The present book introduces new functions named as dominating functions, sequential functions and dominating sequential functions, which dominates all most all classical elementary functions. Following the tradition each new function has been divided into four types as trigonometric, hyperbolic, exponential and logarithmic functions. In the book we find that the classical elementary functions are the particular case of the new functions.
This is the reason of calling them dominating functions. In fact dominating functions behave like superset of the classical one as will be seen in the book. These functions solve the lack of mathematical notations for many nonelementary functions generated from indefinite integrations. Some of the examples have been given in the present book. The students, teachers and researchers will find new functions for further research in different areas of mathematics, physics, etc.
Acknowledgement
The content of this book has been discussed in short in the thesis of my doctoral research
work submitted in 2012 in the University Department of Mathematics, Vinoba Bhave
University, Hazaribag, Jharkhand under the supervision of Dr. D. K. Sen. The thesis has been
published as an e-book by GRIN Verlag Open Publishing and Lap Lambert Publishing. The
thesis has been divided into two parts: Six Conjectures on Integration and Dominating
Sequential Function. First part has been published by GRIN Verlag. The second part is in
your hand with major modifications including many properties, which were left in the thesis
and thesis book.
Dharmendra Kumar Yadav
Preface
The present book introduces new functions named as dominating functions, sequential
functions and dominating sequential functions, which dominates all most all classical
elementary functions. Following the tradition each new function has been divided into four
types as trigonometric, hyperbolic, exponential and logarithmic functions. In the book we
find that the classical elementary functions are the particular case of the new functions. This
is the reason of calling them dominating functions. In fact dominating functions behave like
superset of the classical one as will be seen in the book. These functions solve the lack of
mathematical notations for many nonelementary functions generated from indefinite
integrations. Some of the examples have been given in the present book. The students,
teachers and researchers will find new functions for further research in different areas of
mathematics, physics, etc.
Dharmendra Kumar Yadav
CONTENTS
Chapter-I 1-6
Elementary Function
Function 1
Elementary Function 1
Indefinite Integration 2
Range and Difficulty of Problem of Indefinite Integration 2
Nonintegrable Functions 3
Existence of Integrals and Lack of Notations of Functions 3
Nonelementary Functions Due to Lack of Notations 4
Faddeeva Functions 4
Plasma Dispersion Function 4
Jackson Function 4
Dawson Integral 5
Frensel Function 5
Frensel Integrals 5
Sitenko Function 5
Fried and Cante Function 5
Gordeyev Integral 5
Error Function 6
Complementary Error Function 6
Exponential Integral 6
Sine and Cosine Integrals 6
Logarithmic Integral 6
Chapter-II 7-22
Dominating Function
Introduction 7
Derivation of Dominating Function 7
Singular Dominating Function 9
Regular Dominating Function 9
Particular Case of Dominating Function 9
Dominatable Function 10
Power Series of Trigonometric Functions 10
Power Series of Hyperbolic Functions 11
Power Series of Exponential Functions 12
Power Series of Logarithmic Functions 13
Convergence of Dominating Function 13
Convergence of Regular Dominating Function 14
Points of Convergence of Regular Dominating Function 14
Convergence Theorem for Regular Dominating Function 14
Necessary Condition for Convergence of Regular Dominating Function 15
Convergence of Singular Dominating Function 15
Points of Convergence of Singular Dominating Function 15
Convergence Theorem for Singular Dominating Function 16
Necessary Condition for Convergence of Singular Dominating Function 16
Convergence Theorem for Dominating Function 16
Conjecture on Dominating Function 17
Necessary Condition for Convergence of Dominating Function 17
Points of Convergence of Dominating Function 17
Interval and Radius of Convergence of Dominating Function 17
Interval and Radius of Convergence of Regular Dominating Function 17
Interval and Radius of Convergence of Singular Dominating Function 18
Interval of Divergence 19
Conclusion of Interval and Radius of Convergence of Dominating Function 19
Continuity, Differentiability and Integrability of Dominating Function 21
Chapter-III 23-44
Types of Dominating Functions
Introduction 23
Dominating Trigonometric Functions 23
Relations Between Dominating Trigonometric Functions 25
Dominating Trigonometric Functions in Terms of Exponential Functions 27
Why Do We Call Dominating Trigonometric Functions? 28
Dominating Trigonometric Functions with Constant Coefficients 28
Series Expansion of Dominating Trigonometric Functions 29
Derivatives of Dominating Trigonometric Functions 30
Dominating Hyperbolic Functions 32
Relations Between Dominating Hyperbolic Functions 33
Dominating Hyperbolic Functions in Terms of Exponential Functions 36
Why Do We Call Dominating Hyperbolic Functions? 36
Dominating Hyperbolic Functions with Constant Coefficients 37
Series Expansion of Dominating Hyperbolic Functions 38
Derivatives of Dominating Hyperbolic Functions 39
Relations Between D-Trigonometric, D-Hyperbolic and Trigonometric Functions 40
Dominating Exponential Functions with base a 42
Dominating Exponential Functions with base e 42
Derivatives of Dominating Exponential Functions 43
Dominating Logarithmic Functions 44
Derivatives of Dominating Logarithmic Functions 44
Chapter-IV 45-55
Dominating Sequential Functions
Introduction 45
Sequential Trigonometric Functions 45
Dominating Sequential Trigonometric Functions 48
Sequential Hyperbolic Functions 50
Dominating Sequential Hyperbolic Functions 51
Sequential Exponential Functions 53
Dominating Sequential Exponential Functions 54
Sequential Logarithmic Functions 54
Dominating Sequential Logarithmic Functions 55
Chapter-V 56-95
Applications and Indefinite Integrals of Dominating Sequential Functions
Introduction 56
Indefinite Integrals of Dominating Trigonometric Functions 56
Particular Case 65
Indefinite Integrals of Dominating Hyperbolic Functions 68
Particular Case 78
Indefinite Integrals of Dominating Exponential Functions 81
Indefinite Integrals of Dominating Logarithmic Functions 86
Indefinite Integrals of Classical Nonelementary Functions 90
Indefinite Integrals of Trigonometric Nonelementary Functions 90
Indefinite Integrals of Hyperbolic Nonelementary Functions 93
Indefinite Integrals of Exponential Nonelementary Functions 95
Chapter-VI 96-98
Existence Theorems on Indefinite Integrability
Introduction 96
Existence Theorems on Indefinite Integrability of a Function 96
Necessary Condition for Indefinite Integrability 97
Sufficient Condition for Indefinite Integrability 97
Necessary and Sufficient Condition for Indefinite Integrability 97
Comparative Study of Existence Theorems on Integrability 97
Modified Sufficient Condition of Integrability 98
Existence Theorem of Integrability (Modified Form) 98
References 99-106
1
Chapter-I
Elementary Functions
A function has a very close relationship with various phenomena of the real world. We know
that the area A of a circle of radius r is
2
r
A
=
. In this formula A is called a function of r,
where r is known as an independent variable and A is the dependent variable. We call the set
over which independent variable r varies the domain and the set over which A varies the
range of the function. Formally we define it as:
Function: Let A and B be two non-empty sets, then a rule f which associates each element
of A with a unique element of B is called a mapping or function from A to B. If f is a function
from A to B, we write
B
A
f
:
read as f is a function from A to B. If f associates x of A to
y of B, then we say that y is the image of the element x under the function f and denote it by
y=f(x). If the domain and range of a function f are subsets of the set R of real numbers, then
f is said to be a real valued function.
As modern calculus cannot be studied without functions, the present book cannot be studied
without elementary functions. An elementary function is one of the foundational notions of
Pre-calculus and Calculus courses. We cannot expect serious understanding of central
calculus, theorems such as theorems of continuity, differentiability, and integrability, without
detailed comprehension of the notion of elementary functions. Actually, one can say that the
traditional Calculus course in general is the mathematical analysis of elementary functions.
In the same time there is a strange paradox that some Calculus textbook do not contain even
the term "elementary functions". Generally it is defined as:
Elementary Function: A function is called an elementary function if it can be written as an
equation in the form of y=f(x), with f representing an expression formed by combining a
2
finite collection of powers of x, trigonometric functions, hyperbolic functions, exponentials,
logarithms, inverse trigonometric functions, and inverse hyperbolic functions together
through additions, subtractions, multiplications, divisions, powers, and compositions.
Not all functions are elementary. The most common example of a non-elementary function
is a piecewise-defined function, which uses several different equations to define the rule based
on where in the domain an input variable is selected. Other types of nonelementary functions
arise in calculus as "anti-derivatives" of elementary functions.
Indefinite integration can be considered as a machine to produce nonelementary functions.
Recently
Yadav has introduced six conjectures on nonelementary functions, from which
infinite number of nonelementary functions can be derived. How nonelementary functions
have been derived from integration, can be better understood by defining them at a glance as
follows:
Indefinite Integration: Gottfried Leibniz (1684) considered an indefinite integration of an
elementary function f(x) as a solution F(x), composed of elementary functions, such that
F'(x) f (x)
=
. In mathematical symbol, we denote it by
x
a
F(x)
f (t)dt
f (x)dx k
=
=
+
which on differentiation gives us
F'(x) f (x),
=
where the constant of integration k
corresponds to the value of the integral for the lower limit a.
Range and Difficulty of Problem of Indefinite Integration
Let us suppose that f(x) belongs to some special class of functions S. Then we may ask
whether F(x) is itself a member of S, or can be expressed, according to some simple standard
mode of expression, in terms of functions which are members of S. The range and difficulty
of our problem will depend upon our choice of:
(i) a class of functions, and
3
(ii) a standard mode of expression.
We shall take S to be the class of elementary functions, and our mode of expression to be that
of explicit expression in finite terms.
Nonelementary Functions
An indefinite integral of an elementary function is either an elementary function or can be
expressed in terms of elementary functions in finite number of steps. Those functions whose
indefinite integrals are neither elementary nor can be expressed in terms of elementary
functions are classically known as nonelementary or nonintegrable functions.
Yadav has
called them indefinite nonintegrable functions. If we say that an indefinite integral f (x)dx
is elementary (or integrable) it means that its integral exists and can be expressed in terms of
elementary functions in closed form.
Existence of Integrals and Lack of Notations of Functions
We know that the integral of an elementary function is the assertion of the Fundamental
Theorem of Calculus: Every continuous function has an anti-derivative. Although it is not
closely bound up with the assumption that the integrand is continuous, it may be extended to
wide classes of functions with discontinuities.
But there is no guarantee that we can find a formula for an anti-derivative in terms of
elementary functions like sine, cosine, logarithm, and so forth. There are elementary
functions which have anti-derivatives but they cannot be expressed in terms of elementary
functions due to the lack of notations of those functions.
For example, the error function
2
x
e dx
-
, the exponential integral
x
e dx
x
, the sine integral
sin x dx
x
, the cosine integral cos x dx
x
, etc.
4
Nonelementary Functions due to Lack of Notations
We know that the integral of every continuous function exists and is itself a continuous
function of the upper limit, and this fact has nothing to do with the question whether the
integral can be expressed in terms of elementary functions or not.
In this sense the process of
integration forms a basis for the generation of new functions. Following are the list of
nonelementary functions originated due to the lack of notations:
Faddeeva Function w(x) introduced by Fadeeva and Terent'ev (1954). It is also called
complex error function or probability integral by Weideman (1994), Baumjohann and
Treumann (1997). Some Russian authors Mikhailovskiy (1975), Bogdanov et al. (1976) call
it Complex Cramp function. It is given by
+
=
-
x
0
t
x
dt
e
i
2
1
e
)
x
(
w
2
2
Plasma Dispersion Function Z(x) given by Fried and Conte (1961) also known as Fried-
Conte function is given by
-
-
-
=
dt
x
t
e
1
)
x
(
Z
2
t
Jackson Function G(x) propounded by Jackson (1960) and does not (yet) have a name [to
the knowledge of Lehtinen, 2010] is given by
-
-
-
=
x
t
dt
te
1
)
x
(
G
2
t
5
Dawson Integral presented by Abramowitz and Stegun (1965) also sometimes called
Dawson's function is given by
-
=
x
0
y
x
dy
e
e
)
x
(
F
2
2
Frensel Functions C(x) and S(x) introduced by Abramowitz and Stegun (1965) are denoted
and defined by
=
+
x
0
2
t
i
dt
e
)
x
(
iS
)
x
(
C
2
Frensel integrals are alternatively denoted and defined by
( )
=
t
0
2
dx
x
cos
)t
(
x
, and,
( )
=
t
0
2
dx
x
sin
)t
(
y
Sitenko Function (x) given by Sitenko (1982) and defined by
dt
e
xe
2
)
x
(
x
0
t
x
2
2
-
=
Fried and Conte Function Y(x) (1961) is defined by
-
=
x
0
t
x
dt
e
x
e
)
x
(
Y
2
2
Gordeyev Integral
)
,
w
(
G
presented by Gordeyev (
1952) is defined by
(
)
dt
e
w
)
,
w
(
G
0
2
t
t
cos
1
iwt
2
-
-
-
=
6
Error or Cramp Function introduced by Gauss. Some Russian authors Mikhailovskiy
(1975), Bogdanov et al. (1976) call erf(x) a
Cramp function. It is given by
-
=
dx
e
2
)
x
(
erf
2
x
or
-
=
z
0
t
dt
e
2
)
z
(
erf
2
Complementary Error Function is given by
)
x
(
erf
1
dt
e
2
)
x
(
erfc
x
t
2
-
=
=
-
Exponential Integral is denoted and defined by
=
dx
x
e
)
x
(
Ei
x
Sine and Cosine Integrals are defined by
=
dx
x
x
sin
)
x
(
Si
=
dx
x
x
cos
)
x
(
Ci
Logarithmic Integral is denoted and defined by
=
x
0
dt
t
log
1
)
x
(
li
or
=
x
log
dx
)
x
(
li
7
Chapter-II
Dominating Function
The topic of domination was given formal mathematical definition by
Berge (1958) and Ore
(
1962). The earliest ideas of dominating sets date back to the origin of game of chess in India.
John Van Neumann considered domination in digraphs while solving problems in game
theory. The importance of this term can be better understood by `
Five Queen's Problem'.
In 1850, a question was raised namely what is the minimum number of queens to be placed
in a chess board so that all the squares are either attached by a queen or occupied by a queen?
It was found that 5 is that number.
Following the concept of domination theory, a natural question arises that what can be the
form of a function or a mathematical formula, which dominates all most all the elementary
functions in the sense that the elementary functions are the particular case of that functions
or formula for some particular condition. For this a two sided infinite series have been
developed
1
2
2
1
0
1
2
2
1
...
...
+
+
+
+
+
+
c
c
b
b x
b x
x
x
and called it a
Dominating Function (D-function) or Dominating Series (D-series).
Derivation of Dominating Function
We know from Laurent's Series that " if f(z) is an analytic function within and on the
boundary of the ring shaped region R bounded by two concentric circles C
1
and C
2
of radii
R
1
and R
2
(R
1
R
2
) respectively, having centre at the point z=a, then for all z in R
1
2
2
1
0
1
2
2
1
b
b
f (z) ...
a
a (z a) a (z a) ...
(z a)
(z a)
=
+
+
+
+
-
+
-
+
-
-
1
0
(
)
(
)
-
=
=
=
-
+
-
n
n
n
n
n
n
b z a
a z a
, n N
(1)
8
Replace z by x in Laurent's series (1) by putting y=0 in z=x+iy without considering the values
a
n
and b
n
, we get
1
0
( )
(
)
(
)
n
n
n
n
n
n
f x
b x a
a x a
-
=
=
=
-
+
-
0
1
2
2
1
0
1
2
2
1
{...
} { (
)
(
)
(
) ...}
(
)
(
)
-
-
=
+
+
+
-
+
-
+
-
+
-
-
a
a
a x a
a x a
a x a
x a
x a
Again replace (x-a) by (ax+b) in the above series, we get
0
1
2
2
1
0
1
2
2
1
( ) {...
} { (
)
(
)
(
) ...}
(
)
(
)
a
a
f x
a ax b
a ax b
a ax b
ax b
ax b
-
-
=
+
+
+
+
+
+
+
+
+
+
+
(2)
The above series suggests that there exists a two sided infinite series of the form
1
0
( )
(
)
(
)
n
n
n
n
n
n
B
f x
A ax b
ax b
=
=
=
+
+
+
(3)
where B
n
, A
n
are real or complex numbers for all n. This series (3) can be named as a two-
sided infinite power series in (ax+b) and can be written in single form as
( )
(
)
n
n
n
f x
C ax b
=-
=
+
(4)
where C
n
=B
n
for all n]-, -1] and C
n
=A
n
for all n[0, [.
Obviously the series
1
(
)
n
n
n
B
ax b
=
+
is one-sided infinite series of negative power and the other
series
0
(
)
n
n
n
A ax b
=
+
is one-sided infinite series of positive power. We can call them one-
sided finite series of negative and positive powers if they have finite number of terms.
Now let us write the series (3) into two parts as follows:
1
0
( )
(
)
(
)
n
n
n
n
n
n
B
f x
A ax b
ax b
=
=
=
+
+
+
= f
SD
+ f
RD
(5)
9
Here the second series f
RD
is continuous and differentiable at every point, whereas the first
series f
SD
is neither continuous nor differentiable at the point x=(-b/a).
We call the function f(x) given by (5) a
Dominating Function (D-Function) or Dominating
Series (D-Series). We call the first series f
SD
a
Singular Dominating Series or function
(S.D. Series or function) and the second series f
RD
a
Regular Dominating Series or function
(R.D. Series or function).
The two series can be named as Regular Dominating Finite (R.D.F.), Regular Dominating
Infinite (R.D.I.), Singular Dominating Finite (S.D.F.) or Singular Dominating Infinite
(S.D.I.) Series according to the finite and infinite number of terms in the series.
Examples 1: For the existence of D-function, we can study the following functions in series:
·
Polynomials of finite degree are R.D.F. series.
·
Trigonometric functions sinx, cosx, secx, tanx are R.D.I. series.
·
Trigonometric functions cotx, cosecx, sin(1/x), tan(1/x) are S.D.I. series.
·
Hyperbolic functions sinhx, coshx, tanhx, sechx are R.D.I. series.
·
Exponential functions
2
2
x
x
x
e ,e ,e
-
are R.D.I. series.
·
Logarithmic functions log(1+x), log(1-x
2
) are R.D.I. series.
·
Inverse circular functions sin
-1
x, cos
-1
x are R.D.I. series.
·
Inverse hyperbolic function tanh
-1
x is R.D.I. series, etc.
Particular Case of Dominating Function
To study the properties of a D-function, we write the series (5) as
1
0
( )
(
)
(
)
n
n
n
SD
RD
n
n
n
c
f x
b x a
f
f
x a
=
=
=
+
-
=
+
-
Now for a=0, we have
10
1
0
( )
=
=
=
+
n
n
n
n
n
n
c
f x
b x
x
1
2
2
1
0
1
2
2
1
...
...
=
+
+ + +
+
+
c
c b b x b x
x
x
This particular form of D-function will play a major role in developing new functions.
Dominatable Functions
A function which can be expressed in the form of dominating function f(x) can be named as
Dominatable Function (d-able Function). Similarly if a function can be expressed in the
form of R-D function or S-D function can be called
R. D.-able and S. D.-able functions
respectively.
Example 2: All the basic elementary functions are dominatable functions.
Proof: Using Taylor's series expansion about the point x=0 and Laurent's series of different
functions, we find their power series expansion as follows:
Power Series of Trigonometric Functions
2n 1
n
n 0
x
sin x
( 1)
(2n 1)!
+
=
=
-
+
3
5
7
x
x
x
x
...
3! 5! 7!
= -
+
-
+
2n
n
n 0
x
cos x
( 1)
(2n)!
=
=
-
2
4
6
x
x
x
1
...
2! 4! 6!
= -
+
-
+
2n
2n
n 1
2n 1
2n
n 1
2 (2
1)B
tan x
( 1)
x
(2n)!
-
-
=
-
=
-
3
5
7
1
2
17
x
x
x
x ...
3
15
315
= +
+
+
+
2n
n 1 2n
2n
n 1
1
x
cot x
1
( 1) 2 B
x
(2n)!
-
=
=
-
-
n 2n
2n 1
2n
n 0
( 1) 2 B x
(2n)!
-
=
-
=
3
5
1 1
1
2
x
x
x ...
x 3
45
945
= -
-
-
-
11
2n
n
2n
n 0
x
sec x
( 1) E
(2n)!
=
=
-
2
4
6
1
5
61
1
x
x
x ...
2
24
720
= +
+
+
+
n 1
2n 1
2n 1
2n
n 0
( 1) 2(2
1)B
cosecx
x
(2n)!
+
-
-
=
-
-
=
3
5
1 1
7
31
x
x
x ...
x 6
360
15120
= +
+
+
+
2n 1
1
2n
2
n 0
(2n)! x
sin x
2 (n!) 2n 1
+
-
=
=
+
3
5
7
1 x
1.3 x
1.3.5 x
x
...
2 3
2.4 5
2.4.6 7
= +
+
+
+
2n 1
1
2n
2
n 0
(2n)! x
cos x
2
2 (n!) 2n 1
+
-
=
= -
+
3
5
7
1 x
1.3 x
1.3.5 x
x
...
2
2 3 2.4 5 2.4.6 7
= -
+
+
+
+
1
tan x
-
3
5
7
3
5
x
x
x
x
..., x 1
3
5
7
if x 1
1
1
1
...,
if x
1
2 x 3x
5x
-
+
-
+
=
+
± - +
-
+
-
-
1
cot x
-
3
5
7
3
5
x
x
x
x
... , x 1
2
3
5
7
p 0 if x 1
1
1
1
p
...,
p 1if x
1
x 3x
5x
-
-
+
-
+
=
=
+ -
+
+
=
-
(2n 1)
1
2n
2
n 0
(2n)! x
sec x
2
2 (n!) (2n 1)
-
+
-
=
= -
+
3
5
7
1
1 x
1.3 x
1.3.5 x
x
...
2
2 3
2.4 5
2.4.6 7
-
-
-
-
= -
+
+
+
+
(2n 1)
1
2n
2
n 0
(2n)! x
cosec x
2 (n!) (2n 1)
-
+
-
=
=
+
3
5
7
1
1 x
1.3 x
1.3.5 x
x
...
2 3
2.4 5
2.4.6 7
-
-
-
-
=
+
+
+
+
Power Series of Hyperbolic Functions
2n 1
n 0
x
sinh x
(2n 1)!
+
=
=
+
3
5
7
x
x
x
x
...
3! 5! 7!
= +
+
+
+
2n
n 0
x
cosh x
(2n)!
=
=
2
4
6
x
x
x
1
...
2! 4! 6!
= +
+
+
+
2n
2n
2n 1
2n
n 1
2 (2
1)B
tanh x
x
(2n)!
-
=
-
=
3
5
7
1
2
17
x
x
x
x ...
3
15
315
= -
+
-
+
12
2n
2n
2n
n 1
1
x
coth x
1
2 B
x
(2n)!
=
=
+
3
5
1 1
1
2
x
x
x ...
x 3
45
945
= +
-
+
+
2n
2n
n 0
x
sec hx
E
(2n)!
=
=
2
4
6
1
5
61
1
x
x
x ...
2
24
720
= -
+
-
+
2n 1
2n 1
2n
n 1
2(1 2
)B
1
cosechx
x
x
(2n)!
-
-
=
-
= +
3
5
1 1
7
31
x
x
x ...
x 6
360
15120
= -
+
-
+
n
2n 1
1
2n
2
n 0
( 1) (2n)! x
sinh x
2 (n!) (2n 1)
+
-
=
-
=
+
3
5
7
2
4
6
1 x
1.3 x
1.3.5 x
x
..., x 1
2 3 2.4 5 2.4.6 7
if x 1
1
1.3
1.3.5
log 2x
...
if x
1
2.2x
2.4.4x
2.4.6.6x
-
+
-
+
=
+
±
+
-
+
-
-
-
1
cosh x
-
1
2
4
6
1
ifcosh x 0, x 1
1
1.3
1.3.5
log 2x
...
2.2x
2.4.4x
2.4.6.6x
ifcosh x 0, x 1
-
-
+
= ±
-
+
+
+
-
1
tanh x
-
3
5
7
x
x
x
x
...
3
5
7
= +
+
+
+
1
coth x
-
3
5
7
1
1
1
1
...
x 3x
5x
7x
= +
+
+
+
Power Series of Exponential Functions
n
x
n 0
x
e
n!
=
=
2
3
4
x x
x
x
1
...
1! 2! 3! 4!
= + +
+
+
+
2
3
x
xloga
x log a (x log a)
(x log a)
a
e
1
...
1!
2!
3!
=
= +
+
+
+
n
n
3
5
6
x
2
n
( 2) sin
x
x
x
x
4
e sin x x x
...
...
3 30 90
n!
= +
+
-
-
+
+
+
n
n
3
4
x
n
( 2) cos
x
x
x
4
e cos x 1 x
...
...
3
6
n!
= + -
-
+
+
+
13
Power Series of Logarithmic Functions
n
2
3
4
n 1
n 1
x
x
x
x
log(1 x)
( 1)
x
...
n
2
3
4
-
=
+
=
-
= -
+
-
+
n
n 1
x
log(1 x)
n
=
-
= -
3
5
x
1
x
1
x
log(a x) log a 2
... ,a 0, a
2a x 3 2a x
5 2a x
+
=
+
+
+
+
-
+
+
+
2
3
3
5
2
3
(x 1)
(x 1)
(x 1)
...,0 x 2
2
3
x 1
1 x 1
1 x 1
log x
2
... , x 0
x 1
3 x 1
5 x 1
x 1
1 x 1
1 x 1
1
..., x
x
2
x
3
x
2
-
-
- -
+
-
-
-
-
=
+
+
+
+
+
+
-
-
-
+
+
+
From above it is clear that all most all the basic elementary functions are dominatable
function.
Example 3: The functions sin x
x
, sinx, coshx,
2
x
e are R.D.-able and
2
tan x
x
,
x
e
x
are S.D.-able.
Convergence of Dominating function
Since we have
0
(
)
n
RD
n
n
f
A ax b
=
=
+
=
0
0
(
)
(
)
n
n
n
n
n
n
n
b
a A x
x p
a
=
=
+
=
-
,
where
n
= a
n
A
n
and p=-b/a; and
1
1
1
(
)
(
)
(
)
n
n
n
SD
n
n
n
n
n
n
n
B
B
f
b
ax b
x p
a x
a
=
=
=
=
=
=
+
-
+
, where
n
= B
n
/a
n
.
Therefore to study the convergence of D-function, we write it as
14
1
0
( )
(
)
(
)
n
n
n
SD
RD
n
n
n
c
f x
b x a
f
f
x a
=
=
=
+
-
=
+
-
Convergence of Regular Dominating Function f
RD
For f
RD
we have U
n
=b
n
(x-a)
n
. Applying Cauchy's root test, we find that f
RD
is convergent for
1/
1
n
n
x a
b
+
and divergent for
1/
1
n
n
x a
b
+
.
Points of Convergence of f
RD
We have
0
(
)
n
RD
n
n
f
b x a
=
=
-
; let b
n
0
We ignore the first term b
0
and take U
n
= b
n
(x-a)
n
.
Applying D'Alembert's ratio test and then Cauchy's root test, we find that f
RD
is convergent
for
1
n
n
b
x a
b
+
+
, and divergent for
1
n
n
b
x a
b
+
+
. For
1
n
n
b
x a
b
+
= +
, it is convergent if
1
1
1
n
n
n
n
b
b
+
+
, and divergent if
1
1
1
n
n
n
n
b
b
+
+
. For
1
1
1
n
n
n
n
b
b
+
+
=
, the nature of f
RD
depends on b
n
.
Convergence theorem for f
RD
The regular dominating function f
RD
converges if and only if for every 0, there exists an
integer N0 such that
(
)
m
k
k
k n
b x a
=
-
(6)
for sufficiently very small values of ; if mnN.
In particular by taking m=n, the inequality (6) becomes
15
(
)
n
n
b x a
-
, nN (
)
.
n
n
x a
x a
b
-
Necessary Condition for Convergence of f
RD
If f
RD
converges, then
lim (
)
0
n
n
n
b x a
-
= , i. e., xa, as n
The condition b
n
(x-a)
n
0 is not, however, sufficient to ensure the convergence of f
RD
.
Convergence of Singular Dominating Function f
SD
For
f
SD
we have
(
)
n
n
n
c
U
x a
=
-
. Applying Cauchy's root test, we find that f
SD
is convergent
for
1/n
n
x a c
+
and divergent for
1/n
n
x a c
+
.
Points of Convergence of f
SD
We have
1
(
)
n
SD
n
n
c
f
x a
=
=
-
; let c
n
0.
Here we have
(
)
n
n
n
c
U
x a
=
-
. Applying D'Alembert's ratio test and then Cauchy's root test,
we find that f
SD
is convergent for
1
n
n
c
x a
c
+
+
, divergent for
1
n
n
c
x a
c
+
+
and for
1
n
n
c
x a
c
+
= +
, it is convergent for
1
1
1
n
n
n
n
c
c
+
+
, divergent for
1
1
1
n
n
n
n
c
c
+
+
and for
1
1
1
n
n
n
n
c
c
+
+
=
, its nature depends
on c
n
.
Convergence theorem for f
SD
16
f
SD
converges if and only if for every 0, there exists an integer N0 such that
(
)
m
k
k
k n
c
x a
=
-
(7)
for sufficiently very large values of ; if mnN and n0.
In particular, by taking m=n, the inequality (7) becomes
(
)
n
n
c
x a
-
; (nN)
1/
(
) ( )
n
n
c
x a
x
a
-
.
Necessary Condition for Convergence of f
SD
If f
SD
converges, then
lim
0
(
)
n
n
n
c
x a
=
-
i.e., (x-a)
n
xa as n.
The condition (x-a)
n
is not, however, sufficient to ensure the convergence of f
SD
.
Combining the above results, we get the following properties:
Convergence theorem for D-function
The dominating function f
D
= F
SD
+f
RD
converges if and only if for every
0
0 (for
sufficiently small values of
0
) and
0 (for sufficiently large values of
), there exists
an integer N0 (may be different for both cases), such that
(
)
m
k
k
k n
b x a
=
-
0
and
(
)
m
k
k
k n
c
x a
=
-
simultaneously, if mnN.
In particular, by taking m = n, they imply x a
and x a
for f
RD
and f
SD
respectively. But
it is impossible to find x which satisfies both conditions simultaneously. Also it is clear that
x = a is not a point of convergence. Therefore the D-function cannot be convergent at any
point.
17
To make it convergent, we have to restrict the number of terms in either of the series f
RD
and
f
SD
.
Conjecture: The D-function f
D
cannot converge if it is two sided infinite power series.
For example:
( )
1
x
0
0
1
f x
e
e
!
. !
n
x
n
n
n
x
n
x n
=
=
=
+
=
+
Necessary Condition for Convergence of D-function
The necessary condition for the convergence of D-function is same as of f
RD
and f
SD
functions, provided that one of the series f
RD
or f
SD
has finite number of terms, i.e., if f
D
is a
convergent series (or function), then either of the two series f
RD
and f
SD
has finite number of
terms. In other words, f
D
converges if and only if, either of the series f
RD
or f
SD
has finite
number of terms, i.e., it is one sided infinite series and another sided finite series.
Points of Convergence of D-function
If the function f
D
satisfies the necessary condition of convergence, then x = a is the only point
of non-convergence, i.e., f
D
is convergent for all xR under the constraint of necessary
condition, except at the point x=a.
Interval and Radius of Convergence of Dominating Function
Let us consider the interval and radius of convergence of the D-function, for which we shall
first consider these for f
RD
and f
SD
. Then we shall combine the results:
Interval and Radius of Convergence of f
RD
We have
0
(
)
n
RD
n
n
f
b x a
=
=
-
18
Here U
n
=
(
) .
n
n
b x a
-
Applying Ratio Test, we get
n
n
n
n
n
n
n
n
b
a
x
b
n
a
x
b
a
x
b
n
U
U
n
)
(
lim
)
(
)
(
lim
lim
1
1
1
1
-
=
-
-
=
+
+
+
+
1
1
(
) 1
n
n
n
n
b x a
b
x a
b
b
+
+
-
-
1
1
)
(
+
+
-
-
n
n
n
n
b
b
a
x
b
b
1
1
n
n
n
n
b
b
a
x a
b
b
+
+
-
+
(8)
which is the interval of convergence of f
RD
and the radius of convergence of f
RD
is
1
n
n
b
b
+
. The
end-points in (8) are included in the interval of convergence if f
RD
converges at these points.
Here a is the centre of convergence.
Interval and Radius of Convergence of f
SD
We have
1
(
)
n
SD
n
n
c
f
x a
=
=
-
.
Taking
(
)
n
n
n
c
U
x a
=
-
and applying Ratio test, we get
1
1
1
1
lim
lim
lim
(
)
(
)
(
)
n
n
n
n
n
n
n
n
U
c
c
x a
n
n
n
U
x a
c
c x a
+
+
+
+
-
=
=
-
-
1
1
1
;
.(
)
n
n
n
n
c
c
x a
x a
c x a
c
+
+
-
-
1
(
)
n
n
c
x a
c
+
-
-
and
1
(
)
n
n
c
x a
c
+
-
1
n
n
c
x a
c
+
-
and
1
n
n
c
x a
c
+
+
i.e.,
1
1
]
,
[ ]
, [
n
n
n
n
c
c
x
a
a
c
c
+
+
- -
+
or,
1
1
[
,
]
n
n
n
n
c
c
x R a
a
c
c
+
+
- -
+
which is the interval of convergence of f
SD
.
19
In this case, we cannot find the centre of convergence. So let us introduce a new term the
interval of divergence beyond which a function is convergent.
Therefore, the interval of divergence of f
SD
is
1
1
n
n
n
n
c
c
a
x a
c
c
+
+
-
+
Conclusion of Interval and Radius of Convergence of f
D
We have
0
1
(
)
(
)
n
n
D
n
n
n
n
c
f
b x a
x a
=
=
=
-
+
-
=
1
[ (
)
]
(
)
n
n
n
n
n
c
b x a
x a
=
-
+
-
Ignoring the first tern b
0
and taking
(
)
(
)
n
n
n
n
n
c
U
b x a
x a
=
-
+
-
.
Applying Ratio Test, we get
1
1
lim
+
Un
Un
n
1
1
1
1
(
)
lim
(
)
1
(
)
(
)
n
n
n
n
n
n
n
n
c
b x a
x a
c
n
b x a
x a
+
+
+
+
-
+
-
-
+
-
Taking b
n
0, c
n
0, we have
1
1
1
1
lim
lim
(
)
(
)
(
)
(
)
n
n
n
n
n
n
n
n
c
c
b x a
b x a
n
n
x a
x a
+
+
+
+
-
+
-
+
-
-
There arises six cases for the convergence of f
D
:
20
Case-I: When (x-a)=0, i.e., x = a; f
D
is divergent if
lim
0
n
c
n
. But if x = a and
lim
0
n
c
n
=
, f
D
is convergent. In this case
n
b
n
lim
does not matter.
Case-II: When (x-a) 1, i.e., x a+1, then f
D
is convergent if
0
lim
=
n
b
n
and
1
n
n
b
b
+
,
i.e., b
n
is a monotonically decreasing sequence. In this case
lim
n
c
n
does not matter.
Case-III: When (x-a) -1, i.e., x a-1, then f
D
is convergent if
0
lim
=
n
b
n
and
1
n
n
b
b
+
, i.e., b
n
is a decreasing sequence. In this case also,
lim
n
c
n
does not matter.
Case-IV: When -1 (x-a) 1, i.e., a-1 x a+1, then f
D
is convergent if
lim
0
n
c
n
=
and
1
n
n
c
c
+
, i.e., c
n
is a decreasing sequence. In this case
n
b
n
lim
does not matter.
Case-V: When (x-a) = 1, i.e., x = a+1, then U
n
= b
n
+ c
n
. In this case the convergence of f
D
depends on the sequences b
n
and c
n
. If both are convergent, f
D
is convergent, otherwise
divergent.
Case-VI: When (x-a) = -1, i.e., x = a-1, then U
n
= (b
n
+ c
n
).(-1)
n
. In this case, the convergence
of f
D
depends on the sequences (-1)
n
b
n
and (-1)
n
.
c
n.
If both are convergent, f
D
is convergent,
otherwise divergent.
21
Here it is clear that we cannot find the unique radius of convergence of f
D
. The above
conditions are only necessary conditions for the convergence of f
D
and they must not be
treated as sufficient conditions.
Continuity, Differentiability Integrability of Dominating Function
We know that for the function f(x) given in a power series as
0
( )
.(
)
n
n
n
f x
a x c
=
=
-
if the interval of convergence is ]c-R, c+R[ (plus possible endpoints), then f(x) is continuous,
differentiable, and integrable on that interval.
To obtain the derivative or the integral of f(x), we pass the continuity, derivative or integral
through the summation sign , known as term by term differentiation and integration. In other
words if a function is given as a power series, it is differentiable and integrable in
the interior of the interval of convergence, which can be differentiated and integrated quite
easily by treating every term separately.
We know also from the term-wise integration of a Fourier series that: the integral of any
function f(x) satisfying Dirichlet conditions on the interval
l
x
l
-
can be obtained by
term-by-term integration of the Fourier series representation of f(x). So, if f(x) has the Fourier
series representation
=
=
+
+
=
1
1
0
sin
cos
2
)
(
n
n
n
n
l
x
n
b
l
x
n
a
a
x
f
, for
l
x
l
-
Then we have
=
+
=
-
-
-
-
+
+
=
1
1
1
0
)
1
(
cos
sin
)
(
2
)
(
n
n
n
n
n
x
l
l
x
n
n
b
l
l
x
n
n
a
l
l
x
a
dx
x
f
22
Applying the property of the power series and the Fourier series as well as the basic properties
about the continuity and differentiability of a polynomial and a rational function as well as
making the results of Bernoulli's conjecture, Laplace's theorem, Abel's theorem and
Liouville's theorem on indefinite integrability, we can easily find a property on continuity,
differentiability and integrability of D-function as follows:
Property 1: In the interval of convergence of the function
1
0
2
(
)
(
)
(
)
=
=
=
-
+
+
-
-
n
n
D
n
n
n
n
c
c
f
b x a
x a
x a
It is continuous, differentiable, and integrable on that interval and to obtain the derivative, or
the integral of f
D
, we pass the derivative or integral through the summation sign .
In other words,
=
=
+
-
-
-
+
-
=
1
1
1
1
)
(
)
(
n
n
n
n
n
n
D
a
x
nc
a
x
nb
f
dx
d
and
1
1
1
0
1
(
)
log(
)
1
(1 )(
)
+
-
=
=
-
=
+
+
-
+
-
-
n
n
D
n
n
n
n
c
x a
f dx
b
c
x a
n
n x a
From this property we can conclude that:
Property 2: A dominating function is always continuous, differentiable and indefinite
integrable, except at the point of discontinuity.
23
Chapter-III
Types of Dominating Functions
Introduction
In this chapter the concept of d-function has been applied to propound different types of
dominating functions like dominating trigonometric, dominating hyperbolic, dominating
exponential and dominating logarithmic with their power series expansion. These will
dominate all most all the classical trigonometric, hyperbolic, logarithmic and exponential
functions. The error function, complementary error function, Frensel functions, exponential
integral, sine and cosine integrals, etc. will be the particular case of these functions.
Dominating Trigonometric Functions
Let
,rm R
be any real numbers. Then the six dominating trigonometric functions
corresponding to six classical trigonometric functions can be defined and denoted as follows:
=
r
m
x
dsin
m
x
r
x
sin
=
r
m
x
dcos
m
x
r
x
cos
=
r
m
x
dtan
m
x
r
x
tan
=
r
m
x
dcosec
m
x
r
x
cosec
=
r
m
x
dsec
m
x
r
x
sec
24
=
r
m
x
dcot
m
x
r
x
cot
where `d' denotes for dominating function.
These functions will be called as "dominating sine, dominating cosine, dominating tangent,
dominating cosecant, dominating secant and dominating cotangent" respectively.
Example:
2
2
1
2
2
1
3
1
3
sin
cos
tan
sin ,
cos ,
tan , .
x
x
x
d
x
d
x
d
x etc
x
x
x
=
=
=
Note: For non-zero m, we should keep into our mind that
r
m
x
dcot
1
r
m
x
dcos
r
m
x
dsin
r
m
x
dtan
Since we have
=
=
r
x
tan
m
x
r
x
cos
m
x
r
x
sin
r
m
x
dcos
r
m
x
dsin
.
Similarly can prove that
r
m
x
dtan
1
r
m
x
dsin
r
m
x
dcos
r
m
x
dcot
r
m
x
dsin
1
r
m
x
dcosec
r
m
x
dcosec
1
r
m
x
dsin
25
r
m
x
dcos
1
r
m
x
dsec
r
m
x
dsec
1
r
m
x
dcos
Relations Between Dominating Trigonometric Functions
1.
m
2
x
1
r
m
x
dcosec
.
r
m
x
dsin
=
For m=0 and r=1, we have
1
ecx
cos
.
x
sin
0
x
1
1
0
x
dcosec
.
1
0
x
dsin
=
=
2.
m
2
x
1
r
m
x
dcos
.
r
m
x
dsec
=
For m=0 and r=1, we have
1
x
cos
.
x
sec
0
x
1
1
0
x
dcos
.
1
0
x
dsec
=
=
3.
m
2
x
1
r
m
x
dcot
.
r
m
x
dtan
=
For m=0 and r=1, we have
1
x
cot
.
x
tan
0
x
1
1
0
x
dcot
.
1
0
x
dtan
=
=
4.
m
2
x
1
r
m
x
2
dcos
r
m
x
2
dsin
=
+
For m=0 and r=1, we have
1
x
2
cos
x
2
sin
0
x
1
1
0
x
2
dcos
1
0
x
2
dsin
=
+
=
+
5.
m
2
x
1
r
m
x
2
dtan
-
r
m
x
2
dsec
=
26
For m=0 and r=1, we have
1
x
2
tan
x
2
sec
0
x
1
1
0
x
2
dtan
1
0
x
2
dsec
=
-
=
-
6.
m
2
x
1
r
m
x
2
dcot
r
m
x
2
dcosec
=
-
For m=0 and r=1, we have
1
x
2
cot
x
2
sec
co
0
x
1
1
0
x
2
dcot
1
0
x
2
dcosec
=
-
=
-
7.
( )
( )
r
b
r
a
x
dcos
.
x
2dsin
m
x
r
x
cos
.
r
x
2sin
m
x
r
2x
sin
r
m
2x
dsin
=
=
=
, where a+b=m.
For m=0 and r=1, we have
( )
[
]
( ) ( )
cosx
.
2sinx
x
cos
.
x
2sin
0
x
1
2x
sin
x
2
sin
1
0
2x
dsin
1
1
=
=
=
=
8.
( )
( )
r
2
r
2
m/2
m/2
x
sin
d
x
cos
d
m
x
r
x
2
sin
r
x
2
cos
m
x
r
2x
cos
r
m
2x
dcos
-
=
-
=
=
For m=0 and r=1, we have
( )
[
]
( )
( )
x
sin
x
cos
x
sin
x
cos
0
x
1
2x
cos
x
2
cos
1
0
2x
dcos
2
2
1
2
1
2
-
=
-
=
=
=
9.
-
=
-
=
=
r
x
2
tan
1
r
m
x
2tan
r
x
2
tan
1
m
x
r
x
2tan
m
x
r
2x
tan
r
m
2x
dtan
For m=0 and r=1, we have
( )
[
]
x
2
tan
1
2tanx
1
x
2
tan
1
0
x
1
x
2tan
0
x
1
2x
tan
x
2
tan
1
0
2x
dtan
-
=
-
=
=
=
10.
-
=
r
m/3
x
3
4dsin
r
m
x
3dsin
r
m
3x
dsin
27
For m=0 and r=1, we have
( )
[
]
x
3
4sin
3sinx
r
0
x
3
4dsin
1
0
x
3dsin
x
3
sin
1
0
3x
dsin
-
=
-
=
=
11.
-
=
r
m
x
dcos
3
r
m/3
x
3
4dcos
r
m
3x
dcos
For m=0 and r=1, we have
( )
[
]
cosx
3
x
3
4cos
1
0
x
dcos
3
1
0
x
3
4dcos
x
3
cos
1
0
3x
dcos
-
=
-
=
=
12.
( )
( )
( )
( )
( )
(
)
r
2
r
3
/
m
3
r
m
m
r
r
m
x
tan
3
1
x
tan
d
x
dtan
3
x
3x
tan
3x
dtan
-
-
=
=
For m=0 and r=1, we have
( )
( )
[
]
( )
( )
( )
(
)
( )
( )
( )
x
tan
3
1
x
tan
x
3tan
x
tan
3
1
x
tan
d
x
dtan
3
x
3
tan
3x
dtan
2
3
1
2
1
0
3
1
0
1
0
-
-
=
-
-
=
=
Dominating Trigonometric Functions in Terms of Exponential Functions
( )
m
ix
ix
r
m
2ix
e
e
x
dsin
r
r
-
-
=
( )
m
ix
ix
r
m
2x
e
e
x
dcos
r
r
-
+
=
( )
(
)
r
r
r
r
ix
ix
m
ix
ix
r
m
e
e
ix
e
e
x
dtan
-
-
+
-
=
( )
(
)
(
)
r
r
r
r
ix
ix
m
ix
ix
r
m
e
e
x
e
e
i
x
dcot
-
-
-
+
=
( )
(
)
r
r
ix
ix
m
r
m
e
e
x
i
2
x
dcosec
-
-
=
( )
(
)
r
r
ix
ix
m
r
m
e
e
x
2
x
dsec
-
+
=
28
Why Do We Call Dominating Trigonometric Functions ?
For particular values of m and r (as m=0 and r=1), we get the classical trigonometric functions
as follows:
1
0
sin
sin
d
x
x
=
1
0
cos
cos
d
x
x
=
1
0
tan
tan
d
x
x
=
1
0
cot
cot
d
x
x
=
1
0
cos
cos
d
ecx
ecx
=
1
0
sec
sec
d
x
x
=
Dominating Trigonometric Functions with Constant Coefficients
For every constant `k and p' we define and denote theses functions as follows:
sin( )
sin(
)
r
r
m
m
kx
d
kx
x
=
cos( )
cos(
)
r
r
m
m
kx
d
kx
x
=
tan( )
tan(
)
r
r
m
m
kx
d
kx
x
=
cot( )
cot(
)
r
r
m
m
kx
d
kx
x
=
cos ( )
cos (
)
r
r
m
m
ec kx
d
ec kx
x
=
sec( )
sec(
)
r
r
m
m
kx
d
kx
x
=
and
29
sin(
)
sin(
)
r
r
m
m
d
kx
kx
px
p
=
( )
sin
k r
p m
d
x
=
cos(
)
cos( )
r
r
m
m
d
kx
kx
px
p
=
( )
cos
k r
p m
d
x
=
( )
cot(
)
cot( )
cot
r
r
k r
m
p m
m
d
kx
kx
d
x
px
p
=
=
( )
tan(
)
tan(
)
tan
r
r
k r
m
p m
m
d
kx
kx
d
x
px
p
=
=
( )
cos (
)
cos (
)
cos
r
r
k r
m
p m
m
d
ec kx
ec kx
d
ec x
px
p
=
=
( )
s (
)
s ( )
s
r
r
k r
m
p m
m
d ec kx
ec kx
d ec x
px
p
=
=
where k is a constant goes with x
r
as kx
r
and p is another constant goes with x
m
as px
m
.
Example:
( )
2
2
6 2
3
8 3
3
sin(6 )
sin(6 )
sin
8
8
d
x
x
d
x
x
=
=
,
( )
7
7
3 7
3
5 3
3
tan(3 )
tan(3 )
tan
5
5
d
x
x
d
x
x
=
=
Series Expansion of Dominating Trigonometric Functions
By using Taylor's (and Maclaurin's) expansions, we get the following expansion of
dominating trigonometric functions:
(2 1)
3
5
7
0
( 1)
1
1
1
1
sin
...
(2 1)!
1!
3!
5!
7!
n
n
r m
r
r
r
r
r
m
m
m
m
m
n
x
x
x
x
x
d
x
n
x
x
x
x
+
-
=
-
=
=
-
+
-
+
+
(2 )
0
( 1)
cos
(2 )!
n
n r m
r
m
n
x
d
x
n
-
=
-
=
2
4
6
1
1
1
1 ...
2!
4!
6!
r
r
r
m
m
m
m
x
x
x
x
x
x
x
=
-
+
-
+
1 2
2
(2 1)
2
1
( 1) 2 (2
1)
tan
(2 )!
n
n
n
n
r m
r
n
m
n
B x
d
x
n
-
-
-
=
-
-
=
3
5
7
1
2
17 ...
3
15
315
r
r
r
r
m
m
m
m
x
x
x
x
x
x
x
x
=
+
+
+
+
30
1
2 1
(2 1)
2
0
( 1) 2(2
1)
cos
(2 )!
n
n
n
r m
r
n
m
n
B x
d
ecx
n
+
-
-
-
=
-
-
=
3
5
1
1
7
31
...
6
360
15120
r
r
r
r m
m
m
m
x
x
x
x x
x
x
x
=
+
+
+
+
(2 )
2
0
( 1)
sec
(2 )!
n
n r m
r
n
m
n
E x
d
x
n
-
=
-
=
2
4
6
1
1
5
61 ...
2
24
720
r
r
r
m
m
m
m
x
x
x
x
x
x
x
=
+
+
+
+
2
(2 1)
2
0
( 1) 2
cot
(2 )!
n
n
n
r m
r
n
m
n
B x
d
x
n
-
-
=
-
=
3
5
1
1
1
2
...
3
45
945
r
r
r
r m
m
m
m
x
x
x
x x
x
x
x
=
-
-
-
-
where, B
2n
and E
2n
are Bernoulli numbers and Euler numbers respectively.
Derivatives of Dominating Trigonometric Functions
1. For
( )
m
r
r
m
x
)
sin(x
x
dsin
y
=
=
, we have
2m
r
1
m
r
1
r
m
x
)
sin(x
mx
)
cos(x
rx
x
dx
dy
-
-
-
=
( )
( )
( )
r
m
r
m
1
r
r
m
x
dsin
x
m
x
dcos
rx
x
sin
d'
-
=
-
For m=0 and r=1, we have
( )
( )
( )
cosx
x
dsin
x
0
x
dcos
x
dx
d(sinx)
x
sin
d'
1
0
1
0
0
1
0
=
-
=
=
2. For
( )
m
r
r
m
x
)
cos(x
x
dcos
y
=
=
, we have
2m
r
1
m
r
1
r
m
x
)
cos(x
mx
)
(-1)sin(x
rx
x
dx
dy
-
-
-
=
( )
( )
( )
r
m
r
m
1
r
r
m
x
dcos
x
m
x
dsin
rx
x
cos
d'
-
-
=
-
For m=0 and r=1, we have
( )
x
in
s
dcosx
x
0
dsinx
x
dx
d(cosx)
x
cos
d'
1
0
1
0
0
1
0
-
=
-
-
=
=
3. For
( )
m
r
r
m
x
)
tan(x
x
dtan
y
=
=
, we have
31
2m
r
1
m
r
2
1
r
m
x
)
tan(x
mx
)
(x
sec
rx
x
dx
dy
-
-
-
=
( )
( )
( )
r
m
r
m
2
1
r
r
m
x
dtan
x
m
x
dsec
rx
x
tan
d'
-
=
-
For m=0 and r=1, we have
( )
( )
( )
x
sec
x
dtan
x
0
x
dsec
x
dx
d(tanx)
x
tan
d'
2
1
0
1
0
2
0
1
0
=
-
=
=
4. For
( )
m
r
r
m
x
)
cosec(x
x
dcosec
y
=
=
, we have
2m
r
1
m
r
r
1
r
m
x
)
cosec(x
mx
)
cot(x
)
cosec(x
rx
x
-
dx
dy
-
-
-
=
( )
( ) ( )
( )
r
m
r
b
r
a
1
r
r
m
x
dcosec
x
m
x
dcot
x
dcosec
rx
x
cosec
d'
-
-
=
-
For m=0 and r=1, we have
( )
( ) ( )
( )
1
0
1
0
1
0
0
1
0
x
dcosec
x
0
x
dcot
x
dcosec
x
dx
d(cosecx)
x
cosec
d'
-
-
=
=
cotx
.
cosecx
-
=
5. For
( )
m
r
r
m
x
)
sec(x
x
dsec
y
=
=
, we have
2m
r
1
m
r
r
1
r
m
x
)
sec(x
mx
)
tan(x
)
sec(x
rx
x
dx
dy
-
-
-
=
( )
( ) ( )
( )
r
m
r
b
r
a
1
r
r
m
x
dsec
x
m
x
dtan
x
dsec
rx
x
sec
d'
-
=
-
For m=0 and r=1, we have
( )
1
0
1
0
1
0
0
1
0
dcosx
x
0
dtanx
dsecx
x
dx
d(secx)
x
sec
d'
-
=
=
x tanx
ec
s
=
6. For
( )
m
r
r
m
x
)
cot(x
x
dcot
y
=
=
, we have
2m
r
1
m
r
2
1
r
m
x
)
cot(x
mx
)
(x
cosec
rx
x
-
dx
dy
-
-
-
=
( )
( )
( )
r
m
r
m/2
2
1
r
r
m
x
dcot
x
m
x
dcosec
rx
x
cot
d'
-
-
=
-
32
For m=0 and r=1, we have
( )
( )
( )
x
sec
co
x
dtan
x
0
x
dcosec
x
dx
d(cotx)
x
cot
d'
2
1
0
1
0
2
0
1
0
-
=
-
-
=
=
Dominating Hyperbolic Functions
Let
,rm R
be any real numbers. Then the six dominating hyperbolic functions
corresponding to six classical hyperbolic functions can be defined and denoted as follows:
=
r
m
x
dsinh
m
x
r
x
sinh
=
r
m
x
dcosh
m
x
r
x
cosh
=
r
m
x
dtanh
m
x
r
x
tanh
=
r
m
x
dcosech
m
x
r
x
cosech
=
r
m
x
dsech
m
x
r
x
sech
=
r
m
x
dcoth
m
x
r
x
coth
where d denotes for dominating function.
Example:
2
2
3
3
sinh
sinh
x
d
x
x
=
,
1
1
cosh
cosh
x d
x
x
=
, etc.
Note: For non-zero m, we should keep into our mind that
33
r
m
x
dcoth
1
r
m
x
dcosh
r
m
x
dsinh
r
m
x
dtanh
Since we have
=
=
r
x
tanh
m
x
r
x
cosh
m
x
r
x
sinh
r
m
x
dcosh
r
m
x
dsinh
Similarly we have
r
m
x
dtanh
1
r
m
x
dsinh
r
m
x
dcosh
r
m
x
dcoth
r
m
x
dsinh
1
r
m
x
dcosech
r
m
x
dcosech
1
r
m
x
dsinh
r
m
x
dcosh
1
r
m
x
dsech
r
m
x
dsech
1
r
m
x
dcosh
Relations Between Dominating Hyperbolic Functions
1.
m
2
x
1
r
m
x
dcosech
.
r
m
x
dsinh
=
For m=0 and r=1, we have
34
1
echx
cos
.
x
sinh
0
x
1
1
0
x
dcosech
.
1
0
x
dsinh
=
=
2.
m
2
x
1
r
m
x
dcosh
.
r
m
x
dsech
=
For m=0 and r=1, we have
1
x
cosh
.
hx
sec
0
x
1
1
0
x
dcosh
.
1
0
x
dsech
=
=
3.
m
2
x
1
r
m
x
dcoth
.
r
m
x
dtanh
=
For m=0 and r=1, we have
1
x
coth
.
x
tanh
0
x
1
1
0
x
dcoth
.
1
0
x
dtanh
=
=
4.
m
2
x
1
r
m
x
2
dsinh
r
m
x
2
dcosh
=
-
For m=0 and r=1, we have
1
x
2
sinh
x
2
cosh
0
x
1
1
0
x
2
dsinh
1
0
x
2
dcosh
=
-
=
-
5.
m
2
x
1
r
m
x
2
dtanh
r
m
x
2
dsech
=
+
For m=0 and r=1, we have
1
x
2
tanh
x
2
sech
0
x
1
1
0
x
2
dtanh
1
0
x
2
dsech
=
+
=
+
6.
m
2
x
1
r
m
x
2
dcosech
r
m
x
2
dcoth
-
=
-
For m=0 and r=1, we have
1
x
2
cosech
x
2
th
co
0
x
1
1
0
x
2
dcosech
1
0
x
2
dcoth
-
=
-
-
=
-
35
7.
( )
( )
r
b
r
a
x
dcosh
.
x
2dsinh
m
x
r
x
cosh
.
r
x
2sinh
m
x
r
2x
sinh
r
m
2x
dsinh
=
=
=
For m=0 and r=1, we have
( )
[
]
( )
( )
coshx
.
2sinhx
x
cosh
.
x
2sinh
0
x
1
2x
sinh
x
2
sinh
1
0
2x
dsinh
1
1
=
=
=
=
8.
( )
( )
r
2
r
2
m/2
m/2
x
sinh
d
x
cosh
d
m
x
r
x
2
sinh
r
x
2
cosh
m
x
r
2x
cosh
r
m
2x
dcosh
+
=
+
=
=
For m=0 and r=1, we have
( )
[
]
( )
( )
x
sinh
x
cosh
x
sinh
x
cosh
0
x
1
2x
cosh
x
2
cosh
1
0
2x
dcosh
2
2
1
2
1
2
+
=
+
=
=
=
9.
+
=
+
=
=
r
x
2
tanh
1
r
m
x
dtanh
2
r
x
2
tanh
1
m
x
r
x
2tanh
m
x
r
2x
tanh
r
m
2x
dtanh
For m=0 and r=1, we have
( )
[
]
x
2
tanh
1
2tanhx
1
x
2
tanh
1
0
x
1
x
2tanh
0
x
1
2x
tanh
x
2
tanh
1
0
2x
dtanh
+
=
+
=
=
=
10.
( )
( )
( )
( )
( )
( )
r
m/3
3
r
m
m
r
3
r
m
r
r
m
x
4dsinh
x
3dsinh
x
x
4sinh
x
3sinh
x
3x
sinh
3x
dsinh
+
=
+
=
=
For m=0 and r=1, we have
( )
( )
[
]
( )
( )
( )
x
4sinh
3sinhx
x
4sinh
x
3sinh
x
3x
sinh
x
3
sinh
3x
dsinh
3
1
3
1
0
1
1
0
+
=
+
=
=
=
11.
( )
( )
( )
( )
( )
( )
r
m
r
m/3
3
m
r
r
3
m
r
r
m
x
3dcosh
x
4dcosh
x
x
3cosh
x
4cosh
x
3x
cosh
3x
dcosh
-
=
-
=
=
For m=0 and r=1, we have
36
( )
( )
[
]
( )
( )
( )
3coshx
x
4cosh
x
x
3cosh
x
4cosh
x
3x
cosh
x
3
cosh
3x
dcosh
3
0
1
1
3
0
1
1
0
-
=
-
=
=
=
12.
( )
( )
( )
( )
( )
(
)
( )
( )
( )
(
)
r
2
r
3
/
m
3
r
m
r
2
m
r
3
r
m
r
r
m
x
tanh
3
1
x
tanh
d
x
dtanh
3
x
tanh
3
1
x
x
tanh
x
3tanh
x
3x
tanh
3x
dtanh
+
+
=
+
+
=
=
For m=0 and r=1, we have
( )
( )
[
]
( )
( )
( )
(
)
( )
( )
( )
(
)
x
tanh
3
1
x
tanh
x
3tanh
x
tanh
3
1
x
tanh
d
x
dtanh
3
x
3
tanh
3x
dtanh
2
3
1
2
1
0
3
1
0
1
0
+
+
=
+
+
=
=
Dominating Hyperbolic Functions in Terms of Exponential Functions
( )
m
x
x
r
m
2x
e
e
x
dsinh
r
r
-
-
=
( )
m
x
x
r
m
2x
e
e
x
dcosh
r
r
-
+
=
( )
(
)
r
r
r
r
x
x
m
x
x
r
m
e
e
x
e
e
x
dtanh
-
-
+
-
=
( )
(
)
(
)
r
r
r
r
x
x
m
x
x
r
m
e
e
x
e
e
x
dcoth
-
-
-
+
=
( )
(
)
r
r
x
x
m
r
m
e
e
x
2
x
dcosech
-
-
=
( )
(
)
r
r
x
x
m
r
m
e
e
x
2
x
dsech
-
+
=
Why Do We Call Dominating Hyperbolic Functions ?
For m=0 and r=1, we get the classical hyperbolic functions as follows:
1
0
sinh
sinh
d
x
x
=
1
0
cosh
cosh
d
x
x
=
1
0
tanh
tanh
d
x
x
=
37
1
0
coth
coth
d
x
x
=
1
0
cos
cos
d
echx
echx
=
1
0
sec
sec
d
hx
hx
=
Dominating Hyperbolic Functions with Constant Coefficient
For every constant `k' we define and denote it as follows:
sinh( )
sinh(
)
r
r
m
m
kx
d
kx
x
=
cosh( )
cosh(
)
r
r
m
m
kx
d
kx
x
=
tanh( )
tanh(
)
r
r
m
m
kx
d
kx
x
=
coth( )
coth(
)
r
r
m
m
kx
d
kx
x
=
cos
( )
cos
(
)
r
r
m
m
ech kx
d
ech kx
x
=
sec ( )
sec (
)
r
r
m
m
h kx
d
h kx
x
=
and for two constants k and p, we denote them as follows:
sinh(
)
sinh(
)
r
r
m
m
d
kx
kx
px
p
=
( )
sinh
k r
p m
d
x
=
cosh(
)
cosh( )
r
r
m
m
d
kx
kx
px
p
=
( )
cosh
k r
p m
d
x
=
( )
tanh(
)
tanh(
)
tanh
r
r
k r
m
p m
m
d
kx
kx
d
x
px
p
=
=
( )
coth(
)
coth(
)
coth
r
r
k r
m
p m
m
d
kx
kx
d
x
px
p
=
=
38
( )
cos h(
)
cos h(
)
cos h
r
r
k r
m
p m
m
d
ec kx
ec kx
d
ec
x
px
p
=
=
( )
s h(
)
s h(
)
s h
r
r
k r
m
p m
m
d ec kx
ec kx
d ec
x
px
p
=
=
Example:
( )
1
5 1
3
9 3
3
sinh(5 )
sinh(5 )
sinh
9
9
d
x
x
d
x
x
=
=
,
( )
4
4
12 4
3
9 3
3
tanh(12 )
tanh(12 )
tanh
9
9
d
x
x
d
x
x
=
=
Series Expansion of Dominating Hyperbolic Functions
We get the following series by applying Taylor's (and Maclaurin's) series on dominating
hyperbolic functions:
(2 1)
0
sinh
(2 1)!
n
r m
r
m
n
x
d
x
n
+
-
=
=
+
3
5
7
1
1
1 ...
3!
5!
7!
r
r
r
r
m
m
m
m
x
x
x
x
x
x
x
x
=
+
+
+
+
2
0
cosh
(2 )!
nr m
r
m
n
x
d
x
n
-
=
=
2
4
6
1
1
1
1 ...
2!
4!
6!
r
r
r
m
m
m
m
x
x
x
x
x
x
x
=
+
+
+
+
1 2
2
(2 1)
2
1
( 1) 2 (2
1)
tanh
(2 )!
n
n
n
n
r m
n
r
m
n
B x
d
x
n
-
-
-
=
-
-
=
, where
2n
B is modulus of
2n
B
3
5
7
1
2
17 ...
3
15
315
r
r
r
r
m
m
m
m
x
x
x
x
x
x
x
x
=
-
+
-
+
2 1
(2 1)
2
0
( 1) 2 (2
1)
cos
(2 )!
n
n
n
r m
n
r
m
n
B x
d
echx
n
-
-
-
=
-
-
=
3
1
1
7
...
6
360
r
r
r m
m
m
x
x
x x
x
x
=
-
+
-
2
2
0
( 1)
sec
(2 )!
n
nr m
n
r
m
n
E x
d
hx
n
-
=
-
=
2
4
6
1
1
5
61 ...
2
24
720
r
r
r
m
m
m
m
x
x
x
x
x
x
x
=
-
+
-
+
2
(2 1)
2
1
1
2
1
coth
(2 )!
n
n
r m
r
n
m
m
n
B x
d
x
x
n
-
-
+
=
=
+
3
5
1
1
1
2
...
3
45
945
r
r
r
r m
m
m
m
x
x
x
x x
x
x
x
=
+
-
+
+
where B
2n
and E
2n
are Bernoulli numbers and Euler numbers respectively.
39
Derivatives of Dominating Hyperbolic Functions
1. For
( )
m
r
r
m
x
)
sinh(x
x
dsinh
y
=
=
, we have
2m
r
1
m
r
1
r
m
x
)
sinh(x
mx
)
cosh(x
rx
x
dx
dy
-
-
-
=
( )
( )
( )
r
m
r
m
1
r
r
m
x
dsinh
x
m
x
dcosh
rx
x
sinh
d'
-
=
-
For m=0 and r=1, we have
( )
( )
( )
coshx
x
dsinh
x
0
x
dcosh
x
dx
d(sinhx)
x
sinh
d'
1
0
1
0
0
1
0
=
-
=
=
2. For
( )
m
r
r
m
x
)
cosh(x
x
dcosh
y
=
=
, we have
2m
r
1
m
r
1
r
m
x
)
cosh(x
mx
)
sinh(x
rx
x
dx
dy
-
-
-
=
( )
( )
( )
r
m
r
m
1
r
r
m
x
dcosh
x
m
x
dsinh
rx
x
cosh
d'
-
=
-
For m=0 and r=1, we have
( )
x
inh
s
dcoshx
x
0
dsinhx
x
dx
d(coshx)
x
cosh
d'
1
0
1
0
0
1
0
=
-
=
=
3. For
( )
m
r
r
m
x
)
tanh(x
x
dtanh
y
=
=
, we have
2m
r
1
m
r
2
1
r
m
x
)
tanh(x
mx
)
(x
sech
rx
x
dx
dy
-
-
-
=
( )
( )
( )
r
m
r
m/2
2
1
r
r
m
x
dtanh
x
m
x
dsech
rx
x
tanh
d'
-
=
-
For m=0 and r=1, we have
( )
( )
( )
x
sech
x
dtanh
x
0
x
dsech
x
dx
d(tanhx)
x
tanh
d'
2
1
0
1
0
2
0
1
0
=
-
=
=
4. For
( )
m
r
r
m
x
)
cosech(x
x
dcosech
y
=
=
, we have
2m
r
1
m
r
r
1
r
m
x
)
cosech(x
mx
)
coth(x
)
cosech(x
rx
x
-
dx
dy
-
-
-
=
40
( )
( )
( )
( )
r
m
r
b
r
a
1
r
r
m
x
dcosech
x
m
x
dcoth
x
dcosech
rx
x
cosech
d'
-
-
=
-
For m=0 and r=1, we have
( )
( )
( )
( )
1
0
1
0
1
0
0
1
0
x
dcosech
x
0
x
dcoth
x
dcosech
x
dx
d(cosechx)
x
cosech
d'
-
-
=
=
cothx
.
cosechx
-
=
5. For
( )
m
r
r
m
x
)
sech(x
x
dsech
y
=
=
, we have
2m
r
1
m
r
r
1
r
m
x
)
sech(x
mx
)
tanh(x
)
sech(x
rx
x
-
dx
dy
-
-
-
=
( )
( )
( )
( )
r
m
r
b
r
a
1
r
r
m
x
dsech
x
m
x
dtanh
x
dsech
rx
x
sech
d'
-
-
=
-
For m=0 and r=1, we have
( )
1
0
1
0
1
0
0
1
0
dcoshx
x
0
dtanhx
dsechx
x
dx
d(sechx)
x
sech
d'
-
-
=
=
x tanhx
ech
s
-
=
6. For
( )
m
r
r
m
x
)
coth(x
x
dcoth
y
=
=
, we have
2m
r
1
m
r
2
1
r
m
x
)
coth(x
mx
)
(x
cosech
rx
x
-
dx
dy
-
-
-
=
( )
( )
( )
r
m
r
m/2
2
1
r
r
m
x
dcoth
x
m
x
dcosech
rx
x
coth
d'
-
-
=
-
For m=0 and r=1, we have
( )
( )
( )
x
sech
co
x
dtanh
x
0
x
dcosech
x
dx
d(cothx)
x
coth
d'
2
1
0
1
0
2
0
1
0
-
=
-
-
=
=
Relations Between Dominating Trigonometric, Dominating Hyperbolic and
Trigonometric Functions
1.
( )
( )
( )
( )
( )
r
m
m
x
x
m
ix
i
ix
i
m
r
r
m
x
idsinh
2ix
e
e
2ix
e
e
x
ix
sin
ix
dsin
r
r
r
r
=
-
=
-
=
=
-
-
41
For m=0 and r=1, we have
( )
[
]
( )
( )
( )
( )
x
isinh
2i
e
e
2i
e
e
x
ix
sin
)
ix
sin(
ix
dsin
x
x
ix
i
ix
i
0
1
1
0
=
-
=
-
=
=
=
-
-
2.
( )
( )
( )
( )
( )
r
m
m
x
x
m
ix
i
ix
i
m
r
r
m
x
dcosh
2x
e
e
2x
e
e
x
ix
os
c
ix
dcos
r
r
r
r
=
+
=
+
=
=
-
-
For m=0 and r=1, we have
( )
[
]
( )
( )
( )
( )
x
cosh
2
e
e
2
e
e
x
ix
os
c
)
ix
cos(
ix
dcos
x
x
ix
i
ix
i
0
1
1
0
=
+
=
+
=
=
=
-
-
3.
( )
( )
( )
( )
( )
( )
( )
r
m
m
r
r
m
r
r
m
r
r
m
x
dtanh
i
x
x
cosh
x
sinh
i
x
ix
cos
ix
sin
x
ix
tan
ix
dtan
=
=
=
=
For m=0 and r=1, we have
( )
[
]
( )
( )
( )
( )
( )
( )
x
tanh
i
x
cosh
x
sinh
i
ix
cos
ix
sin
x
ix
tan
)
ix
tan(
ix
dtan
0
1
0
=
=
=
=
=
4.
( )
( )
( )
( )
( )
( )
r
m
m
r
m
ix
ix
m
r
r
m
x
dsin
i
x
x
sin
i
2x
e
e
x
ix
sinh
ix
dsinh
r
r
=
=
-
=
=
-
For m=0 and r=1, we have
( )
( )
( )
( )
( )
( )
x
sin
i
1
x
sin
i
2
e
e
x
ix
sinh
)]
ix
sinh(
[
ix
dsinh
ix
ix
0
1
0
=
=
+
=
=
=
-
5.
( )
( )
( )
( )
( )
( )
r
m
m
r
m
ix
ix
m
r
r
m
x
dcos
x
x
cos
2x
e
e
x
ix
cosh
ix
dcosh
r
r
=
=
+
=
=
-
For m=0 and r=1, we have
( )
( )
( )
( )
( )
( )
x
cos
1
x
cos
2
e
e
x
ix
cosh
)]
ix
cosh(
[
ix
dcosh
ix
ix
0
1
0
=
=
+
=
=
=
-
6.
( )
( )
( )
( )
( )
( )
r
m
m
r
r
m
r
m
r
r
m
x
dtan
i
x
x
tan
i
ix
cosh
2x
ix
sinh
x
ix
tanh
ix
dtanh
=
=
=
=
For m=0 and r=1, we have
( )
( )
( )
( )
( )
( )
x
tan
i
1
x
tan
i
ix
cosh
ix
sinh
x
ix
tanh
)]
ix
tanh(
[
ix
dtanh
0
1
0
=
=
=
=
=
42
Dominating Exponential Functions with Base a
By combining the concepts of dominating functions and exponential functions, we get the
following dominating exponential functions with bases `a' and `e' with the same conditions
on `a' as in the classical exponential function:
(i)
(
)
0
0
1
( log )
(log )
!
!
r
r
m
x
r
n
n
x
rn m
m
m
n
n
a
x
a
a
da
x
x
x
n
n
-
=
=
=
=
=
2
2
3
3
1
(log )
(log )
(log )
...
2!
3!
r
r
r
m
m
m
m
x
a x
a x
a
x
x
x
x
=
+
+
+
+
For r=1 and m=0, we have,
1
0
0
(log )
!
n
x
x
n
n
a
da
a
x
n
=
=
=
(ii)
(
)
0
( log )
;
tan
!
r
r
m
kx
n
kx
rn m
m
n
a
k
a
da
x
k cons
t
x
n
-
=
=
=
=
2
2
3
3
1
( log )
( log )
( log )
...
2!
3!
r
r
r
m
m
m
m
x
k
a x
k
a x
k
a
x
x
x
x
=
+
+
+
+
For r=1 and m=0, we have,
1
0
0
( log )
!
n
kx
kx
n
n
k
a
da
a
x
n
=
=
=
(iii)
0
( log )
!
r
k r
p m
kx
n
x
nr m
m
n
a
k
a
da
x
px
n p
-
=
=
=
2
2
3
3
1
( log )
( log )
( log )
...
2!
3!
r
r
r
m
m
m
m
x
k
a
x
k
a
x
k
a
px
px
px
px
=
+
+
+
+
Dominating Exponential Functions with Base e
Changing the base by `e' in the above functions, we get the following functions:
(i)
0
!
r
r
m
x
nr m
x
m
n
e
x
de
x
n
-
=
=
=
2
3
1
1
1
...
2!
3!
r
r
r
m
m
m
m
x
x
x
x
x
x
x
=
+
+
+
+
43
For r=1 and m=0, we have,
1
0
0
!
n
x
x
n
x
de
e
n
=
=
=
(ii)
0
;
tan
!
r
r
m
kx
n nr m
kx
m
n
e
k x
de
k cons
t
x
n
-
=
=
=
=
2
2
3
3
1
...
2!
3!
r
r
r
m
m
m
m
kx
k x
k x
x
x
x
x
=
+
+
+
+
For r=1 and m=0, we have,
1
0
kx
kx
de
e
=
(iii)
cos
sin
r
m
ix
r
r
m
m
de
d
x
id
x
=
+
For m=0, we have,
0
cos
sin
cos
sin
r
r
ix
r
r
r
r
ix
de
d
x id
x
x i
x
e
=
+
=
+
=
For r=1 and m=0, we have,
1
0
cos
sin
cos
sin
ix
ix
de
d
x id
x
x i
x e
=
+
=
+
=
(iv)
cos
sin
r
m
ix
r
r
m
m
de
d
x
id
x
-
=
-
For m=0, we have,
0
cos
sin
cos
sin
r
r
ix
r
r
r
r
ix
de
d
x id
x
x i
x
e
-
-
=
-
=
-
=
For r=1 and m=0, we have,
1
0
cos
sin
cos
sin
ix
ix
de
d
x id
x
x i
x e
-
-
=
-
=
-
=
(v)
0
!
r
k r
p m
kx
n
x
nr m
m
n
e
k
de
x
px
n p
-
=
=
=
2
2
3
3
1
...
2!
3!
r
r
r
m
m
m
m
kx
k x
k x
px
px
x
px
=
+
+
+
+
Derivatives of Dominating Exponential Functions
1. For
m
)
a
(ln
x
m
x
x
x
e
x
a
da
y
r
r
r
m
=
=
=
, we have
m
2
)
a
(ln
x
1
m
1
r
)
a
(ln
x
m
x
e
mx
a
ln
rx
e
x
dx
dy
r
r
-
-
-
=
-
=
-
-
m
2
1
m
1
r
m
)
a
(ln
x
x
mx
a
ln
rx
x
e
r
-
=
+
-
1
m
m
1
r
x
x
m
x
x
)
a
ln
r
(
a
r
44
For m=0 and r=1, we have
( )
a
ln
a
x
0
x
x
)
a
(ln
a
dx
)
a
(
d
a
'
d
x
1
0
0
x
x
x
1
1
0
=
-
=
=
2. For
m
x
x
x
e
de
y
r
r
m
=
=
, we have
m
2
x
1
m
1
r
x
m
x
e
mx
rx
e
x
dx
dy
r
r
-
-
-
=
-
=
-
-
m
2
1
m
1
r
m
x
x
mx
rx
x
e
r
-
=
+
-
1
m
m
1
r
x
x
m
x
x
r
e
r
For m=0 and r=1, we have
( )
x
1
0
0
x
x
x
e
x
0
x
x
e
dx
)
e
(
d
e
'
d
1
1
0
=
-
=
=
Dominating Logarithmic Functions
Applying the concepts of dominating functions on logarithmic functions, we get the
following dominating logarithmic functions
(i)
log
log
r
r
m
m
x
d
x
x
=
(ii)
( )
( )
log
log
;
r
r
m
m
kx
d
kx
k constant
x
=
=
(iii)
(
)
(
)
1
1
log 1
log 1
( 1)
r
nr m
r
n
m
m
n
x
x
d
x
x
n
-
+
=
+
+
=
=
-
2
3
4
1
1
1
...
2
3
4
r
r
r
r
m
m
m
m
x
x
x
x
x
x
x
x
=
-
+
-
+
(iv)
(
)
(
)
1
log 1
log 1
( 1)
r
nr m
r
m
m
n
x
x
d
x
x
n
-
=
-
-
=
=
-
2
3
4
1
1
1
...
2
3
4
r
r
r
r
m
m
m
m
x
x
x
x
x
x
x
x
= -
-
-
-
-
Derivatives of Dominating Logarithmic Functions
For
( )
m
m
r
r
m
x
)
x
ln(
r
x
)
x
ln(
x
ln
d
y
=
=
=
, we have
1
m
1
m
m
2
1
m
m
x
)
x
ln(
mr
x
r
x
mx
)
x
ln(
r
)
x
/
1
(
rx
dx
dy
+
+
-
-
=
-
=
For m=0 and r=1, we have
( )
x
1
x
0
x
1
dx
)
x
ln(
d
x
ln
'
d
1
0
1
0
1
0
=
-
=
=
+
+
45
Chapter-IV
Dominating Sequential Functions
Introduction
In this chapter some new functions have been introduced originated from the classical
trigonometric, hyperbolic, exponential, logarithmic and all dominating functions by taking
their inner product with an arbitrary sequence. The functions originated from the classical
functions have been called sequential functions and those originated from the dominating
functions have been called dominating sequential functions. In developing these functions
we have expressed them in series in the -vectors form by taking each term of the series in
a sequence and then applied the concept of inner product of two n-vectors. Thereafter we
have given them particular symbols according to their classical symbols.
Sequential Trigonometric Functions
Let us consider a sequence
n
n
a
b
and the sine function with their series expansion as
3
1
2
1
2
3
1
, , ,..., ,.........
n
n
n
n
n
a
a
a
a a
b
b b b
b
=
3
1
1
2
1
2
3
1
0
, , ,...,
,...
n
n
n
a
a
a a
b b b
b
+
+
=
and
3
5
3
5
2 1
0
( 1)
sin
...
,
, ,...,
,...
3! 5!
3! 5!
(2 1)!
n
n
n
x
x
x x
x
x x
x
n
+
-
= -
+
-
=
-
+
Taking their inner product as
.sin
n
n
a
x
b
=
3
1
1
2
1
2
3
1
, , ,...,
,...
n
n
a
a
a a
b b b
b
+
+
.
3
5
2 1
( 1)
,
, ,...,
,...
3! 5!
(2 1)!
n
n
x x
x
x
n
+
-
-
+
3
5
2 1
3
1
1
2
1
2
3
1
( 1)
...
...
3!
5!
(2 1)!
n
n
n
n
a
a
a
a x
x
x
x
b
b
b
b
n
+
+
+
-
=
-
+
- +
+
+
2 1
1
0
1
( 1)
(2 1)!
n
n
n
n
n
a
x
b
n
+
+
=
+
=
-
+
1
1
sin
n
n
a
x
b
+
+
=
46
Similarly we can write
)!
1
n
2
(
x
b
a
)
1
(
x
b
a
sin
1
n
2
n
n
1
n
1
n
n
n
-
-
=
-
=
-
.
Note: In symbol we have taken
1
1
n
n
a
b
+
+
because in it n0. We should keep into our mind that
while writing the sequential form of a series or function, we have to adjust the nth term of
the sequence according as the lower limit of n in the summation sign. If the lower limit is 0,
we have to take the (n+1)
th
term of the sequence in the series of inner product and if it is 1
then we take the nth term.
Thus by taking the inner product of a sequence and the terms of different trigonometric
functions, we get the following functions named as
Sequential Trigonometric Functions:
(i)
2 1
3
5
1
1
3
1
2
0
1
1
1
2
3
sin
( 1)
...
(2 1)!
3!
5!
n
n
n
n
n
n
n
a
a
a
a
a
x
x
x
x
x
b
b
n
b
b
b
+
+
+
=
+
+
=
-
=
-
+
-
+
3
5
3
1
2
1
2
3
, , ,...
,
, ,...
3! 5!
a
a a
x x
x
b b b
-
=
Similarly we have
)!
1
n
2
(
x
b
a
)
1
(
x
b
a
sin
1
n
2
n
n
1
n
1
n
n
n
-
-
=
-
=
-
(ii)
2
2
4
1
1
3
1
2
0
1
1
1
2
3
cos
( 1)
...
(2 )!
2!
4!
n
n
n
n
n
n
n
a
a
a
a a
x
x
x
x
b
b
n
b b
b
+
+
=
+
+
=
-
=
-
+
-
2
4
3
1
2
1
2
3
, , ,... 1,
, ,...
2! 4!
a
a a
x x
b b b
-
=
Similarly we have
)!
2
n
2
(
x
b
a
)
1
(
x
b
a
cos
2
n
2
n
n
1
n
1
n
n
n
-
-
=
-
=
-
47
(iii)
2
2
2 1
3
1
5
2
3
1
2
1
1
2
3
2 (2
1)
2
tan
( 1)
...
(2 )!
3
15
n
n
n
n
n
n
n
n
n
n
a
a
B x
a
a
a x
x
x
x
b
b
n
b
b
b
-
-
=
-
=
-
=
+
+
+
3
5
3
1
2
1
2
3
2
, , ,...
, ,
,...
3 15
a
a a
x
x
x
b b b
=
Similarly we have
(
)
)!
2
n
2
(
x
B
1
2
2
b
a
)
1
(
x
b
a
tan
1
n
2
2
n
2
2
n
2
2
n
2
1
n
1
n
0
n
n
n
n
+
-
-
=
+
+
+
+
+
+
=
(iv)
2
1
1
2
0
1
1
( 1)
sec
(2 )!
n
n
n
n
n
n
n
n
a
a
E x
x
b
b
n
+
+
=
+
+
-
=
2
4
6
3
1
2
4
1
2
3
4
1
5
61
...
2
24
720
a
a a
a
x
x
x
b b
b
b
=
+
+
+
+
2
4
6
3
1
2
4
1
2
3
4
1
5
61
, , , ,... 1,
,
,
,...
2
24
720
a
a a
a
x
x
x
b b b b
=
Similarly we have
)!
2
n
2
(
x
E
b
a
)
1
(
x
b
a
sec
2
n
2
2
n
2
n
n
1
n
1
n
n
n
-
-
=
-
-
=
-
(v)
1
2 1
2 1
1
1
2
0
1
1
( 1) 2(2
1)
cos
(2 )!
n
n
n
n
n
n
n
n
n
a
a
B x
ec
x
b
b
n
+
-
-
+
+
=
+
+
-
-
=
3
5
3
1
2
4
1
2
3
4
1
1
7
31
...
6
360
15120
a
a
a
a
x
x
x
b x b
b
b
=
+
+
+
+
3
5
3
1
2
4
1
2
3
4
1 1
7
31
, , , ,...
,
,
,
,...
6 360
15120
a
a a
a
x
x
x
b b b b
x
=
Similarly we have
(
)
)!
2
n
2
(
x
B
1
2
2
)
1
(
b
a
x
b
a
ec
cos
3
n
2
2
n
2
3
n
2
n
1
n
n
n
n
n
-
-
-
=
-
-
-
=
(vi)
2
2 1
1
1
2
0
1
1
( 1) 2
cot
(2 )!
n
n
n
n
n
n
n
n
n
a
a
B x
x
b
b
n
-
+
+
=
+
+
-
=
3
5
3
1
2
4
1
2
3
4
1
1
1
2
...
3
45
945
a
a
a
a
x
x
x
b x b
b
b
=
-
-
-
-
48
3
5
3
1
2
4
1
2
3
4
1 1
1
2
, , , ,...
,
,
,
,...
3
45
945
a
a a
a
x
x
x
b b b b
x
-
-
-
=
Similarly we have
)!
2
n
2
(
x
B
2
)
1
(
b
a
x
b
a
cot
3
n
2
2
n
2
2
n
2
1
n
1
n
n
n
n
n
-
-
=
-
-
-
-
=
Examples:
(i)
3
5
2
1
2
5
1
... sin
1
2 3! 3 5!
1
x
x
n
x
x
n
+
-
+
-
=
+
2
n
2
1
n
1
n
2
1
n
2
n
0
n
2
x
)
1
(
n
1
)
1
n
(
x
)
1
(
1
n
1
n
-
-
=
+
=
-
+
-
=
-
+
+
=
(ii)
2
4
3
2 3
4
2
... cos
1 2 2! 9 4!
1
x
x
n
x
n
+
-
+
-
=
+
)!
2
n
2
(
x
)
1
(
1
)
1
n
(
1
n
)!
n
2
(
x
)
1
(
1
n
2
n
2
n
2
1
n
1
n
3
n
2
n
0
n
3
-
-
+
-
+
=
-
+
+
=
-
-
=
=
Similarly we have
3
2
5
1
4
9 2
... tan
3
5 3 7 15
2 1
x
n
x
x
x
n
+
+
+
=
+
(
)
(
)
)!
n
2
(
x
B
1
2
2
)
1
(
1
n
2
n
)!
2
n
2
(
x
B
1
2
2
)
1
(
3
n
2
)
1
n
(
1
n
2
n
2
n
2
n
2
1
n
1
n
2
1
n
2
2
n
2
2
n
2
2
n
2
n
0
n
2
-
-
=
+
+
+
+
=
-
-
+
=
+
-
-
+
+
=
Dominating Sequential Trigonometric Functions
Combining the concepts of dominating functions and sequential trigonometric functions, we
get dominating sequential trigonometric functions defined and denoted as follows:
(i)
1
(2 1)
1
1
1
0
1
1
sin
( 1)
sin
(2 1)!
r
n
n
n
r m
n
r
n
n
m
m
n
n
n
a
x
b
a
a
x
d
x
b
x
b
n
+
+
-
+
+
+
=
+
+
-
=
=
+
)!
1
n
2
(
x
)
1
(
b
a
m
r
)
1
n
2
(
1
n
1
n
n
n
-
-
=
-
-
-
=
49
(ii)
1
2
1
1
1
0
1
1
cos
( 1)
cos
(2 )!
r
n
n
nr m
n
r
n
n
m
m
n
n
n
a
x
b
a
a
x
d
x
b
x
b
n
+
-
+
+
+
=
+
+
-
=
=
)!
2
n
2
(
x
)
1
(
b
a
m
r
)
2
n
2
(
1
n
1
n
n
n
-
-
=
-
-
-
=
(iii)
1 2
2
(2 1)
2
1
tan
( 1) 2 (2
1)
tan
(2 )!
r
n
n
n
n
n
r m
n
r
n
n
n
m
m
n
n
n
a x
b
a
a
B x
d
x
b
x
b
n
-
-
-
=
-
-
=
=
)!
2
n
2
(
x
B
)
1
2
(
2
)
1
(
b
a
m
r
)
1
n
2
(
2
n
2
2
n
2
2
n
2
n
0
n
1
n
1
n
+
-
-
=
-
+
+
+
+
=
+
+
(iv)
1
2
1
1
1
2
0
1
1
sec
( 1)
sec
(2 )!
r
n
n
nr m
n
r
n
n
n
m
m
n
n
n
a
x
b
a
a
E x
d
x
b
x
b
n
+
-
+
+
+
=
+
+
-
=
=
)!
2
n
2
(
x
E
)
1
(
b
a
m
r
)
2
n
2
(
2
n
2
1
n
1
n
n
n
-
-
=
-
-
-
-
=
(v)
1
1
2 1
(2 1)
1
1
1
2
0
1
1
cos
( 1) 2(2
1)
cos
(2 )!
r
n
n
n
n
r m
n
r
n
n
n
m
m
n
n
n
a
ec
x
b
a
a
B x
d
ec
x
b
x
b
n
+
+
-
-
-
+
+
+
=
+
+
-
-
=
=
)!
2
n
2
(
x
B
)
1
2
(
2
)
1
(
b
a
m
r
)
3
n
2
(
2
n
2
3
n
2
n
1
n
n
n
-
-
-
=
-
-
-
-
=
(vi)
1
2
(2 1)
1
1
1
2
0
1
1
cot
( 1) 2
cot
(2 )!
r
n
n
n
n
r m
n
r
n
n
n
m
m
n
n
n
a
x
b
a
a
B x
d
x
b
x
b
n
+
-
-
+
+
+
=
+
+
-
=
=
)!
2
n
2
(
x
E
2
)
1
(
b
a
m
r
)
3
n
2
(
2
n
2
2
n
2
n
1
n
n
n
-
-
=
-
-
-
-
=
Example:
2
2
1
3
2
1 2 1 5
1
...
sin
2 3! 3 5!
1
x
n
d
x
x
n
+
-
+
-
=
+
1
3
2
x
n
1
)
1
n
(
sin
d
+
-
=
)!
1
n
2
(
x
)
1
(
n
1
)
1
n
(
)!
1
n
2
(
x
)
1
(
1
n
1
n
m
r
)
1
n
2
(
1
n
1
n
2
m
r
)
1
n
2
(
n
0
n
2
-
-
+
-
=
+
-
+
+
=
-
-
-
=
-
+
=
50
Sequential Hyperbolic Functions
The inner product of an arbitrary sequence and the sequence formed by terms of different
hyperbolic functions in series give the following sequential hyperbolic functions:
(i)
2 1
3
5
3
1
2
1
1
2
3
sinh
...
(2 1)!
3!
5!
n
n
n
n
n
n
a
a
a
a
a
x
x
x
x
x
b
b
n
b
b
b
-
=
=
=
+
+
+
-
)!
1
n
2
(
x
b
a
)
1
n
2
(
0
n
1
n
1
n
+
=
+
=
+
+
3
5
3
1
2
1
2
3
, , ,...
, , ,...
3! 5!
a
a a
x x
x
b b b
=
(ii)
2 2
2
4
3
1
2
1
1
2
3
cosh
...
(2
2)!
2!
4!
n
n
n
n
n
n
a
a
a
a a
x
x
x
x
b
b
n
b b
b
-
=
=
=
+
+
+
-
)!
n
2
(
x
b
a
n
2
0
n
1
n
1
n
=
+
+
=
2
4
3
1
2
1
2
3
, , ,... 1, , ,......
2! 4!
a
a a
x x
b b b
=
(iii)
2
2
2 1
2
1
1
2 (2
1)
tanh
( 1)
(2 )!
-
-
=
-
=
-
n
n
n
n
n
n
n
n
n
n
B x
a
a
x
b
b
n
)!
2
n
2
(
x
B
)
1
2
(
2
)
1
(
b
a
)
1
n
2
(
2
n
2
2
n
2
2
n
2
n
0
n
1
n
1
n
+
-
-
=
+
+
+
+
=
+
+
3
5
7
3
1
2
4
1
2
3
4
2
17
...
3
15
315
=
-
+
-
+
a
a
a
a
x
x
x
x
b
b
b
b
3
5
3
1
2
1
2
3
2
, , ,...
,
,
,...
3 15
a
a a
x
x
x
b b b
-
=
(iv)
2
2
1
1
0
1
1
( 1)
sec
(2 )!
n
n
n
n
n
n
n
n
E x
a
a
h
x
b
b
n
+
+
=
+
+
-
=
)!
2
n
2
(
x
E
)
1
(
b
a
)
2
n
2
(
2
n
2
1
n
1
n
n
n
-
-
=
-
-
-
=
2
4
6
3
1
2
4
1
2
3
4
1
5
61
...
2
24
720
a
a a
a
x
x
x
b b
b
b
=
-
+
-
+
2
4
6
3
1
2
4
1
2
3
4
1
5
61
, , , ,... 1,
,
,
,...
2
24
720
a
a a
a
x
x
x
b b b b
-
-
=
51
(v)
2 1
2 1
2
1
1
0
1
1
( 1) 2 (2
1)
cos
(2 )!
n
n
n
n
n
n
n
n
n
B x
a
a
ech
x
b
b
n
-
-
+
+
=
+
+
-
-
=
)!
2
n
2
(
x
B
)
1
2
(
2
)
1
(
b
a
)
3
n
2
(
2
n
2
3
n
2
1
n
1
n
n
n
-
-
-
=
-
-
-
-
=
3
5
3
1
2
4
1
2
3
4
1
1
7
31
...
6
360
15120
a
a
a
a
x
x
x
b x b
b
b
=
-
+
-
+
3
5
3
1
2
4
1
2
3
4
1 1
7
31
, , , ,...
,
,
,
,...
6 360
15120
a
a a
a
x
x
x
b b b b
x
-
-
=
(vi)
2
2 1
1
2
1
1
1
1
2
1
coth
(2 )!
n
n
n
n
n
n
n
n
a
a
B x
a
x
b
b x
b
n
-
+
=
+
=
+
)!
n
2
(
x
B
2
b
a
x
1
b
a
)
1
n
2
(
2
n
2
2
n
2
0
n
2
n
2
n
1
1
+
+
+
=
+
+
+
=
3
5
3
1
2
4
1
2
3
4
1
1
1
2
...
3
45
945
a
a
a
a
x
x
x
b x b
b
b
=
+
-
+
+
3
5
3
1
2
4
1
2
3
4
1 1
1
2
, , , ,...
, ,
,
,...
3 45
945
a
a a
a
x
x
x
b b b b
x
-
=
Example: Taking n=1, 2, 3, ...
3
5
2
1
4
9
... sinh
4
7 3! 10 5!
3 1
x
x
n
x
x
n
+
+
+
=
+
2
4
3
2
9
28
1
1
... cosh
5 2! 10 4!
1
x
x
n
x
n
+
+
+
+
=
+
{
}
2
2
2
2
4
6
2
(1!)
(2!)
(3!)
1
... cosh (( 1)!)
2!
4!
6!
x
x
x
n
x
+
+
+
+
=
-
Dominating Sequential Hyperbolic Functions
By applying the concepts of dominating functions and the sequential hyperbolic functions,
we get the following dominating sequential hyperbolic functions:
52
(i)
(2 1)
1
sinh
sinh
(2 1)!
r
n
n
r m
n
r
n
n
m
m
n
n
n
a x
b
a
a x
d
x
b
x
b
n
-
-
=
=
=
-
)!
1
n
2
(
x
b
a
m
r
)
1
n
2
(
0
n
1
n
1
n
+
=
-
+
=
+
+
(ii)
(2 2)
1
cosh
cosh
(2
2)!
r
n
n
r m
n
r
n
n
m
m
n
n
n
a x
b
a
a x
d
x
b
x
b
n
-
-
=
=
=
-
)!
n
2
(
x
b
a
m
nr
2
0
n
1
n
1
n
-
=
+
+
=
(iii)
1 2
2
(2 1)
2
1
tanh
( 1) 2 (2
1)
tanh
(2 )!
r
n
n
n
n
n
r m
n
n
r
n
n
m
m
n
n
n
a x
b
B x
a
a
d
x
b
x
b
n
-
-
-
=
-
-
=
=
)!
2
n
2
(
x
B
)
1
2
(
2
)
1
(
b
a
m
r
)
1
n
2
(
2
n
2
2
n
2
2
n
2
n
0
n
1
n
1
n
+
-
-
=
-
+
+
+
+
=
+
+
(iv)
1
2
1
2
1
1
0
1
1
sec
( 1)
sec
(2 )!
r
n
n
nr m
n
n
r
n
n
m
m
n
n
n
a
h
x
b
E x
a
a
d
h
x
b
x
b
n
+
-
+
+
+
=
+
+
-
=
=
)!
2
n
2
(
x
E
)
1
(
b
a
m
r
)
2
n
2
(
2
n
2
1
n
1
n
n
n
-
-
=
-
-
-
-
=
(v)
1
2 1
(2 1)
2
1
1
1
0
1
1
cos
( 1) 2 (2
1)
cos
(2 )!
r
n
n
n
n
r m
n
n
r
n
n
m
m
n
n
n
a
ech
x
B x
b
a
a
d
ech
x
b
x
b
n
+
-
-
-
+
+
+
=
+
+
-
-
=
=
)!
2
n
2
(
x
B
)
1
2
(
2
)
1
(
b
a
m
r
)
3
n
2
(
2
n
2
3
n
2
1
n
1
n
n
n
-
-
-
=
-
-
-
-
-
=
(vi)
2
(2 1)
1
2
1
1
1
1
coth
2
1
coth
(2 )!
r
n
n
n
r m
n
r
n
n
n
m
m
m r
n
n
n
a x
b
a
a
B x
a
d
x
b
x
b x
b
n
-
-
+
+
=
+
=
=
+
)!
2
n
2
(
x
B
2
b
a
x
1
b
a
m
r
)
1
n
2
(
2
n
2
2
n
2
0
n
2
n
2
n
r
m
1
1
+
+
=
-
+
+
+
=
+
+
+
Example: Taking n=1, 2, 3, ...
53
3
2
1
2
1 1 4
9
...
sinh
4
7 3! 10 5!
3 1
x
x
n
d
x
x
n
+
+
+
=
+
2
3
1
2
2
2
1 9 1 28
1
...
cosh
5 2! 10 4!
1
x
n
d
x
x
n
+
+
+
+
=
+
Sequential Exponential Functions
By taking the inner product of an arbitrary sequence and a sequence formed by the terms of
an exponential series, we get the following sequential exponential functions:
(i)
1
1
2
3
1
3
1
2
4
0
1
1
2
3
4
...
!
1!
2!
3!
n
n
a
n
x
b
n
n
n
a
a
a
a
a
x
x
x
x
e
b
n
b
b
b
b
+
+
+
=
+
=
=
+
+
+
+
)!
1
n
(
x
b
a
1
n
1
n
n
n
-
=
-
=
2
3
3
1
2
4
1
2
3
4
, , , ,... 1, , , ,........
1! 2! 3!
a
a a
a
x x x
b b b b
=
(ii)
1
1
1
1
( )
;
tan
n
n
n
n
a
a
k
x
kx
b
b
e
e
k cons
t
+
+
+
+
=
=
(iii)
1
1
1
1
( )
1
2 1
2 2
0
1
2 1
2 2
( )
cos
sin
!
n
n
n
n
a
a
n
i
x
ix
b
b
n
n
n
n
n
n
n
a
a
a
ix
e
e
x i
x
b
n
b
b
+
+
+
+
+
+
+
=
+
+
+
=
=
=
+
)!
1
n
(
)
ix
(
b
a
1
n
1
n
n
n
-
=
-
=
(iv)
1
1
1
1
(
)
1
2 1
2 2
0
1
2 1
2 2
( )
cos
sin
!
n
n
n
n
a
a
n
i
x
ix
b
b
n
n
n
n
n
n
n
a
a
a
ix
e
e
x i
x
b
n
b
b
+
+
+
+
-
-
+
+
+
=
+
+
+
-
=
=
=
-
)!
1
n
(
)
ix
(
b
a
1
n
1
n
n
n
-
-
=
-
=
Example:
2
( !)
( 1)
2
3
4
2
3
4
2!
3!
4!
1
...
2 3
4
5
n
n
x
n
x
x
x
x
e
+
+ +
+
+
+
=
{
}
!(2 1)
2
3
1 3 5
7
...
n n
x
x
x
x
e
+
+
+
+
+
=
{
}
1
2
3
( 1)
2
3
2
3
4
1
...
2! 3!
4!
n
n
x
x
x
x
e
-
+
+
+
+
+
=
54
1
2
3
( 1)!
1
2
3
1 2
3
4
...
2 3
4
5
n
n
n
x
n
x
x
x
e
-
-
+
+
+
+
+
=
Dominating Sequential Exponential Functions
Combining the concepts of dominating functions and the sequential exponential functions,
we get the following dominating sequential exponential functions:
(i)
1
1
1
1
1
0
1
!
r
n
r
n
n
m
n
a
x
a
b
nr m
x
b
n
m
n
n
a
e
x
de
x
b
n
+
+
+
+
-
+
=
+
=
=
)!
1
n
(
x
b
a
m
nr
1
n
n
n
-
=
-
=
(ii)
1
1
1
1
1
0
1
!
r
n
r
n
n
m
n
a
kx
a
b
n nr m
k
x
b
n
m
n
n
a
e
k x
de
x
b
n
+
+
+
+
-
+
=
+
=
=
)!
1
n
(
x
k
b
a
m
nr
m
1
n
n
n
-
=
-
=
(iii)
1
1
2 1
2
2 1
2
cos
sin
r
n
m
n
a
i
x
b
r
r
n
n
m
m
n
n
a
a
de
d
x
id
x
b
b
+
+
-
-
=
+
(iv)
1
1
2 1
2
2 1
2
cos
sin
r
n
m
n
a
i
x
b
r
r
n
n
m
m
n
n
a
a
de
d
x
id
x
b
b
+
+
-
-
-
=
-
Sequential Logarithmic Functions
The inner product of an arbitrary sequence and a sequence formed by the terms of logarithmic
function in series gives sequential logarithmic functions as:
(i)
2
3
1
3
1
2
1
1
2
3
log
(1 )
...
( 1)
1
2
3
n
n
n
n
n
n
n
a
a
a
a
a
x
x
x
x
x
b
b
b
b
b n
+
=
+
=
-
+
-
=
-
1
n
x
)
1
(
b
a
1
n
n
0
n
1
n
1
n
+
-
=
+
=
+
+
2
3
3
1
2
1
2
3
, , ,...
,
, ,...
1
2 3
a
a a
x
x x
b b b
=
-
(ii)
1
log
(1 )
log
(1 )
- = -
-
n
n
n
n
a
a
x
b
b
x
2
3
3
1
2
1
1
2
3
...
( 1)
1
2
3
=
= -
+
+
+
=
-
n
n
n
n
a
a
a
a
x
x
x
x
b
b
b
b n
1
n
x
)
1
(
b
a
1
n
0
n
1
n
1
n
+
-
=
+
=
+
+
55
2
3
3
1
2
1
2
3
, , ,...
,
,
,...
1
2
3
a
a a
x
x
x
b b b
=
-
-
-
Example:
2
3
2
2
4
6
2
... log
(1
)
2 1 5 2 7 3
1
x
x
x
n
x
n
-
+
-
=
+
+
3 2
3 3
3 4
4
4
2
2
2
2
2
2
2
3
4
( 1)
1
( 1)
... log
log
(1 )
1 1 2 1 3 1 4 1
( 1) 1 (1 )
( 1) 1
x
x
x
x
n
n
x
n
x
n
+
+
+
+
+
+
=
= -
-
+
+
+
+
+
+
-
+
+
2
3
3
3.6
3.6.9
3.6.9...(3 )
1
... log
.
7
7.10
7.10.13
7.10.13...(3
4)
(1 )
n
x
x
x
n
n
x
+
+
+
=
+
-
( )
2
3
1
1.2
1.2.3
1
... log
3
3.5
3.5.7
(2 1) (1 )
n n
x
x
x
n
x
+
+
+
=
+
-
2
4
9
2
3
1
2
3
1
... log
2
3
4
1
(1 )
n
n
x
x
x
n
n
x
+
+
+
=
+
-
2
3
4
2
2
2
1
... log
(1
)
2
3
4
x
x
x
x
x
n
-
+
-
+
=
+
Dominating Sequential Logarithmic Functions
Applying the concepts of dominating function on sequential logarithmic functions, we get
the following dominating sequential logarithmic functions:
(i)
1
1
log
(1
)
log
(1
)
( 1)
r
n
nr m
n
r
n
n
n
m
m
n
n
n
a
x
b
a
a x
d
x
b
x
b
n
-
+
=
+
+
=
=
-
1
n
x
)
1
(
b
a
m
nr
n
0
n
1
n
1
n
+
-
=
-
=
+
+
(ii)
(
)
1
log
1
log
(1
)
( 1)
r
n
nr m
n
r
n
n
m
m
n
n
n
a
x
b
a
a x
d
x
b
x
b
n
-
=
-
-
=
=
-
1
n
x
)
1
(
b
a
m
nr
0
n
1
n
1
n
+
-
=
-
=
+
+
Example:
(
)
1
2
2
1 4 1 6
2
...
log
1
5 2 7 3
1
x
n
d
x
x
n
-
+
-
=
+
+
56
Chapter-V
Applications and Indefinite Integrals of
Dominating Sequential Functions
Introduction
The two principal inventors of calculus Newton and Leibniz had very distinct approaches to
integration. Newton allowed for infinite series solutions and evaluated integrals by expanding
functions in power series and integrating term-by-term. In contrast, Leibniz favored solutions
in finite terms and searched for closed form expressions of integrals.
Well into the 18
th
century, mathematicians expressed different preferences (finite vs. infinite series) for
representations of indefinite integrals.
Bernoulli's conjecture, Laplace's theorem and Abel's theorem contain the possible form of
an indefinite integral of a function. Liouville's theorem characterizes the possible form of
elementary anti-derivatives. It asserts also that if an elementary function f(x) is integrable in
elementary terms then there are severe constraints on the possible form of an elementary anti-
derivative of f(x) as has been mentioned in strong Liouville's theorem.
Keeping these concepts into our mind, we can apply the term-by-term integration of a power
series and of a dominating function to generate the general indefinite integrals of different
dominating functions, sequential functions and dominating sequential functions, which will
finally solve the problem of the standard mode of expression of some classical nonintegrable
functions in terms of dominating sequential functions.
Indefinite Integrals of Dominating Trigonometric Functions
In finding the general indefinite integrals of dominating trigonometric functions, we first
express the given functions in terms of particular case of d-function or in power series and
then we integrate each term of the series separately. Finally we denote them in terms of
different dominating sequential functions, if possible, as follows:
57
Indefinite Integral of
r
r
m
m
sin x dx dsinx dx ; m,r R
x
=
n (2n 1)r
m
n 0
1
( 1) x
dx
x
(2n 1)!
+
=
-
=
+
n
(2n 1)r m
n 0
( 1)
x
dx
(2n 1)!
+
-
=
-
=
+
n
(2n 1)r m 1
n 0
n
n
(2n 1)r m 1
m r 1
m r 1
n 0
n
n
2r
2r
( 1)
x
m r 1
k; if n
(2n 1)!{(2n 1) m 1}
2r
( 1)
( 1)
x
m r 1
log x
k; if n
(2n 1)!
(2n 1)!{(2n 1) m 1}
2r
+
- +
=
+
- +
- -
- -
=
=
-
- -
+
+
+ - +
=
-
-
- -
+
+
=
+
+
+ - +
r
m 1
n
r
m 1
m r 1
m r 1
n
n
2r
2r
1
m r 1
dsin
x
k; if n
(2n 1)r m 1
2r
( 1)
1
m r 1
log x
dsin
x
k; if n
(2n 1)!
(2n 1)r m 1
2r
-
-
- -
- -
=
- -
+
+
- +
=
-
- -
+
+
=
+
+
- +
Example: We have sin x dx
x
1
1
dsin x dx
=
[Here m=1, r=1 and
m r 1 1 1 1
1
n
2r
2
2
- -
- -
-
]
r
m 1
1
dsin
x
k
(2n 1)r m 1
-
=
+
+
- +
1
0
1
dsin
x
k
(2n 1) 1 1
=
+
+ - +
1
sin
x k,n 0.
(2n 1)
=
+
+
Example: We have
2
sin x dx
x
1
2
dsin x dx
=
[Here m=2, r=1 and
m r 1 2 1 1
n
0
2r
2
- -
- -
=
=
=
]
58
n
r
m 1
m r 1
m r 1
n
n
2r
2r
( 1)
1
log x
dsin
x
k
(2n 1)!
(2n 1)r m 1
-
- -
- -
=
-
=
+
+
+
+
- +
0
1
1
n 0
n 0
( 1)
1
log x
dsin
x
k
(0 1)!
2n
=
-
=
+
+
+
1
1
n 0
1
log x
dsin
x
k
2n
=
+
+
.
Indefinite Integral of
r
r
m
m
cos x dx dcosx dx
x
=
n 2nr
n
2nr m
m
n 0
n 0
1
( 1) x
( 1)
dx
x
dx
x
(2n)!
(2n)!
-
=
=
-
-
=
=
n
2nr m 1
n 0
n
n
2nr m 1
m 1
m 1
n 0
n
n
2r
2r
( 1)
x
m 1
k; if n
(2n)! {2nr m 1}
2r
( 1)
( 1)
x
m 1
log x
k; if n
(2n)!
(2n)! {2nr m 1}
2r
- +
=
- +
-
-
=
=
-
-
+
- +
=
-
-
-
+
+
=
- +
r
m 1
n
r
m 1
m 1
m 1
n
n
2r
2r
1
m 1
d cos
x
k; if n
2nr m 1
2r
( 1)
1
m 1
log x
d cos
x
k; if n
(2n)!
2nr m 1
2r
-
-
-
-
=
-
+
- +
=
-
-
+
+
=
- +
Example: We have cos x dx
x
1
1
d cos x dx
=
[Here m=1, r=1 and n=0]
n
r
m 1
m 1
m 1
n
n
2r
2r
( 1)
1
log x
d cos
x
k
(2n)!
2nr m 1
-
-
-
=
-
=
+
+
- +
59
0
1
0
n 0
n 0
( 1)
1
log x
d cos
x
k
(0)!
2n
=
-
=
+
+
1
0
n 0
1
log x
d cos
x
k
2n
=
+
+
.
Example: We have
1
2
2
cos x dx dcosx dx
x
=
[Here m=2, r=1 and
1
n
2
]
r
m 1
1
d cos
x
k
2nr m 1
-
=
+
- +
1
1
1
d cos
x k
2n 1
=
+
-
.
Indefinite Integral of
r
r
m
m
tan x dx dtanx dx
x
=
n 1 2n
2n
(2n 1)r
2n
m
n 1
( 1) 2 (2
1)B
1
x
dx
x
(2n)!
-
-
=
-
-
=
n 1 2n
2n
(2n 1)r m
2n
n 1
( 1) 2 (2
1)B
x
dx
(2n)!
-
- -
=
-
-
=
{
}
{
}
n 1 2n
2n
(2n 1)r m 1
2n
n 1
n 1 2n
2n
2n
m r 1
n
2r
n 1 2n
2n
(2n 1)r m 1
2n
m r 1
n 1
n
2r
( 1) 2 (2
1)B
x
m r 1
k;n
(2n)!
(2n 1)r m 1
2r
( 1) 2 (2
1)B logx
(2n)!
( 1) 2 (2
1)B
x
m r 1
k;n
(2n)!
(2n 1)r m 1
2r
-
-
- +
=
-
+ -
=
-
-
- +
+ -
=
-
-
+ -
+
-
- +
-
-
=
+
-
-
+ -
+
=
-
- +
60
r
m 1
n 1 2n
2n
2n
m r 1
n
2r
r
m 1
m r 1
n
2r
1
m r 1
d tan
x
k;n
(2n 1)r m 1
2r
( 1) 2 (2
1)B logx
(2n)!
1
m r 1
d tan
x
k;n
(2n 1)r m 1
2r
-
-
+ -
=
-
+ -
+ -
+
-
- +
-
-
=
+
+ -
+
=
-
- +
Example: We have
1
1
tan x dx dtanx dx
x
=
[Here m=1, r=1 and
1
n
2
]
=
1
0
1
d tan
x
k
(2n 1)
+
-
=
1
d tan
x k
(2n 1)
+
-
=
1
tan
x k
(2n 1)
+
-
.
Example: We have
1
2
2
tan x dx dtanx dx
x
=
[Here m=2, r=1 and n=1]
n 1 2n
2n
1
2n
1
n 1
n 1
( 1) 2 (2
1)B
1
log x
d tan
x
k
(2n)!
(2n 2)
-
=
-
-
=
+
+
-
0 2
2
1
2
1
n 1
( 1) 2 (2 1)B
1
log x
d tan
x
k
(2)!
(2n 2)
-
-
=
+
+
-
[
]
1
2
1
n 1
1
6B log x
d tan
x
k
(2n 2)
=
+
+
-
1
1
n 1
1
log x
d tan
x
k
(2n 2)
=
+
+
-
61
Example: We have
1
1
x tan xdx
d tan x dx
-
=
[Here m=-1, r=1 and
1
n
2
-
]
1
2
1
d tan
x
k
(2n 1)
-
=
+
+
Indefinite Integral of
r
r
m
m
cot x dx dcotx dx
x
=
n 2n
(2n 1)r
2n
m
n 0
( 1) 2 B
1
x
dx
x
(2n)!
-
=
-
=
n 2n
(2n 1)r m
2n
n 0
( 1) 2 B
x
dx
(2n)!
- -
=
-
=
{
}
{
}
n 2n
(2n 1)r m 1
2n
n 0
n 2n
2n
m r 1
n
2r
n 2n
(2n 1)r m 1
2n
m r 1
n 0
n
2r
( 1) 2 B
x
m r 1
k;n
(2n)!
(2n 1)r m 1
2r
( 1) 2 B logx
(2n)!
( 1) 2 B
x
m r 1
k;n
(2n)!
(2n 1)r m 1
2r
-
- +
=
+ -
=
-
- +
+ -
=
-
+ -
+
-
- +
-
=
+
-
+ -
+
=
-
- +
r
m 1
n 2n
r
2n
m 1
m r 1
m r 1
n
n
2r
2r
1
m r 1
d cot
x
k;n
(2n 1)r m 1)
2r
( 1) 2 B
1
log x
d cot
x
(2n)!
(2n 1)r m 1)
m r 1
k;n
2r
-
-
+ -
+ -
=
+ -
+
-
- +
-
=
+
-
- +
+ -
+ =
Example: We have
1
1
cot x dx dcotx dx
x
=
62
[Here m=1, r=1 and
1
n
2
]
=
1
0
1
d cot
x
k
(2n 1)
+
-
=
1
cot
x k
(2n 1)
+
-
.
Example: We have
1
2
2
cot x dx dcotx dx
x
=
[Here m=2, r=1 and n=1]
1 2
1
2
1
n 1
( 1) 2 B
1
log x
d cot
x
k
(2)!
(2n 2)
-
=
+
+
-
1
1
n 1
1
1
log x
d cot
x
k
3
(2n 2)
= -
+
+
-
Indefinite Integral of
r
r
m
m
cosecx dx dcosecx dx
x
=
( 2n 1) r
n 1
2n 1
2n
m
n 0
( 1) 2(2
1)B
1
x
dx
x
(2n)!
-
+
-
=
-
-
=
n 1
2n 1
(2n 1)r m
2n
n 0
( 1) 2(2
1)B
x
dx
(2n)!
+
-
-
-
=
-
-
=
{
}
{
}
n 1
2n 1
(2n 1)r m 1
2n
n 0
n 1
2n 1
2n
m r 1
n
2r
n 1
2n 1
(2n 1)r m 1
2n
m r 1
n 0
n
2r
( 1) 2(2
1)B
x
m r 1
k;n
(2n)!
(2n 1)r m 1
2r
( 1) 2(2
1)B logx
(2n)!
( 1) 2(2
1)B
x
m r 1
k;n
(2n)!
(2n 1)r m 1
2r
+
-
-
- +
=
+
-
+ -
=
+
-
-
- +
+ -
=
-
-
+ -
+
-
- +
-
-
=
+
-
-
+ -
+
=
-
- +
63
r
m 1
n 1
2n 1
2n
m r 1
n
2r
r
m 1
m r 1
n
2r
1
m r 1
d cosec
x
k;n
(2n 1)r m 1
2r
( 1) 2(2
1)B logx
(2n)!
1
m r 1
d cosec
x
k;n
(2n 1)r m 1
2r
-
+
-
+ -
=
-
+ -
+ -
+
-
- +
-
-
=
+
+ -
+
=
-
- +
Example: We have
1
1
cosecxdx dcosecx dx
x
=
[Here m=1, r=1 and
1
n
2
]
1
0
1
d cosec
x
k
(2n 1)
=
+
-
1
cosec
x k
(2n 1)
=
+
-
.
Example: We have
1
2
2
cosecxdx dcosecx dx
x
=
[Here m=2, r=1 and n=1]
2
2 1
1
2
1
n 1
( 1) 2(2
1)B
1
log x
d cosec
x
k
(2)!
(2n 2)
-
-
-
=
+
+
-
1
1
n 1
1
1
log x
d cosec
x
k
6
(2n 2)
=
+
+
-
.
Indefinite Integral of
r
r
m
m
sec x dx dsecx dx
x
=
n
2nr
2n
m
n 0
( 1) E
1
x dx
x
(2n)!
=
-
=
n
2nr m
2n
n 0
( 1) E
x
dx
(2n)!
-
=
-
=
64
{
}
{
}
n
2nr m 1
2n
n 0
n
n
2nr m 1
2n
2n
m 1
m 1
n 0
n
n
2r
2r
( 1) E
x
m 1
k;n
(2n)!
2nr m 1
2r
( 1) E
( 1) E
x
m 1
log x
k;n
(2n)!
(2n)!
2nr m 1
2r
- +
=
- +
-
-
=
=
-
-
+
- +
=
-
-
-
+
+
=
- +
r
m 1
n
r
2n
m 1
m 1
m 1
n
n
2r
2r
1
m 1
dsec
x
k;n
(2nr m 1)
2r
( 1) E
1
m 1
log x
dsec
x
k;n
(2n)!
(2nr m 1)
2r
-
-
-
-
=
-
+
- +
=
-
-
+
+
=
- +
Example: We have
1
1
sec xdx dsecx dx
x
=
[Here m=1, r=1 and n=0]
0
1
0
0
n 0
n 0
( 1) E
1
log x
dsec
x
k
(0)!
2n
=
-
=
+
+
1
0
n 0
1
log x
dsec
x
k
2n
=
+
+
n 0
1
log x
sec
x
k
2n
=
+
+
Example: We have
1
2
2
sec xdx dsecx dx
x
=
[Here m=2, r=1 and
1
n
2
]
1
1
1
dsec
x k
(2n 1)
=
+
-
65
Particular Cases: Putting m=0 and r=1, we get the indefinite integrals of classical
trigonometric functions as follows:
Indefinite Integral of
1
1
0
0
sin x dx dsinx dx
x
=
[Here m=0, r=1 and n-1]
1
1
1
dsin
x
k
(2n 1) 1
-
=
+
+ +
1
1
1
dsin
x
k
(2n 2)
-
=
+
+
1
1
1
sin
x k
x
2n 2
-
=
+
+
1
x sin
x k
2n 2
=
+
+
3
5
1
1 x
1 x
x
x
... k
2
4 3! 6 5!
=
-
+
-
+
2
4
6
x
x
x ... k
2! 4! 6!
=
-
+
-
+
2
4
6
x
x
x
1
... c;c k 1
2! 4! 6!
= - +
-
+
-
+
= +
2
4
6
x
x
x
1
... c
2! 4! 6!
= - -
+
-
+
+
cos x c
= -
+
sin xdx
=
Indefinite Integral of
1
1
0
0
cos x dx dcosx dx
x
=
[Here m=0, r=1 and n
-12
]
66
1
1
1
d cos
x
k
(2n 1)
-
=
+
+
1
1
1
cos
x k
x
2n 1
-
=
+
+
1
x cos
x k
2n 1
=
+
+
2
4
1 1 x
1 x
x 1
... k
1 3 2! 5 4!
=
-
+
-
+
3
5
x
x
x
... k
3! 5!
=
-
+
-
+
sin x k
=
+
cos xdx
=
Indefinite Integral of
1
1
0
0
tan x dx dtanx dx
x
=
[Here m=0, r=1 and n=0]
1
1
1
d tan
x
k
2n
-
=
+
1
1
1
tan
x k
x
2n
-
=
+
1
x tan
x k
2n
=
+
3
5
1
1 1
1 2
x
x
x
x ... k
2
4 3
6 15
=
+
+
+
+
2
4
6
x
1 x
2 x ... k
2 3 4 15 6
=
+
+
+
+
3
5
x 1
2
x
x ... dx
1 3
15
=
+
+
+
tan xdx
=
67
Indefinite Integral of
1
1
0
0
cot x dx dcotx dx
x
=
[Here m=0, r=1 and n=0]
0 0
1
0
1
( 1) 2 B
1
log x d cot
x
k
0!
2n
-
-
=
+
+
1
1
1
log x
cot
x k
x
2n
-
=
+
+
1
log x x cot
x k
2n
=
+
+
3
5
1 1
1 1
1 2
log x x
x
x
x ... k
2 3
4 45
6 945
-
=
+
-
-
-
+
2
4
6
1 1
1 1
2 1
log x
x
x
x ... k
3 2
45 4
945 6
=
-
-
-
-
+
3
5
1 1
1
2
x
x
x ... dx
x 3
45
945
=
-
-
-
-
cot xdx
=
Indefinite Integral of
1
1
0
0
cosecx dx dcosecx dx
x
=
[Here m=0, r=1 and n=0]
1
1
0
1
n 0
n 0
( 1)2(2
1)B
1
log x
d cosec
x
k
0!
2n
-
-
=
-
-
=
+
+
1
n 0
1
1
log x
cosec
x
k
x
2n
-
=
+
+
n 0
1
log x
x cosec
x
k
2n
=
+
+
3
5
1 1
1 7
1 31
log x x
x
x
x ... k
2 6
4 360
6 15120
=
+
+
+
+
+
68
2
4
6
1 1
7 1
31 1
log x
x
x
x ... k
6 2
360 4
15120 6
=
+
+
+
+
+
3
5
1 1
7
31
x
x
x ... dx
x 6
360
15120
=
+
+
+
+
cosecxdx
=
Indefinite Integral of
1
1
0
0
sec x dx dsecx dx
x
=
[Here m=0, r=1 and n
-12
]
1
1
1
dsec
x
k
2n 1
-
=
+
+
1
1
1
sec
x k
x
2n 1
-
=
+
+
1
x sec
x k
2n 1
=
+
+
2
4
1 1
1 5
x 1
x
x ... k
3 2
5 24
=
+
+
+
+
3
5
1 1
5 1
x
x
x ... k
2 3
24 5
= +
+
+
+
2
4
1
5
1
x
x ... dx
2
24
=
+
+
+
sec xdx
=
Indefinite Integrals of Dominating Hyperbolic Functions
In finding the general indefinite integrals of dominating hyperbolic functions, we first express
the given functions in terms of particular case of d-function or in power series and then we
integrate each term of the series separately. Finally we denote them in terms of different
dominating sequential functions, if possible, as follows:
69
Indefinite Integral of
r
r
m
m
sinh x dx dsinhx dx ; m,r R
x
=
(2n 1)r
m
n 0
1
x
dx
x
(2n 1)!
+
=
=
+
(2n 1)r m
n 0
1
x
dx
(2n 1)!
+
-
=
=
+
{
}
{
}
(2n 1)r m 1
n 0
m r 1
n
2r
(2n 1)r m 1
m r 1
n 0
n
2r
1
x
m r 1
k;n
(2n 1)! (2n 1)r m 1
2r
1
log x
(2n 1)!
1
x
m r 1
k;n
(2n 1)! (2n 1)r m 1
2r
+
- +
=
- -
=
+
- +
- -
=
- -
+
+
+
- +
=
+
+
- -
+
=
+
+
- +
r
m 1
m r 1
n
2r
r
m 1
m r 1
n
2r
1
m r 1
dsinh
x
k; if n
(2n 1)r m 1
2r
1
log x
(2n 1)!
1
m r 1
dsinh
x
k; if n
(2n 1)r m 1
2r
-
- -
=
-
- -
- -
+
+
- +
=
+
+
- -
+
=
+
- +
Example: We have sinh x dx
x
1
1
dsinh x dx
=
[Here m=1, r=1 and
m r 1 1 1 1
1
n
2r
2
2
- -
- -
-
]
r
m 1
1
dsinh
x
k
(2n 1)r m 1
-
=
+
+
- +
1
0
1
dsinh
x
k
(2n 1) 1 1
=
+
+ - +
1
sinh
x k
(2n 1)
=
+
+
.
70
Example: We have
2
sinh x dx
x
1
2
dsinh x dx
=
[Here m=2, r=1 and
m r 1 2 1 1
n
0
2r
2
- -
- -
=
=
=
]
1
1
n 0
n 0
1
1
log x
dsinh
x
k
(2n 1)!
(2n 1) 2 1
=
=
+
+
+
+ - +
1
1
n 0
1
1
log x
dsinh
x
k
1!
2n
=
+
+
1
1
n 0
1
log x
dsinh
x
k
2n
=
+
+
.
Indefinite Integral of
r
r
m
m
cosh x dx dcoshx dx
x
=
2nr
2nr m
m
n 0
n 0
1
x
1
dx
x
dx
x
(2n)!
(2n)!
-
=
=
=
=
{
}
{
}
2nr m 1
n 0
2nr m 1
m 1
m 1
n 0
n
n
2r
2r
1
x
m 1
k; if n
(2n)! 2nr m 1
2r
1
1
x
m 1
log x
k; if n
(2n)!
(2n)! 2nr m 1
2r
- +
=
- +
-
-
=
=
-
+
- +
=
-
+
+
=
- +
r
m 1
r
m 1
m 1
m 1
n
n
2r
2r
1
m 1
d cosh
x
k; if n
2nr m 1
2r
1
1
m 1
log x
d cosh
x
k; if n
(2n)!
2nr m 1
2r
-
-
-
-
=
-
+
- +
=
-
+
+
=
- +
Example: We have cosh x dx
x
1
1
d cosh x dx
=
[Here m=1, r=1 and n=0]
71
1
0
n 0
n 0
1
1
log x
d cosh
x
k
(0)!
2n
=
=
+
+
n 0
1
log x
cosh
x
k
2n
=
+
+
.
Example: We have
1
2
2
cosh x dx dcoshx dx
x
=
[Here m=2, r=1 and
1
n
2
]
1
1
1
d cosh
x k
2n 1
=
+
-
1
1
cos
x k
x
2n 1
=
+
-
.
Indefinite Integral of
r
r
m
m
tanh x dx dtanhx dx
x
=
n 1 2n
2n
2n
(2n 1)r
m
n 1
( 1) 2 (2
1) B
1
x
dx
x
(2n)!
-
-
=
-
-
=
n 1 2n
2n
2n
(2n 1)r m
n 1
( 1) 2 (2
1) B
x
dx
(2n)!
-
- -
=
-
-
=
72
{
}
{
}
n 1 2n
2n
(2n 1)r m 1
2n
n 1
n 1 2n
2n
2n
m r 1
n
2r
n 1 2n
2n
(2n 1)r m 1
2n
m r 1
n 1
n
2r
( 1) 2 (2
1) B
x
m r 1
k;n
(2n)!
(2n 1)r m 1
2r
( 1) 2 (2
1) B
log x
(2n)!
( 1) 2 (2
1) B
x
m r 1
k;n
(2n)!
(2n 1)r m 1
2r
-
-
- +
=
-
+ -
=
-
-
- +
+ -
=
-
-
+ -
+
-
- +
-
-
=
+
-
-
+ -
+
=
-
- +
r
m 1
n 1 2n
2n
2n
m r 1
n
2r
r
m 1
m r 1
n
2r
1
m r 1
d tanh
x
k;n
(2n 1)r m 1
2r
( 1) 2 (2
1) B
log x
(2n)!
1
m r 1
d tanh
x
k;n
(2n 1)r m 1
2r
-
-
+ -
=
-
+ -
+ -
+
-
- +
-
-
=
+
+ -
+
=
-
- +
Example: We have
1
1
tanh x dx dtanhx dx
x
=
[Here m=1, r=1 and
1
n
2
]
1
0
1
d tanh
x
k
(2n 1)
=
+
-
1
tanh
x k
(2n 1)
=
+
-
.
Example: We have
1
2
2
tanh x dx dtanhx dx
x
=
[Here m=2, r=1 and n=1]
73
0 2
2
2
1
1
n 1
n 1
( 1) 2 (2 1) B
1
log x
d tanh
x
k
(2)!
(2n 2)
=
-
-
=
+
+
-
1
1
n 1
4.3 1
1
log x
d tanh
x
k
2 6
(2n 2)
=
+
+
-
1
1
n 1
1
log x
d tanh
x
k
(2n 2)
=
+
+
-
.
Indefinite Integral of
r
r
m
m
coth x dx dcothx dx
x
=
2n
(2n 1)r
2n
m
r
n 1
2 B
1 1
x
dx
x
x
(2n)!
-
=
=
+
2n
(2n 1)r m
2n
m r
n 1
2 B
1
x
dx
x
(2n)!
- -
+
=
=
+
{
}
{
}
[
]
{
}
2n
1 (m r)
(2n 1)r m 1
2n
m r 1
n 1
m r 1
n
2r
2n
(2n 1)r m 1
2n
m r 1
m r 1
n 1
n
2r
1 (m r)
2 B
x
x
1 (m r)
(2n)! (2n 1)r m 1
m r 1
k;m r 1 n
2r
2 B
x
log x
(2n)! (2n 1)r m 1
m r 1
k;m r 1 n
2r
x
1 (m
-
+
-
- +
+ -
=
+
-
- +
+ =
+ -
=
-
+
+
-
+
-
- +
+ -
+
+
+
-
- +
=
+ -
+
+ =
-
{
}
{
}
2n
2n
(2n 1)r m 1
2n
2n
m r 1
m r 1
n 1
n
m r 1
n
2r
2r
2 B
2 B
x
log x
r)
(2n)!
(2n)! (2n 1)r m 1
m r 1
k;m r 1 n
2r
-
- +
+ -
+ -
=
=
+
+
+
+
-
- +
+ -
+
+
=
74
{
}
{
}
[
]
{
}
{
}
2n
(2n 1)r m 1
2n
(m r) 1
m r 1
n 1
m r 1
n
2r
2n
(2n 1)r m 1
2n
m r 1
m r 1
n 1
n
2r
2 B
1
1
x
1 (m r) x
(2n)! (2n 1)r m 1
m r 1
k;m r 1 n
2r
2 B
x
log x
(2n)! (2n 1)r m 1
m r 1
k;m r 1 n
2r
1
1
1 (m r) x
-
- +
+ -
+ -
=
+
-
- +
+ =
+ -
=
+
-
+
-
- +
+ -
+
+
+
-
- +
=
+ -
+
+ =
-
+
{
}
2n
2n
(2n 1)r m 1
2n
2n
(m r) 1
m r 1
m r 1
n 1
n
m r 1
n
2r
2r
2 B
2 B
x
log x
(2n)!
(2n)! (2n 1)r m 1
m r 1
k;m r 1 n
2r
-
- +
+ -
+ -
+ -
=
=
+
+
+
-
- +
+ -
+
+
=
[
]
{
}
r
m 1
2n
(2n 1)r m 1
2n
m r 1
m r 1
n 1
n
2r
2n
r
2n
m 1
m r 1
n
2r
1
m r 1
d coth
x
k;m r 1 n
,n 0
(2n 1)r m 1
2r
2 B
x
log x
k;
(2n)! (2n 1)r m 1
m r 1
m r 1 n
2r
2 B
1
log x
d coth
x
(2n)!
(2n 1)r m 1
-
-
- +
+ =
+ -
=
-
+ -
=
+ -
+
+
-
- +
+
+
-
- +
+ -
=
+ =
+
-
- +
m r 1
n
2r
k;
m r 1
m r 1 n
2r
+ -
+
+ -
+
=
Example: We have
1
1
coth x dx dcothx dx
x
=
[Here m=1, r=1, m+r=2 and
1
n
2
]
1
0
1
d coth
x
k,n 0
(2n 1)
=
+
-
1
coth
x k,n 0.
(2n 1)
=
+
-
75
Example: We have
1
2
2
coth x dx dcothx dx
x
=
[Here m=2, r=1, m+r=3 and n=1]
2
1
2
1
n 1
n 1
2 B
1
log x
d coth
x
k
(2)!
(2n 2)
=
=
+
+
-
1
1
n 1
1
1
log x
d coth
x
k.
3
(2n 2)
=
+
+
-
Example: We have
1
0
0
coth xdx dcothx dx
x
=
[m=0, r=1, m+r=1 and n0]
{ }
2n
2n
2n
n 1
2 B
x
log x
k.
(2n)! 2n
=
=
+
+
Indefinite Integral of
r
r
m
m
cosechx dx dcosechx dx
x
=
(2n 1)r
n
2n 1
2n
m
n 0
( 1) 2 2
1 B
1
x
dx
x
(2n)!
-
-
=
-
-
=
n
2n 1
2n
(2n 1)r m
n 0
( 1) 2 2
1 B
x
dx
(2n)!
-
-
-
=
-
-
=
76
{
}
{
}
n
2n 1
(2n 1)r m 1
2n
n 0
n
2n 1
2n
m r 1
n
2r
n
2n 1
(2n 1)r m 1
2n
n 0
m r 1
n
2r
( 1) 2 2
1 B
x
m r 1
k;n
(2n)!
(2n 1)r m 1
2r
( 1) 2 2
1 B
log x
(2n)!
( 1) 2 2
1 B
x
m r 1
k;n
(2n)!
(2n 1)r m 1
2r
-
-
- +
=
-
+ -
=
-
-
- +
=
+ -
-
-
+ -
+
-
- +
-
-
=
+
-
-
+ -
+
=
-
- +
r
m 1
n
2n 1
2n
m r 1
n
2r
r
m 1
m r 1
n
2r
1
m r 1
d cosech
x
k;n
(2n 1)r m 1
2r
( 1) 2 2
1 B
log x
(2n)!
1
m r 1
d cosech
x
k;n
(2n 1)r m 1
2r
-
-
+ -
=
-
+ -
+ -
+
-
- +
-
-
=
+
+ -
+
=
-
- +
Example: We have
1
1
cosechxdx dcosechx dx
x
=
[Here m=1, r=1 and
1
n
2
]
1
0
1
d cosech
x
k
(2n 1)
=
+
-
1
cosech
x k
(2n 1)
=
+
-
.
Example: We have
1
2
2
cosechxdx dcosechx dx
x
=
[Here m=2, r=1 and n=1]
77
1
2 1
2
1
1
n 1
n 1
( 1) 2 2
1 B
1
log x
d cosech
x
k
(2)!
(2n 2)
-
-
-
=
+
+
-
1
1
n 1
1
1
log x
d cosech
x
k
6
(2n 2)
-
=
+
+
-
, n0
Indefinite Integral of
r
r
m
m
sechx dx dsechx dx
x
=
n
2n
2nr
m
n 0
( 1) E
1
x dx
x
(2n)!
=
-
=
n
2n
2nr m
n 0
( 1) E
x
dx
(2n)!
-
=
-
=
{
}
{
}
n
2nr m 1
2n
n 0
n
n
2nr m 1
2n
2n
m 1
m 1
n 0
n
n
2r
2r
( 1) E
x
m 1
k;n
(2n)!
2nr m 1
2r
( 1) E
( 1) E
x
m 1
log x
k;n
(2n)!
(2n)!
2nr m 1
2r
- +
=
- +
-
-
=
=
-
-
+
- +
=
-
-
-
+
+
=
- +
r
m 1
n
2n
r
m 1
m 1
m 1
n
n
2r
2r
1
m 1
dsech
x
k;n
(2nr m 1)
2r
( 1) E
1
m 1
log x
dsech
x
k;n
(2n)!
(2nr m 1)
2r
-
-
-
-
=
-
+
- +
=
-
-
+
+
=
- +
Example: We have
1
1
sechxdx dsechx dx
x
=
[Here m=1, r=1 and n=0]
0
0
1
0
n 0
n 0
( 1) E
1
log x
dsec h
x
k
(0)!
2n
=
-
=
+
+
n 0
1
log x
sec h
x
k
2n
=
+
+
78
Example: We have
1
2
2
sechxdx dsechx dx
x
=
[Here m=2, r=1 and
1
n
2
]
1
1
1
dsech
x k.
2n 1
=
+
-
Particular Cases: Putting m=0 and r=1, we get the indefinite integrals of classical hyperbolic
functions as follows:
Indefinite Integral of
1
1
0
0
sinh x dx dsinhx dx
x
=
[Here m=0, r=1 and n-1]
1
1
1
dsinh
x
k
(2n 2)
-
=
+
+
1
x sinh
x k,n 0
2n 2
=
+
+
3
5
1
1 x
1 x
x
x
... k
2
4 3! 6 5!
=
+
+
+
+
2
4
6
x
1 x
1 x ... k
2 4 3! 6 5!
=
+
+
+
+
2
4
6
x
x
x
1
... c;c k 1
2! 4! 6!
= +
+
+
+
+
= -
cosh x c
=
+
sinh xdx.
=
Indefinite Integral of
1
1
0
0
cosh x dx dcoshx dx
x
=
[Here m=0, r=1 and n
-12
]
79
1
1
1
d cosh
x
k,n 0
(2n 1)
-
=
+
+
1
x cosh
x k
2n 1
=
+
+
2
4
6
1 1 x
1 x
1 x
x 1
... k
1
3 2! 5 4! 7 6!
=
+
+
+
+
+
3
5
7
x
x
x
x
... k
3! 5! 7!
=
+
+
+
+
+
sinh x k
=
+
cosh xdx
=
Indefinite Integral of
1
1
0
0
tanh x dx dtanhx dx
x
=
[Here m=0, r=1 and n0]
1
1
1
d tanh
x
k
2n
-
=
+
1
x tanh
x k,n 1
2n
=
+
3
5
7
1
1 1
1 2
1 17
x
x
x
x
x ... k
2
4 3
6 15
8 315
=
-
+
-
+
+
2
4
6
8
x
1 x
2 x
17 x ... k
2 3 4 15 6 315 8
=
-
+
-
+
+
3
5
7
x 1
2
17
x
x
x ... dx
1 3
15
315
=
-
+
-
+
tanh xdx
=
80
Indefinite Integral of
1
1
0
0
coth x dx dcothx dx
x
=
[Here m=0, r=1 and n0]
[
]
{ }
2n
2n
2n
n 0
n 1
n 0
2 B
x
log x
k
(2n)! 2n
=
=
=
+
+
6
2
4
2
4
6
6
2
4
2 B
2 B
2 B
x
x
x
log x
... k
2! 2
4! 4
6! 6
=
+
+
+
+
+
2
4
6
1 1
1 1
2 1
log x
x
x
x ... k
3 2
45 4
945 6
=
+
-
+
-
+
3
5
1 1
1
2
x
x
x ... dx
x 3
45
945
=
+
-
+
-
coth xdx
=
Indefinite Integral of
1
1
0
0
cosechx dx dcosechx dx
x
=
[Here m=0, r=1 and n=0]
0
1
0
1
1
n 0
n 0
( 1) 2 2
1 B
1
log x
d cosech
x
k
0!
2n
-
-
=
-
-
=
+
+
n 0
1
log x
x cosech
x
k
2n
=
+
+
3
5
1 1
1 7
1 31
log x x
x
x
x ... k
2 6
4 360
6 15120
-
=
+
+
-
+
+
2
4
6
1 1
7 1
31 1
log x
x
x
x ... k
6 2
360 4
15120 6
=
-
+
-
+
+
3
5
1 1
7
31
x
x
x ... dx
x 6
360
15120
=
-
+
-
+
cosechxdx
=
.
81
Indefinite Integral of
1
1
0
0
sechx dx dsechx dx
x
=
[Here m=0, r=1 and n
-12
]
1
1
1
dsech
x
k,n 0
2n 1
-
=
+
+
1
1
1
sech
x k
x
2n 1
-
=
+
+
1
x sech
x k
2n 1
=
+
+
2
4
6
1 1
1 5
1 61
x 1
x
x
x ... k
3 2
5 24
7 720
=
-
+
-
+
+
3
5
7
1 1
5 1
61 1
x
x
x
x ... k
2 3
24 5
720 7
= -
+
-
+
+
2
4
6
1
5
61
1
x
x
x ... dx
2
24
720
=
-
+
-
+
sechxdx
=
Indefinite Integrals of Dominating Exponential Functions
In finding the general indefinite integrals of dominating exponential functions, we first
express the given functions in terms of particular case of d-function or in power series and
then we integrate each term of the series separately. Finally we denote them in terms of
different dominating sequential functions, if possible, as follows:
Indefinite Integral of
r
r
xm
x
m
a
da dx
dx
x
=
n
nr m
n 0
(log x) x dx
n!
-
=
=
82
{
}
{
}
n
nr m 1
n 0
n
n
nr m 1
m 1
m 1
n 0
n
n
r
r
(log x)
x
m 1
k;n
n!
nr m 1
r
(log x)
(log x)
x
m 1
log x
k;n
n!
n!
nr m 1
r
- +
=
- +
-
-
=
=
-
+
- +
=
-
+
+
=
- +
r
m 1
r
m 1
1
x
nr m 1
1
n
x
nr m 1
m 1
m 1
n
n
r
r
m 1
da
k;n
r
(log x)
m 1
log x
da
k;n
n!
r
-
-
- +
- +
-
-
=
-
+
=
-
+
+
=
Example: We have
1
0
x
x
x
0
a
da dx
dx
a dx
x
=
=
[Here m=0, r=1 and n-1]
1
1
1 x
n 1
da
k
-
+
=
+
1 x
n 1
xa
k
+
=
+
1 xloga
n 1
xe
k
+
=
+
0
2
1 1
1 1
x (x log a)
(x log a)
(x log a) ... k
1! 2
2! 3
=
+
+
+
+
2
3
1
1 1
1 1
(x log a)
(x log a)
(x log a) ... k
(log a)
1! 2
2! 3
=
+
+
+
+
xloga
1 e
k
(log a)
=
+
x
a
k
log a
=
+
Example: We have
1
1
x
x
a
da dx
d
x
=
83
[Here m=1, r=1 and n=0]
1
0
1
0
x
n
n 0
n 0
(log a) logx
da
k
0!
=
=
+
+
1 x
n
n 0
log x
a
k,n 0
=
+
+
Indefinite Integral of
r
r
kxm
kx
m
a
da dx
dx
x
=
n
nr m
n 0
(k log x) x dx
n!
-
=
=
n
nr m
n 0
(k log x) x dx
n!
-
=
=
{
}
{
}
n
nr m 1
n 0
n
n
nr m 1
m 1
m 1
n 0
n
n
r
r
(k log x)
x
m 1
c;n
n!
nr m 1
r
(k log x)
(k log x)
x
m 1
log x
c;n
n!
n!
nr m 1
r
- +
=
- +
-
-
=
=
-
+
- +
=
-
+
+
=
- +
r
m 1
r
m 1
1
k
x
nr m 1
1
n
k
x
nr m 1
m 1
m 1
n
n
r
r
m 1
da
c;n
r
(k log x)
m 1
log x
da
c;n
n!
r
-
-
- +
- +
-
-
=
-
+
=
-
+
+
=
Example: We have
1
0
kx
kx
kx
0
a
da dx
d
a dx
x
=
=
[Here m=0, r=1 and n-1]
1
1
1 kx
n 1
da
c
-
+
=
+
84
1 kx
n 1
xa
c
+
=
+
1
(kloga)
x
n 1
xe
c
+
=
+
1 (kloga)x
n 1
xe
c
+
=
+
{
}
2
(k log a)x
1 (k log a)x 1
x 1
... c
2
1!
3
2!
=
+
+
+
+
{
}
{
}
2
3
(k log a)x
(k log a)x
1
1
1
(xk log a)
... c
(k log a)
2
1!
3
2!
=
+
+
+
+
xkloga
1
e
c
(k log a)
=
+
kx
a
c
(k log a)
=
+
Indefinite Integral of
r
r
xm
x
m
e
de dx
dx
x
=
nr m
n 0
1 x dx
n!
-
=
=
nr m
n 0
1 x dx
n!
-
=
=
{
}
{
}
nr m 1
n 0
nr m 1
m 1
m 1
n 0
n
n
r
r
1
x
m 1
k;n
n! nr m 1
r
1
1
x
m 1
log x
k;n
n!
n! nr m 1
r
- +
=
- +
-
-
=
=
-
+
- +
=
-
+
+
=
- +
r
m 1
r
m 1
1
x
nr m 1
1
x
nr m 1
m 1
m 1
n
n
r
r
m 1
de
k;n
r
1
m 1
log x
de
k;n
n!
r
-
-
- +
- +
-
-
=
-
+
=
-
+
+
=
85
Example: We have
1
0
x
x
0
e
de dx
dx
x
=
[Here m=0, r=1 and n-1]
1
1
1 x
n 1
de
c
-
+
=
+
1 x
n 1
xe
c
+
=
+
2
3
1 x 1 x
1 x
x 1
... c
2 1! 3 2! 4 3!
=
+
+
+
+
+
2
3
4
1 x
1 x
1 x
x
... c
1! 2 2! 3 3! 4
=
+
+
+
+
+
2
3
1
1
1
1
x
x
x ... dx c
1!
2!
3!
=
+
+
+
+
+
x
x
e dx e c
=
=
+
Indefinite Integral of
r
r
kxm
kx
m
e
de dx
dx
x
=
, where k is any constant.
n
nr m
n 0
k x dx
n!
-
=
=
n
nr m
n 0
k x dx
n!
-
=
=
{
}
{
}
n
nr m 1
n 0
n
n
nr m 1
m 1
m 1
n 0
n
n
r
r
k
x
m 1
c;n
n! nr m 1
r
k
k
x
m 1
log x
c;n
n!
n! nr m 1
r
- +
=
- +
-
-
=
=
-
+
- +
=
-
+
+
=
- +
r
m 1
r
m 1
1
kx
nr m 1
1
n
kx
nr m 1
m 1
m 1
n
n
r
r
m 1
de
c;n
r
k
m 1
log x
de
c;n
n!
r
-
-
- +
- +
-
-
=
-
+
=
-
+
+
=
86
Example: We have
1
0
x
de dx
[Here m=0, r=1 and n
12
]
1
1
1 x
n 1
de
c
-
+
=
+
1 x
n 1
xe
c
+
=
+
x
x
e dx e c
=
=
+
Indefinite Integrals of Dominating Logarithmic Functions
In finding the general indefinite integrals of dominating logarithmic functions, we first
express the given functions in terms of particular case of d-function or in power series and
then we integrate each term of the series separately. Finally we denote them in terms of
different dominating sequential functions, if possible, as follows:
Indefinite Integral of
r
r
m
m
log x
d log x dx
dx
x
=
Case-I: When r m 1
= , we have
r
r
m
log x dx, Putlogx z
x
=
1 m z
m
1 e
zdz
m
-
=
(
)
1 m z
m
m
m
e
z
k
m 1 m
(1 m)
-
=
-
+
-
-
(1 m)
m
m
m
x
log x
k
m(1 m)
(1 m)
-
=
-
+
-
-
r
m 1
m 1
1
m
1
d log x
k
(1 m)
(1 m) x
-
-
=
-
+
-
-
Case-II: When r m,m 1
, we have
87
r
r
m
log x dx, Putlogx z
x
=
1 m z
r
1 ze
dz
r
-
=
1 m z
r
1
r
e
z
k
(1 m)
(1 m)
-
=
-
+
-
-
r
m 1
m 1
1
log x
r
1
k
(1 m) x
(1 m) x
-
-
=
-
+
-
-
r
m 1
m 1
1
r
1
d log x
k
(1 m)
(1 m) x
-
-
=
-
+
-
-
Case-III: When m=1, we have
r
r
r
r
1
m
log x
log x
dx
d log x dx[Put log x
z]
x
x
=
=
=
1 zdz
r
=
2
1 z
k
r 2
=
+
r 2
1 (log x )
k
r
2
=
+
2
r (logx) k
2
=
+
Indefinite Integral of
r
r
m
m
log(1 x )
d log(1 x )dx
dx
x
+
+
=
n 1
nr m
n 1
( 1) x dx
n
+
-
=
-
=
n 1
nr m
n 1
( 1)
x
dx
n
+
-
=
-
=
88
{
}
{
}
n 1
nr m 1
n 1
n 1
n 1
nr m 1
m 1
m 1
n 1
n
n
r
r
( 1)
x
m 1
k;n
n
nr m 1
r
( 1)
( 1)
x
m 1
log x
k;n
n
n
nr m 1
r
+
- +
=
+
+
- +
-
-
=
=
-
-
+
- +
=
-
-
-
+
+
=
- +
r
m 1
n 1
r
m 1
m 1
m 1
n
n
r
r
1
m 1
d log
(1 x ) k;n
nr m 1
r
( 1)
1
m 1
log x
d log
(1 x )
k;n
n
nr m 1
r
-
+
-
-
-
=
-
+
+
- +
=
-
-
+
+
+
=
- +
Example: We have
1
0
d log(1 x )dx
+
[Here m=0, r=1 and n
-12
]
1
1
1
d log
(1 x ) k
n 1
-
=
+
+
+
1
x log
(1 x) k
n 1
=
+ +
+
2
3
4
1
1 x
1 x
1 x
x
x
... k
2
3 2 4 3 5 4
=
-
+
-
+
+
3
4
5
2
1
1 x
1 x
1 x
x
... k
2
2 3 3 4 4 5
=
-
+
-
+
+
2
3
4
1
1
1
x
x
x
x ... dx
2
3
4
=
-
+
-
+
log(1 x)dx
=
+
Example: We have
1
2
2
log(1 x)
d log(1 x )dx
dx
x
+
+
=
[Here m=2, r=1 and n=1]
89
2
1
1
n 1
n 1
( 1)
1
log x
d log
(1 x )
k
1
n 1
=
-
=
+
+
+
-
1
1
n 1
1
log x
d log
(1 x )
k
n 1
=
+
+
+
-
.
Indefinite Integral of
r
r
m
m
log(1 x )
d log(1 x )dx
dx
x
-
-
=
nr m
n 1
( 1)x dx
n
-
=
-
=
nr m
n 1
( 1) x dx
n
-
=
-
=
{
}
{
}
nr m 1
n 1
nr m 1
m 1
m 1
n 1
n
n
r
r
( 1) x
m 1
k;n
n nr m 1
r
( 1)
( 1) x
m 1
log x
k;n
n
n nr m 1
r
- +
=
- +
-
-
=
=
-
-
+
- +
=
-
-
-
+
+
=
- +
r
m 1
r
m 1
m 1
m 1
n
n
r
r
1
m 1
d log
(1 x ) k;n
nr m 1
r
( 1)
1
m 1
log x
d log
(1 x )
k;n
n
nr m 1
r
-
-
-
-
=
-
-
+
- +
=
-
-
+
-
+
=
- +
Example: We have
1
0
d log(1 x )dx
-
[Here m=0, r=1 and n-1]
1
1
1
d log
(1 x ) k
n 1
-
=
-
+
+
1
x log
(1 x) k
n 1
=
- +
+
2
3
1
1 x
1 x
x
x
... k
2
3 2 4 3
= -
+
+
+
+
3
4
2
1
1 x
1 x
x
... k
2
3 2 4 3
= -
+
+
+
+
3
4
2
1
1 x
1 x
x
... k
2
2 3 3 4
= -
+
+
+
+
2
3
1
1
x
x
x ... dx
2
3
= -
+
+
+
log(1 x)dx
=
-
90
Indefinite Integrals of Classical Nonelementary Functions
In this section we have found the general integrals of some classical nonelementary functions
related to trigonometric, hyperbolic, exponential, and hyperbolic functions as follows:
Indefinite Integrals of Trigonometric Nonelementary Functions:
In finding the general indefinite integrals of trigonometric nonelementary functions, we first
express the given functions in terms of particular case of d-function or in power series and
then we integrate each term of the series separately. Finally we denote them in terms of
different dominating sequential functions, if possible, as follows:
Indefinite Integral of sin x dx
x
2n 1
n
n 0
1
x
( 1)
dx
x
(2n 1)!
+
=
=
-
+
n
2n
n 0
( 1)
x dx
(2n 1)!
=
-
=
+
n
2n 1
n 0
( 1)
x
k
(2n 1)! 2n 1
+
=
-
=
+
+
+
(
)
n 2n 1
n 0
1
( 1) x
k
(2n 1)
2n 1 !
+
=
-
=
+
+
+
1
sin
x k
(2n 1)
=
+
+
1
1
2
1
d cos
x
k
(2n 1)
-
=
+
+
2
1
x cos
x k
(2n 1)
=
+
+
Indefinite Integral of cos x dx
x
91
2n
n
n 0
1
x
( 1)
dx
x
(2n)!
=
=
-
2
4
6
1 1 x
x
x ... dx
x 1 2! 4! 6!
=
-
+
-
+
3
5
1 x x
x ... dx
x 2! 4! 6!
=
-
+
-
+
2
4
6
x
x
x
log x
... k
2!2 4!4 6!6
=
-
+
-
+
+
n 0
1
log x
cos
x
k
2n
=
+
+
Indefinite Integral of tan x dx
x
n 1 2n
2n
2n 1
2n
n 1
( 1) 2 (2
1)B
1
x
dx
x
(2n)!
-
-
=
-
-
=
n 1 2n
2n
2n 2
2n
n 1
( 1) 2 (2
1)B
x
dx
(2n)!
-
-
=
-
-
=
n 1 2n
2n
2n 1
2n
n 1
( 1) 2 (2
1)B
x
k
(2n)!
(2n 1)
-
-
=
-
-
=
+
-
1
tan
x k
(2n 1)
=
+
-
Indefinite Integral of cot x dx
x
n 2n
2n 1
2n
n 0
( 1) 2 B
1
x
dx
x
(2n)!
-
=
-
=
n 2n
2n 2
2n
n 0
( 1) 2 B
x
dx
(2n)!
-
=
-
=
92
n 2n
2n 1
2n
n 0
( 1) 2 B
x
k
(2n)!
(2n 1)
-
=
-
=
+
-
1
cot
x k
(2n 1)
=
+
-
Indefinite Integral of sec x dx
x
n
2n
2n
n 0
( 1) E
1
x dx
x
(2n)!
=
-
=
2
4
6
1 1 x
5
61
x
x ... dx
x 1 2 24
720
=
+
+
+
+
3
5
1 x 5
61
x
x ... dx
x 2 24
720
=
+ +
+
+
2
4
6
1 x
1 5
1 61
log x
x
x ... k
2 2 4 24
6 720
=
+
+
+
+
+
n 0
1
log x
sec
x
k
2n
=
+
+
Indefinite Integral of cosecx dx
x
n 1
2n 1
2n 1
2n
n 0
( 1) 2(2
1)B
1
x
dx
x
(2n)!
+
-
-
=
-
-
=
n 1
2n 1
2n 2
2n
n 0
( 1) 2(2
1)B
x
dx
(2n)!
+
-
-
=
-
-
=
n 1
2n 1
2n 1
2n
n 0
( 1) 2(2
1)B x
k
(2n)!
2n 1
+
-
-
=
-
-
=
+
-
1
cosec
x k
2n 1
=
+
-
93
Indefinite Integrals of Hyperbolic Nonelementary Functions
In finding the general indefinite integrals of hyperbolic nonelementary functions, we first
express the given functions in terms of particular case of d-function or in power series and
then we integrate each term of the series separately. Finally we denote them in terms of
different dominating sequential functions, if possible, as follows:
Indefinite Integral of sinh x dx
x
2n 1
n 0
1
x
dx
x
(2n 1)!
+
=
=
+
2n
n 0
1
x dx
(2n 1)!
=
=
+
2n 1
n 0
1
x
k
(2n 1)! (2n 1)
+
=
=
+
+
+
(
)
2n 1
n 0
1
x
k
(2n 1) 2n 1 !
+
=
=
+
+
+
1
sinh
x k
(2n 1)
=
+
+
Indefinite Integral of cosh x dx
x
2n
n 0
1
x dx
x
(2n)!
=
=
2
4
6
1 1 x
x
x ... dx
x 1 2! 4! 6!
=
+
+
+
+
3
5
1 x x
x ... dx
x 2! 4! 6!
=
+
+
+
+
2
4
6
x
x
x
log x
... k
2!2 4!4 6!6
=
+
+
+
+
+
2
4
6
x
x
x
log x
... k
2!2 4!4 6!6
=
+
+
+
+
+
n 0
1
log x
cosh
x
k
2n
=
+
+
Indefinite Integral of tanh x dx
x
94
2n
2n
2n 1
2n
n 1
2 (2
1)B
1
x
dx
x
(2n)!
-
=
-
=
2n
2n
2n 2
2n
n 1
2 (2
1)B
x
dx
(2n)!
-
=
-
=
2n
2n
2n 1
2n
n 1
2 (2
1)B
x
k
(2n)!
(2n 1)
-
=
-
=
+
-
1
tanh
x k
(2n 1)
=
+
-
Indefinite Integral of coth x dx
x
2n
2n 1
2n
n 1
2 B
1 1
x
dx
x x
(2n)!
-
=
=
+
2n
2n 2
2n
2
n 1
2 B
1
x
dx
x
(2n)!
-
=
=
+
2n
2n 1
2n
n 1
2 B
1
x
k
x
(2n)! (2n 1)
-
=
= - +
+
-
2n
2n 1
2n
n 1
2 B
2
1
x
k
x
x
(2n)! (2n 1)
-
=
= - +
+
+
-
2
1
coth
x k
x
(2n 1)
= - +
+
-
1
coth
x k,n 0
(2n 1)
=
+
-
Indefinite Integral of sechx dx
x
2n
2n
n 0
E
1
x dx
x
(2n)!
=
=
2
4
6
1 1 x
5
61
x
x ... dx
x 1 2 24
720
=
-
+
-
+
3
5
1 x 5
61
x
x ... dx
x 2 24
720
=
- +
-
+
2
4
6
1 x
1 5
1 61
log x
x
x ... k
2 2 4 24
6 720
=
-
+
-
+
+
n 0
1
log x
sec h
x
k
2n
=
+
+
Indefinite Integral of cosechx dx
x
2n 1
2n 1
2n
n 1
2(1 2 )B
1 1
x
dx
x x
(2n)!
-
-
=
-
=
+
2n 1
2n 2
2n
2
n 1
2(1 2 )B
1
x
dx
x
(2n)!
-
-
=
-
=
+
95
2n 1
2n 1
2n
n 1
2(1 2 )B
1
x
k
x
(2n)!
2n 1
-
-
=
-
= - +
+
-
2
1
cosech
x
k
x
(2n 1)
= - +
+
-
Indefinite Integrals of Exponential Nonelementary Functions
In finding the general indefinite integrals of exponential nonelementary functions, we first
express the given functions in terms of particular case of d-function or in power series and
then we integrate each term of the series separately. Finally we denote them in terms of
different dominating sequential functions, if possible, as follows:
Indefinite Integral of
x
e dx
x
2
3
1
x
x
1 x
... dx
x
2! 3!
=
+ +
+
+
2
1
x x
1
... dx
x
2! 3!
=
+ +
+
+
2
3
4
x
x
x
x
log x
... k
1!1 2!2 3!3 4!4
=
+
+
+
+
+
+
1 x
n
n 0
log x
e
k
=
+
+
Indefinite Integral of
x
e dx
x
-
2
3
1
x
x
1 x
... dx
x
2! 3!
=
- +
-
+
2
1
x x
1
... dx
x
2! 3!
=
- +
-
+
2
3
4
x
x
x
x
log x
... k
1!1 2!2 3!3 4!4
=
-
+
-
+
-
+
1 ( x)
n
n 0
log x e
k
-
=
+
+
96
Chapter-VI
Existence Theorems on Indefinite Integrability
Introduction
We know from the textbooks authored by Courant (1935), Apostal (1967), Chatterjee (2002),
Jain et al (2005), etc. that: "Every continuous function has an anti-derivative". But the
condition of continuity of the integrand is only a sufficient condition for the existence of
indefinite integral and not a necessary condition. For example: f(x)=tanx is discontinuous at
infinitely many points, yet it has an indefinite integral logsecx+k. In other words, it is not
closely bound up with the assumption that the integrand is continuous, but that it may be
extended to wide classes of functions with discontinuities.
In this chapter we have propounded a necessary condition, a sufficient condition, a necessary
and sufficient condition, as well as a modified sufficient condition for the existence of
indefinite integral of a function based on d-function theory.
Existence Theorems on Indefinite Integrability of a Function
A deep observation of different indefinite integrable functions mentioned by Hardy (1916),
Courant (1935), Thomas (1966), Apostal (1967), Maron (1988), Strange (1991), Prasad
(1991), Ayers et al (1999), Sharma et al (2000), Narayan (2001), Dhami (2001), James et al
(2001), Chatterjee (2002), Edwards (2006), etc. implies that all most all the indefinite
integrable functions are dominatable functions.
We have showed in chapter-3 section-3.2.2 that all the basic elementary functions are
dominatable functions. Since these elementary functions are indefinite integrable functions,
we can conclude that all most all indefinite integrable functions are d-able functions. We
have also proved that a dominating function is always indefinite integrable. Thus we can
propound the following properties:
97
Necessary Condition for Indefinite Integrability
Every indefinite integrable function is a dominatable function.
Example: e
x
, sinx, sinhx, x
2
+x+3, log(1+x), etc.
Sufficient Condition for Indefinite Integrability
If f(x) be a dominatable function, it is indefinite integrable.
Example: e
x
, cosx, sinhx, etc.
Necessary and Sufficient Condition for Indefinite Integrability
A function f(x) is indefinite integrable if and only if it is dominatable.
In other words, if a function f(x) has an indefinite integral, then it is a d-able function and
conversely if f(x) is a d-able function, it has an indefinite integral.
Example:
2
6
x
2
(x 1) ,e sin x, x cosh x
+
, etc.
Comparative Study of Existence Theorems on Integrability
We have proved earlier that every indefinite integrable function is d-able function and that
every d-function is continuous, which implies that every indefinite integrable function is
continuous.
Thus we see that the condition of continuity for the existence of integrability is included in
the above proposed theorems based on d-function theory. Therefore the above proposed
theorems are superior than the classical sufficient condition for the existence of indefinite
integral of a function.
This is also superior in the sense that the classical theorem does not give any clue to formulate
the nonelementary functions, whereas it gives.
98
Modified Sufficient Condition of Integrability
We have mentioned earlier that the condition of continuity of the integrand is only a sufficient
condition for the existence of indefinite integral and not a necessary condition. We have also
noticed that we can find an indefinite integral of a function, which has some or many
discontinuities. Let us consider a function
=
2
1
,
5
1
0
,
4
)
(
2
x
for
x
x
for
x
x
f
Then we have
3
41
5
4
)
(
)
(
)
(
2
1
2
1
0
2
1
1
0
2
0
=
+
=
+
=
dx
x
xdx
dx
x
f
dx
x
f
dx
x
f
Here f(x) is not continuous but a piecewise continuous function in the interval [0, 2]. We can
have lots of such examples.
This suggests that the concept of continuity can be replaced by piecewise continuity for the
existence of an indefinite integrability of a function as follows:
Existence Theorem of Integrability (Modified Form)
Every continuous or piecewise continuous function is integrable.
From this theorem, we conclude that "even though integration in terms of elementary
functions is not always possible, in all circumstances we are certain at least that the integral
of a continuous or piecewise continuous function exists".
99
References
Allen G. D. (2000), Calculus: Transition of Calculus-I, August, p.5
Apostal T. M. (1967), Calculus, Vol.-I, 2
nd
Edition, John Wiley Sons, Inc., pp.1-11, 75-
77, 120-134, 152-153, 204-205
Awful Truth About Antiderivatives,
http://www.math.unt.edu/Integration_ bee/
AwfulTruth.html
Baccala B. (2006), Risch Integration, University of Maryland,
http://www.free
soft.org/Classes/
Risch2006/, Spring 2006
Baddoura M. J. (1994), Integration in Finite Terms with Elementary Functions and
Dilogarithms,
Ph. D. Thesis, Massachusetts Institute of Technology, January
Barrios J., A Brief History of Elliptic Integral Addition Theorems, NSF Centre,
Undergraduate Research in Mathematics, BYU, pp.1-9
Beaumont J, Bradford R. Davenport J. H. (2003), Better Simplification of Elementary
Functions Through Power Series, Proceeding of ISSAC'03, August 3-6, 2003, Philadelphia,
Pennsylvania, USA, pp.1-7
Boyer C. (1949), "The History of the Calculus and its Conceptual Development", Dover
Publications, New York
Bronstein M. (2000), Symbolic Integration Tutorial, ISSAC'98, Rostock (August 1998) and
Differential Algebra Workshop, Rutgers, November, pp.35
100
Bronstein M. (2005), Symbolic Integration I, Transcendental Functions, Chapter-5,
Springer, www.springer.com/978-3-540-21493-9, pp.129-180
Bronstein M., Integration and Differential Equations in Computer Algebra, Institute of
Scientific Computation, Zurich, Switzerland, pp.1-17
Chatterjee D. (2002), Integral Calculus, T M H Publication, New Delhi , India, pp.1-75
Churchill R. C. (2002), Liouville's Theorem on Integration in Terms of Elementary
Functions, Kolchin Seminar on Differential Algebra, Hunter College, CUNY, pp.1-22
Churchill R. C. (2006), Liouville's Theorem on Integration in Terms of Elementary
Functions, Corrected Extended Version, Kolchin Seminar on Differential Algebra, Hunter
College, CUNY, pp.1-26
Collingwood J., Alexander D. Kleiner A., "Rigor in Analysis: From Newton to Cauchy",
Drake University, Williamsburg, IA, pp. 1-18
Conrad B., Impossibility Theorems for Elementary Integration, University of Michigan,
Ann Arbor, MI 48109-1043, pp.1-13
Courant R. (1935), Differential Integral Calculus, Vol.-I, New Rev. Ed., Inter-science
Publishers, Inc., New York, pp.109-131, 242-245
Dhami H.S. (2001), Integral Calculus, New Age International Publisher, New Delhi, India,
pp.1-113
Edwards C. H. (1979), The Historical Development of the Calculus, Spring-Verlag, New
York
101
Edwards J. (2006), Integral Calculus, AITBS Publishers Distributors, New Delhi, India,
pp.1-134
Fateman R., Symbolic Integration, CS 282, Lecture 14, pp.1-32
Goetz P. (2009), Why certain integrals are "impossible", Sonoma State University, March
11, pp.1-20
Hardy G. H. (1916), The Integration of Functions of a Single Variable, 2
nd
Ed., Cambridge
University Press, London, Reprint 1928, pp.1-62
Hancock H. (1910), Lectures on the Theory of Elliptic Functions, 1
st
Ed., Vol.1, John Wiley
Sons, London
Jain R.K. Iyenger S.R.K. (2005), Advanced Engineering Mathematics, 2
nd
Ed., Narosa
Publishing House, New Delhi, India, pp.30-33
Jeffrey A. (2010), Advanced Engineering Mathematics, Academic Press, An Imprint of
Elsevier, pp. 44-45, 263, 564-565
Kalman D. (2000), Integrability and the glog Function, January 19, pp.1-7
Kaltofen E. (1984), A Note on the Risch Differential Equation, Corrected Version of a paper
Published in the Proceedings EUROSAM'84, Springer Lecture Notes in Computer Science
174, pp.359-366
Kauers M. (2008), Integration of Algebraic Functions: A Simple Heuristic for Finding the
Logarithmic Part, ISSAC'08, Hagenberg, Austria, July, pp.1-8
102
Kowalski, Basic Facts about Functions, pp.1-2
Leerawat U. Laohakosol V. (2002), A Generalization of Liouville's theorem on
Integration in Finite Terms, J. Korean Math. Soc.39(1), pp.13-30
Lehtinen N. G. (2010), Error Functions, April 23, pp.1-10
Marchisotto E. A. Zakeri G. A. (1994), An Invitation to Integration in Finite Terms, The
College Mathematics Journal, Mathematical Association of America, Vol.25, No.4, Sep.,
pp.295-308
Meurer A. (2011), The Risch Algorithm for Symbolic Integration in SymPy, Python
Software Organisation, July
Moses J. (1971), Symbolic Integration: The Stormy Decade, Communications of the ACM,
Vol. 18(8), Aug., pp.548-560
Murphy L. D. (1999), Computer Algebra Systems in Calculus Reform,
http://mste.illinois.edu/murphy
/Papers/CalcReformPaper.html, February
Narayan S. (2001), Integral Calculus, S.Chand Comp. Ltd, India, pp.1-149
Neubacher A. (1992), An Introduction to the Symbolic Integration of Elementary Functions,
Diploma Thesis, Research Institute of Symbolic Computation, Johannes Kepler University,
Linz, Austria, Europe, November, pp.1-97
Prasad L. (1991), Integral Calculus, Student's Friends, Patna, India, pp.1-146
103
Raab C. G., Integration in finite terms for Liouvillian functions, DK Computational
Mathematics, JKU Linz, Austria, pp.1-2
Risch R. H. (1969), The Problem of Integration in Finite Terms, Transactions of the
American Mathematical Society, 139: 167-189
Risch R. H. (1970), The Solution of the Problem of Integration in Finite Terms, Bulletin of
the American Mathematical Society, 76(3), 605-608
Ritt J. F. (1927), On the Integration in Finite Terms of Linear Differential Equations of the
Second Order, p.51-57
Ritt J. F. (1948), Integration in Finite Terms: Liouville's Theory of Elementary Methods,
Columbia University Press, New York, pp.1-98
Rosenlicht M. (1968), Liouville's Theorem on Functions with Elementary Integrals, Pacific
Journal of Mathematics, Vol.24, No.1, p.153-161
Rosenlicht M. (1972), Integration in Finite Terms, The American Mathematical Monthly,
Vol.79, No.9, November, p.963-972
Rothstein M. (1976), Aspects of Symbolic Integration and Simplification of Exponential and
Primitive Functions,
Ph. D. Thesis, University of Wisconsin-Madison, pp.1-104
Sharma G.C. Jain M. (2000), Integral Calculus, Galgotia Publisher, New Delhi, India,
pp. 1-70
104
Singer M. F., Saunders B. D. Caviness B. F. (1985), An Extension of Liouville's
Theorem on Integration in Finite Terms, SIAM J. Comput., Society for Industrial and
Applied Mathematics, Vol.14, No.4, Nov., pp.966-990
Strange G. (1991), Calculus, Wesseley-Cambridge Press,
www.math.gatech. edu
., pp.177-
200
Subbotin I. Y., Hill M., Bilotskii N. N., An algorithmic approach to elementary functions,
National University, USA; Yavneh Hebrew Academy, USA; National Pedagogical
University, Ukrain, pp.1-6
Symbolic Integration,
http://en.wikipedia.org/wiki/Symbolic _integration
, Accessed on 30
th
Jan. 2012
Taylor Series Expansions, Winter-2011
Trager B. M. (1984), Integration of Algebraic Functions, Ph. D. Thesis, Massachusetts
Institute of Technology, September, pp.1-81
Vilenkin N. Ya. Bloch A. Ya. (1978), Elementary functions in the school course of
mathematics, Mathematics in School, Vol.3, p.53-57
Walter J S. (1971), A Unified algorithm for elementary functions, Hewlett-Packard
Company, Palo Alto, California, Spring Joint Computer Conference, p.379-385
Weisstein E. W. (2011), "Indefinite Integral", From Math World--A Wolfram Web
Resource,
http://mathworld.wolfram.com/
IndefiniteIntegral.html , Accessed on 30
th
Dec.,
2011
105
Weisstein E. W. (2011), "Dawson's Integral", From Math World--A Wolfram Web
Resource,
http://mathworld.wolfram.com/Dawson Integral.html
, Accessed on 30
th
Dec.,
2011
Weisstein E. W. (2011), "Frensel Integral", From Math World--A Wolfram Web Resource,
http://mathworld.wolfram.com/ FrenselIntegral.html
, Accessed on 30
th
Dec., 2011
Williams D. P. (1993), Non-elementary Anti-derivatives, Dartmouth College, Hanover,
USA, Dec., pp.1-13
Wolfram Alpha Computational Knowledge Engine, Wolfram Mathematica, A Wolfram
Web Resource,
www.wolframalpha.com
Yadav D. K., Sen D. K. Chauhan R. (2009), "Introduction of a Dominating Function",
International Journal of Mathematical Sciences Engineering Applications, Vol.3, No.III,
121-132, Ascent Publishing House (
www.ascent-journals.com
), Pune, India
Yadav D. K. Sen D. K. (2010), "Properties of Dominating Indefinite Integrable
Functions", International Journal of Mathematical Sciences Engineering Applications,
Vol.4, No.IV, 327-334, Ascent Publishing House (
www.ascent-journals.com
), Pune, India
Yadav D. K. Sen D. K. (2012), "Dominating Function Theory With Reference to Power
Series", International Journal of Mathematical Sciences Engineering Applications, Vol.6,
No.I, 153-159, Ascent Publishing House (
www.ascent-journals.com
), Pune, India
Yadav D. K. Sen D. K. (2012), "Dominating Sequential Functions", International Journal
of Mathematical Sciences Engineering Applications, Vol.6, No.I, 391-401, Ascent
Publishing House (
www.ascent-journals.com
), Pune, India
106
Yadav D. K. Sen D. K. (2012), "General Integrals of Dominating Sequential Functions",
International Journal of Mathematical Sciences Engineering Applications, Vol.6, No. III,
55-62, Ascent Publishing House (
www.ascent-journals.com
), Pune, India
http://mathworld.wolfram.com/IndefiniteIntegral.html
, Accessed on 30
th
Dec., 2011
http://en.wikipedia.org/wiki/trigonometric_functions
http://www.efunda.com/math/taylor_series/hyperbolic.cfm
http://www.efunda.com/math/taylor_series/trig.cfm
Frequently Asked Questions
What are "dominating functions" in mathematics?
Dominating functions are a new class of functions that serve as a superset to classical elementary functions, dominating almost all of them.
What types of dominating functions are introduced?
They are divided into four types: trigonometric, hyperbolic, exponential, and logarithmic functions.
What mathematical problem do these new functions solve?
They provide mathematical notations for many non-elementary functions that are generated during indefinite integration but previously lacked formal representation.
How do elementary functions relate to dominating functions?
Classical elementary functions are shown to be particular cases of these new dominating and sequential functions.
In which fields can these functions be applied?
They are relevant for advanced research in various areas of mathematics, physics, and plasma dispersion studies.
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- Dharmendra Kumar Yadav (Autor:in), 2012, Dominating Sequential Functions: Superset of Elementary Functions, München, GRIN Verlag, https://www.grin.com/document/367784
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