourse Arranged with Special Reverence to
the Needs of Students of Applied mathematics,
1934, c. 404.
3.
G. Boros and V. Moll, Irresistible integrals:
Symbolics, Analysis and Experiments in the
Evaluation of integrals, Cambridge University
Press, Cambridge, 2004, c. 299.
4.
Б.П. Демидович «Сборник задач и
упражнений
по
математическому
анализу», Москва, 2005, 561.
