Excerpt
Acknowledgement
The content of this book has been taken from the first two chapters of my thesis of
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.
In thesis and previous research works, I have called nonelementary functions as
`indefinite nonintegrable functions and nonintegrable functions' because it indicates the
exact meaning of the functions whatever I meant. Although the thesis has been
published as an e-book by GRIN Verlag Open Publishing and Lap Lambert Publishing
earlier but due to its importance for further research, the present book has been
published.
Dharmendra Kumar Yadav
Dedicated To My
Parents
Siri Yadav & Lalmuni Devi
Wife
Sanju Kumari
Daughters
Palak & Phalak
Heartiest Thanks To
Dr. D. S. Lal
Retired Professor of Mathematics, V. B. U. Hazaribag, Jharkhand
Dr. Arun Kumar
Retired Professor of Mathematics, V. B. U. Hazaribag, Jharkhand
Dr. S. K. Aggarwal
Professor of Mathematics, V. B. U. Hazaribag, Jharkhand
Dr. A. Kumar
Professor of Mathematics, V. B. U. Hazaribag, Jharkhand
Dr. R. K. Dwivedi
Associate Professor of Mathematics, V. B. U. Hazaribag, Jharkhand
Dr. P. K. Manjhi
Assistant Professor of Mathematics, V. B. U. Hazaribag, Jharkhand
Dr. S. K. Mishra
Professor of Mathematics, P. G. A. N. College, Patna University, Bihar
Dr. Peeyush Tewari
Associate Professor of Mathematics, Birla Institute of Technology, Ext. Centre Noida
for their valuable support during my doctoral research work.
Warmly Presented To
Dr. Shashi Nijhawan, Principal
Dr. Shiv Kumar Sahdev, Associate Professor, Department of Mathematics
Dr. Babita Gupta, Associate Professor, Department of Mathematics
Dr. Aparna Jain, Associate Professor, Department of Mathematics
Dr. Mridula Buddhraja, Associate Professor, Department of Mathematics
Dr. Surabhi Madan, Associate Professor, Department of Mathematics
Dr. Pankaj Kumar Das, Assistant Professor, Department of Mathematics
Mr. Ashesh Kumar Jharwal, Assistant Professor, Department of Mathematics
Dr. Kumari Priyanka, Assistant Professor, Department of Mathematics
Dr. Vandana Rajpal, Assistant Professor, Department of Mathematics
Dr. Neetu Rani, Assistant Professor, Department of Mathematics
Mr. Uttam Kumar Sinha, Assistant Professor, Department of Mathematics
Dr. Shilpi Verma, Assistant Professor, Department of Mathematics
Mr. Manish Kumar Meena, Assistant Professor, Department of Mathematics
Mr. Jitendra Singh, Assistant Professor, Department of Mathematics
Dr. Jitendra Aggarwal, Assistant Professor, Department of Mathematics
Ms. Deepti, Assistant Professor, Department of Mathematics
Mr. Ankush Kumar, Assistant Professor, Department of Mathematics
Mr. Satish Kumar, Assistant Professor, Department of Mathematics
Mr. Navnit Kumar Yadav, Assistant Professor, Department of Mathematics
Mr. Ajay Kumar, Assistant Professor, Department of Mathematics
Ms. Moon Moon Roy, Assistant Professor, Department of Mathematics
Shivaji College, University of Delhi, Raja Garden, Delhi, India
Contents
Chapters Page No.
1. Introduction 1 6
Function and Elementary Function 1
Range and Difficulty of Problem of Indefinite Integration 2
Existence of Integrals and Lack of Notations 2
Algorithms on Elementary and Nonelementary Functions 3
Strong Liouville's Theorem 4 - 5
2. Six Conjectures 7 - 83
2.1 Conjecture-1 7 24
Examples on Conjecture-1 25 - 28
2.2 Conjecture-2 29 30
Examples on Conjecture-2 30 - 32
2.3 Conjecture-3 33 - 48
Examples on Conjecture-3 48 - 50
2.4 Conjecture-4 51 - 62
Examples on Conjecture-4 62 - 64
2.5 Conjecture-5 65 - 71
Examples on Conjecture-5 71 - 73
2.6 Conjecture-6 74 - 82
Examples on Conjecture-6 82 - 83
Reference 84 - 88
1
Chapter-1
Introduction
Calculus cannot be imagined without functions. The concept of function is very
important because of its 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.
To study the present book we need the knowledge of special type of functions
traditionally known as elementary functions. The concept of elementary functions is
more useful than functions as far as the nonelementary functions are concerned.
Formally we define these two as follows:
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.
Elementary Function: We call a function an elementary function if it can be expressed
as an equation in the form of y=f(x), where f represents an expression formed by the
combination of a finite collection of powers of x, trigonometric functions, inverse
trigonometric functions, hyperbolic functions, inverse hyperbolic functions,
exponentials, and logarithms together through additions, subtractions, multiplications,
divisions, powers, and compositions. All functions are not elementary. A common
example of a non-elementary function is a piecewise-defined function. Other types of
nonelementary functions arise in calculus as "anti-derivatives or indefinite integration"
of elementary functions. The six conjectures are based on this concept.
2
Indefinite Integration: Gottfried Leibniz (1684) defined 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, it is denoted 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 consider that f(x)
belongs to some class of functions S. Then we may ask whether F(x) is a member of S,
or can be expressed, according to some standard mode of expression, in terms of
functions which are members of S. The range and difficulty of the problem depend upon
the choice of:
(i) a class of functions, and
(ii) a standard mode of expression.
We take S as the class of elementary functions, and the mode of expression is taken as
the explicit expression in finite terms.
Note: 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 functions. These
functions can also be named as indefinite nonintegrable or nonintegrable functions. In
other words, if we say that an indefinite integral
f (x)dx
is elementary 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. But it cannot be guaranteed 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-
3
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.
Algorithms on Elementary and Nonelementary Functions:
Many efforts have been done to find the algorithm for elementary and nonelementary
integration of elementary functions in the past. In short some of them are as follows:
John Bernoulli Conjecture (1702): The integral of any rational function is expressible
in terms of other rational functions, trigonometric functions, and logarithmic functions.
Laplace Conclusion (1812): The integral of a rational function of x, e
x
and logx is
either a rational function of those functions or the sum of such a rational function and of
a finite number of constant multiples of logarithms of similar functions.
Laplace's Theorem: A rational function has an anti-derivative and its integral is always
an elementary function. In general it is composed of two parts: one of a rational function
and the other the transcendental part or logarithmic part.
Laplace Conjecture: The integral of an algebraic function need contain only those
algebraic functions which are present in the integrand. This conjecture was proved by
Abel.
N. H. Abel's Theorem (1826, 1829): If y is an algebraic function of x and if the
integral of y is algebraic, the integral is rational in y and x.
Joseph Liouville (1833) based his work on the fact that the derivative of an elementary
function is again an elementary function created a framework for constructive
integration by finding out when indefinite integrals of elementary functions are again
4
elementary functions. The main results on functions with nonelementary integrals began
with Liouville's results. The first problem considered by him in the field of integration
in finite terms deals with the integration of general algebraic functions. He introduced a
theorem, which is reminiscent of Laplace's theorem, now known as Liouville's First
Theorem on Integration or simply Liouville's theorem as stated below:
Liouville's First Theorem on Integration: If an algebraic function is integrable in
finite terms, its anti-derivative is the finite sum of an algebraic function and the
logarithms of algebraic functions.
In mathematical symbols, if f(x) is an algebraic function of x and if
dx
x
f
)
(
is
elementary, then
n
i
i
i
o
U
C
U
dx
x
f
1
)
log(
)
(
where the C
i
's are constants and the U
i
's are algebraic functions of x.
In 1835 he generalized this theorem to several variables and gave Strong Liouville's
theorem, and thereby greatly extended the class of functions one can prove to have
nonelementary integrals.
Strong Liouville's theorem:
(a) If F is an algebraic function of
,
,...,
,
,
2
1
m
y
y
y
x
where
,
,...,
,
2
1
m
y
y
y
are functions of
x whose derivatives
dx
dy
dx
dy
dx
dy
m
..,
,...
,
2
1
are rational functions of
,
,...,
,
,
2
1
m
y
y
y
x
then
dx
y
y
y
x
F
m
,...,
,
,
2
1
is elementary if and only if
n
j
j
j
m
U
C
U
dx
y
y
y
x
F
1
0
2
1
)
log(
,...,
,
,
where the C
j
's are constants, and the U
j
's are algebraic functions of
.
,...,
,
,
2
1
m
y
y
y
x
5
(b) If
m
y
y
y
x
F
,...,
,
,
2
1
is a rational function and
dx
dy
dx
dy
dx
dy
m
..,
,...
,
2
1
are
rational functions of
,
,...,
,
,
2
1
m
y
y
y
x
then the U
j
's in part (a) must be rational functions
of
.
,...,
,
,
2
1
m
y
y
y
x
Thereafter in the same year 1835 he found the special case of this theorem, which gives
the necessary and sufficient conditions for the existence of elementary function of some
special functions.
Strong Liouville's theorem (special case):
If f(x) and g(x) are rational functions with g(x) non-constant, then
dx
e
x
f
x
g )
(
)
(
is elementary if and only if there exists a rational function R(x) such that
).
(
'
)
(
)
(
'
)
(
x
g
x
R
x
R
x
f
For any such R(x),
( )
( )
g x
R x e
is an elementary anti-derivative of f(x).
By applying the above theorems he proved that the following integrals
( )
dx
P x
,
2
x
e dx ,
x
e
dx
x
,
dx
e
2
x
,
dx
x
e
x
,
dx
x
x
sin
,
dx
x
x
cos
,
x
log
dx
cannot be expressed in terms of elementary functions.
In fact, he proved that some integrals such as
2
0
2
( )
;
x
t
Erf x
e dt the error function
1
2
2
0
1
sin
int
1
;
x
st
F x m
m
t
dt the elliptic
egral of the
kind
cannot be expressed in terms of a finite number of elementary functions. They were thus
the first integration problem which seemed impossible to properly solve in the sense of
Leibniz. He showed also that the elliptic integrals of the first and second kinds have no
6
elementary expressions. By 1841, Liouville had developed a theory of integration that
settled the question of integration in finite terms for many important cases.
In an invitation paper E. A. Marchisotto G. A. Zakeri (1994) mentioned two
important examples (4 and 5) obtained from the special case of strong Liouville's
theorem, which are very useful in deciding the elementary and nonelementary functions:
Example 4:
dx
e
x
ax
n
2
2
for n an integer, is non-elementary for
0
a
.
For n=0 and a=-1, this is the error function. By this property, it can be proved that the
following functions are nonelementary:
dt
e
t
dx
x
t
2
2
2
log
,
dt
e
dx
x
t
2
2
log
1
,
dt
e
dx
x
e
at
ax
2
2
Example 5:
dx
e
x
cx
n
for n a positive integer and c a nonzero constant, is
nonelementary.
Using these examples following functions can be proved nonelementary:
dt
t
e
dx
e
t
e
x
,
dt
t
e
dx
x
t
log
1
,
dx
x
x
x
dx
x
log
1
)
log(log
)
log(log
,
.
Im
sin
dx
x
e
dx
x
x
ix
The above theorems and properties are sufficient to study the six conjectures discussed
in the present book. Following the basis of the works by the pioneers of the subject
Yadav has propounded six types of nonelementary functions as conjectures. He used
the term `conjectures' because he has proved them nonelementary only for particular
cases. The proof for higher degree or other possible cases are still open for further
research.
7
Chapter-2
2.1 Conjecture-1
An indefinite integral of the form
dx
x
f
e
x
f
)
(
'
)
(
, where f(x) is a polynomial of degree 2,
or a trigonometric (not inverse trigonometric) function, or a hyperbolic (not inverse
hyperbolic) function is always nonelementary.
Proof: We shall prove it in three different cases:
Case I: When f(x) is a polynomial function of degree 2. Then from strong
Liouville's theorem (special case),
dx
x
f
e
x
f
)
(
'
)
(
is elementary if and only if there exists a rational function R(x) such that
1
R '(x) R(x)f '(x)
f '(x)
(2.1.A)
Let
p(x)
R(x)
q(x)
, where g.c.d.(p(x), q(x))=1. Then we have from (2.1.A)
2
2
f '(x)q(x)p '(x) f '(x)p(x)q '(x) [f '(x)] p(x)q(x) [q(x)]
(2.1.B)
2
f '(x)p(x)q '(x)
f '(x)p '(x) q(x) [f '(x)] p(x)
q(x)
Which implies that
q(x) f '(x) as q(x) cannot divide p(x) and
q '(x) . In this case either
q(x)=k, a constant or a polynomial of degree less than or equal to the degree of
f '(x) .
For q(x)=k, from (2.1.B) we have
2
2
f '(x)kp '(x) [f '(x)] p(x)k k
(2.1.C)
8
Comparing the degrees of x in (2.1.C), we find that degree of x in right hand side is 0
whereas the degree of x in left hand side is greater than or equal to 2, which results out
in a contradiction. So q(x) cannot be a constant.
For q(x) a polynomial of degree less than or equal to the degree of
f '(x) , let us assume
that
f '(x) =q(x).h(x). Then from (2.1.B), we get
2
2
q(x)h(x)q(x)p '(x) q(x)h(x)p(x)q '(x) [q(x)h(x)] p(x)q(x) [q(x)]
2
h(x)q(x)p '(x) h(x)p(x)q '(x) [q(x)h(x)] p(x) q(x)
2
h(x)p(x)q '(x)
h(x)p '(x) 1 q(x)[h(x)] p(x)
q(x)
(2.1.D)
Which implies that q(x)|h(x), since q(x) cannot divide p(x) and
q '(x) . Let
h(x)=q(x).(x). Then from (2.1.D), we have
3
2
q(x) (x)p '(x) p(x)q '(x) (x) [q(x)] [ (x)] p(x) 1
(2.1.E)
Comparing the degrees of x in both sides of (2.1.E), we find that degree of x in right
hand side is 0 whereas the degree of x in left hand side is greater than or equal to 1,
which again results out in a contradiction. Therefore such R(x) does not exist, i. e., the
given function is nonelementary.
Particular Case: Let us take f(x)=x
2
+bx+c. Then
2
f (x)
x
bx c
e
e
dx
dx
f '(x)
2x b
Which is elementary if and only if there exists a rational function R(x) such that
1
R '(x) (2x b)R(x)
(2x b)
(2.1.F)
Let
p(x)
R(x)
q(x)
, where g.c.d.(p(x), q(x))=1.
Then from (2.1.F) we have
9
2
q(x)p '(x) p(x)q '(x) (2x b)p(x)q(x)
1
[q(x)]
(2x b)
(2.1.G)
2
2
(2x b)q(x)p '(x) (2x b)p(x)q '(x) (2x b) p(x)q(x) [q(x)]
2
(2x b)p(x)q '(x)
(2x b)p '(x) q(x) (2x b) p(x)
q(x)
Which implies that q(x)|(2x+b) since q(x) cannot divide p(x) and
q '(x) . In this case
either q(x) is a constant k or k(2x+b).
When q(x)=k, from (2.1.G) we have
2
kp '(x) (2x b)p(x)k
1
k
(2x b)
k
p '(x) (2x b)p(x)
(2x b)
2
(2x b)p '(x) (2x b) p(x) k
(2.1.H)
Comparing the degrees of x in both sides of (2.1.H), we get that the degree of x in right
hand side is 0, whereas the degree of x in left hand side is greater than or equal to 2 for
any polynomial p(x), which results out in a contradiction. So q(x) cannot be a constant.
When q(x)=k(2x+b), then from (2.1.G) we have
2
2
k(2x b)p '(x) p(x)2k k(2x b) p(x)
1
[k(2x b)]
(2x b)
2
(2x b)p '(x) 2p(x) (2x b) p(x) k(2x b)
(2.1.I)
Again comparing the degrees of x in both sides of (2.1.I), we get a contradiction. Thus
we conclude that no such R(x) exists. Therefore the given function is nonelementary for
f(x)=x
2
+bx+c. Similarly we can prove it nonelementary for higher degree polynomial
f(x).
Case II: When f(x) be a trigonometric (not inverse trigonometric) function and
(x) be a polynomial of degree 1. Then
10
2.1.1 For sine function, we have
f (x )
sin (x)
e
dx
e
dx
f '(x)
'(x)cos (x)
On putting sin(x)=z, we get
sin (x)
z
2
2
e
dx
e dz
'(x) cos (x)
[ '(x)] (1 z )
(2.1.1.1)
Sub-case I: When (x)=x+b. Then from (2.1.1.1) we have
z
z
2
2
2
e dz
e dz
[ '(x)] (1 z )
(1 z )
z
z
1
e dz
e dz
2
(1 z)
(1 z)
Where,
z
p
e dz
e
e
dp,[Putting1 z p]
(1 z)
p
From strong Liouville's theorem (special case), it is elementary if and only if there
exists a rational function R(p) such that
1
R '(p) R(p)
p
(2.1.1.1A)
Let
P(p)
R(p)
, where gcd{P(p),Q(p)} 1
Q(p)
Then from (2.1.1.1.A) we have
2
{Q(p)}
pQ(p)P '(p) pP(p)Q '(p) P(p)Q(p)
(2.1.1.1.B)
pP(p)Q'(p)
pP '(p) Q(p) pP(p)
Q(p)
Q(p) p
11
Because Q(x) cannot divide P(p) and Q'(p), which means that Q(p)=k a or Q(p)=kp,
where k is a constant.
When Q(p)=k, from (2.1.1.1.B) we have
pP'(p) pP(p) k
(2.1.1.1.C)
Comparing the degrees of p on both sides of (2.1.1.1.C) results out in a contradiction.
Hence Q(p)k.
When Q(p)=kp, then from (2.1.1.1.B) we have
kp pP'(p) (1 p)P(p)
Which is not true for any polynomial P(p). Thus we conclude that such R(p) does not
exist i. e., this integral is nonelementary.
Similarly
z
p
e dz
1 e
dp,[Putting 1 z p]
(1 z)
e p
is elementary if and only if there exists a rational function R(p) such that
1
R '(p) R(p)
p
Taking
P(p)
R(p)
, where gcd{P(p),Q(p)} 1
Q(p)
We get
pP(p)Q '(p)
pP '(p) Q(p) pP(p)
Q(p)
Giving similar argument for Q(p) as above we find that such R(p) does not exist i. e.,
this integral is also nonelementary.
12
We can also deduce from example 1.4.2 (page 16) that both the integrals
p
p
e
e
dp and
dp
p
p
are nonelementary. Therefore the given function is also nonelementary.
Sub-case II: When (x)=x
2
+bx+c. Then we have from (2.1.1.1)
z
2
2
e dz
[ '(x)] (1 z )
z
2
1
2
1
e dz
b
4c
, where k
4 [sin z k](1 z )
4
z
1
2
2
1
e dz
4 [sin z k] (1 z ) (1 z )
z
2
1
F z, e , 1 z ,sin z dz
1
2
3
F z, y , y , y dz
z
3
1
2
1
2
2
2
2
dy
dy
dy
z
z
1
1
e
y ,
,
dz
dz
y
dz
y
1 z
1 z
Applying strong Liouville's theorem, we find that it is elementary if and only if there
exists an identity of the form
z
n
0
i
i
1
2
i 1
i
dU
U '
e
c
4[sin z k](1 z )
dz
U
where each U
j
is a function of z, y
1
, y
2
, and y
3
. Also by strong Liouville's theorem
(special case), each U
j
must contain e
z
. Considering different forms of U
j
like
1
z
z
1
log[(sin z k)e ],e log(sin z k)
etc. we find that no such U
j
exist. Therefore the given function is nonelementary.
Similarly we can prove it nonelementary for higher degree polynomial (x).
Note: In sub-case-II, we can replace sin
-1
z by its logarithmic representation
1
2
sin z
i log[iz
1 z ]
13
In this case we have
z
2
2
e dz
[ '(x)] (1 z )
z
2
2
2
1
e dz
b
4c
, where k
4
4
[ i log{iz
1 z } k](1 z )
z
2
2
i
e dz
4 [log{iz
1 z } ik](1 z )
z
2
2
F z, e , 1 z , log{iz
1 z } dz
1
2
3
F z, y , y , y dz
z
3
1
2
2
1
2
2
2
2
dy
dy
dy
(iy
z)
z
z
e
y ,
,
dz
dz
y
dz
y (iz y )
1 z
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
z
n
0
i
i
2
2
i 1
e
d
U
C log U
dz
4[log{iz
1 z } ik](1 z )
since left hand side contains e
z
, by strong Liouville's theorem (special case), each U
j
must contain e
z
. Considering different forms of U
j
like
z
2
2
z
e log{log(iz
1 z ) ik},log[{log(iz
1 z ) ik}e }
etc., we find that such U
j
does not exist. Therefore the given function is nonelementary.
2.1.2 For cosine function, we have
f (x )
cos (x )
e
dx
e
dx
f '(x)
'(x) sin (x)
On putting cos(x)=z, we get
cos (x)
z
z
2
2
2
2
e
dx
e dz
e dz
'(x) sin (x)
[
'(x)] (1 z )
[ '(x)] (1 z )
(2.1.2.1)
Sub-case I: When (x)=x+b. Then from (2.1.2.1) we have
14
z
z
2
2
2
e dz
e dz
[ '(x)] (1 z )
(1 z )
which is nonelementary, proved in section 2.1.1 sub-case-I.
Sub-case II: When (x)=x
2
+bx+c. Then from (2.1.2.1) we have
z
z
2
2
1
2
e dz
1
e dz
[ '(x)] (1 z )
4 (cos z k)(1 z )
z
2
1
F[z, e , 1 z , cos z]dz
A similar argument will hold as in section 2.1.1: sub-case-II to prove it nonelementary.
We can prove it nonelementary by taking
1
2
cos z
i log[z
z
1].
2.1.3 For tangent function, we have
f (x )
tan (x)
2
e
dx
e
dx
f '(x)
'(x) sec
(x)
On putting tan(x)=z, we get
tan (x)
z
2
2
2
e
dx
e dz
'(x)sec
(x)
[ '(x)] (1 z )
(2.1.3.1)
Sub-case I: When (x)=x+b. Then from (2.1.3.1) we have
z
z
2
2
2
e dz
e dz
[ '(x)] (1 z )
(1 z )
z
z
1
e dz
e dz
2
(1 iz)
(1 iz)
Now
z
i
ip
e dz
e
e
dp,[on putting (1 iz) p]
(1 iz)
i
p
By strong Liouville's theorem (special case), it is elementary if and only if there exists a
rational function R(x) which satisfies the identity
15
1
R '(p) iR(p)
p
1
R(p) 0 and R '(p)
p
But R(p) cannot be zero, so such R(p) does not exist. Hence it is nonelementary.
Also
z
ip
i
e dz
e
ie
dp,[on putting (1 iz) p]
(1 iz)
p
Again by strong Liouville's theorem (special case), it is elementary if and only if there
exists a rational function R(x) which satisfies the identity
1
R '(p) iR(p)
p
1
R(p) 0 and R '(p)
p
But R(p) cannot be zero, so such R(p) does not exist. Hence it is nonelementary.
Therefore the given function is nonelementary.
Sub-case II: When (x)=x
2
+bx+c. Then from (2.1.3.1) we have
z
z
2
2
2
1
2
e dz
1
e dz
b
4c
,k
[ '(x)] (1 z )
4 [tan z k](1 z )
4
z
2
1
F[z, e , (1 z ), tan z]dz
1
2
3
F[z, y , y , y ]dz
z
3
1
2
1
2
2
dy
dy
dy
1
1
e
y ,
2z,
dz
dz
dz
1 z
y
By strong Liouville's theorem, it is elementary if and only if there exists an identity
containing U
j
, a function of z, y
1
, y
2
, and y
3
of the form
z
n
0
i
i
1
2
i 1
i
dU
U '
e
c
4[tan z k](1 z )
dz
U
Again by strong Liouville's theorem (special case), each U
j
or its derivatives
must
contain e
z
. Considering different forms of U
j
like e
z
log(tan
-1
z+k), log[e
z
(tan
-1
z+k)], etc.
16
we find that no such U
j
exist, i. e., no such identity exist. Hence the given function is
nonelementary. Similarly we can prove it nonelementary for higher degree polynomial
(x).
We can prove it nonelementary by taking
1
i
i z
tan z
log
2
i z
2.1.4 For cotangent function, we have
f (x )
cot (x )
2
e
dx
e
dx
f '(x)
'(x) cos ec (x)
On putting cot(x)=z, we get
cot (x )
z
2
2
2
e
dx
e dz
'(x) cos ec (x)
[ '(x)] (1 z )
(2.1.4.1)
Sub-case-I: When (x)=x+b, we have from (2.1.4.1)
z
z
2
2
2
e dz
e dz
[ '(x)] (1 z )
(1 z )
Which is nonelementary, proved in section 2.1.3, sub-case-I.
Sub-case-II: When (x)=x
2
+bx+c, we have from (2.1.4.1)
z
z
2
2
2
1
2
e dz
1
e dz
b
4c
, k
[ '(x)] (1 z )
4 (cot z k)(1 z )
4
z
2
1
F[z, e , (1 z ), cot z]dz
A similar argument will hold as in section 2.1.3 to prove it nonelementary.
We can prove it nonelementary by taking
1
1
1 iz
cot z
log
2i
1 iz
17
2.1.5 For cosecant function, we have
f (x )
cosec (x )
e
dx
e
dx
f '(x)
'(x)cos ec (x)cot (x)
On putting cosec(x)=z, we get
cosec (x )
z
2 2
2
e
dx
e dz
'(x) cos ec (x)cot (x)
[ '(x)] z (z
1)
(2.1.5.1)
Sub-case I: When (x)=x+b. Then from (2.1.5.1) we have
z
z
2 2
2
2
2
e dz
e dz
[ '(x)] z (z
1)
z (z
1)
z
z
2
2
e dz
e dz
(z
1)
z
Where the first integral
z
2
e dz
(z
1)
is nonelementary as proved in section 2.1.1, sub-case-I and the second integral
z
2 z
2
e dz
z e dz
z
is also nonelementary from example-1.4.2. So the given function is nonelementary.
Sub-case II: When (x)=x
2
+bx+c, then from (2.1.5.1) we have
z
2 2
2
e dz
[ '(x)] z (z
1)
z
2 2
2
e dz
[2x b] z (z
1)
z
2
1
2
2
e dz
b
4c
,k
4[cos ec z k]z (z
1)
4
z
2
1
F[z, e , z
1, cos ec z]dz
1
2
3
F[z, y , y , y ]dz
z
3
1
2
1
2
2
2
2
dy
dy
dy
z
z
1
1
e
y ,
,
dz
dz
y
dz
z y
z
1
z z
1
18
By strong Liouville's theorem, it is elementary if and only if there exists an identity
containg U
j
, a function of z, y
1
, y
2
, and y
3
of the form
z
n
0
i
i
1
2
2
i 1
i
dU
U '
e
c
4[cos ec z k]z (z
1)
dz
U
Again from strong Liouville's theorem (special case), each U
j
and its derivatives must
contain e
z
. Considering different forms of U
j
like e
z
log[cosec
-1
z+k], log[e
z
(cosec
-1
z+k)],
etc., we find that no such U
j
exist. Hence the given function is nonelementary. Similarly
we can prove it nonelementary for higher degree polynomial (x).
We can prove it nonelementary by taking
2
1
z
1 i
cos ec z
i log
z
2.1.6 For secant function, we have
f (x )
sec (x)
e
dx
e
dx
f '(x)
'(x) sec (x) tan (x)
On putting sec(x)=z, we get
sec (x)
z
2 2
2
e
dx
e dz
'(x) sec (x) tan (x)
[ '(x)] z (z
1)
(2.1.6.1)
Sub-case-I: When (x)=x+b, we have from (2.1.6.1)
z
z
2 2
2
2
2
e dz
e dz
[ '(x)] z (z
1)
z (z
1)
Which is nonelementary, proved in section 2.1.5, sub-case-I.
Sub-case-II: When (x)=x
2
+bx+c, we have from (2.1.6.1)
z
z
2
2 2
2
1
2
2
e dz
1
e dz
b
4c
,k
[ '(x)] z (z
1)
4 [sec z k]z (z
1)
4
19
z
2
1
F[z, e , z
1,sec z]dz
It can now be proved nonelementary by the similar procedures as has been applied in
section 2.1.5.
We can prove it nonelementary by taking
2
1
1
1 z
sec z
i log
z
Case III: When f(x) be a hyperbolic (not inverse hyperbolic) function and (x) is a
polynomial of degree 1. Then,
2.1.7 For sine hyperbolic function, we have
f (x )
sinh (x)
e
dx
e
dx
f '(x)
'(x) cosh (x)
On putting sinh(x)=z, we get
sinh (x )
z
2
2
e
dx
e dz
'(x) cosh (x)
[ '(x)] (1 z )
(2.1.7.1)
Sub-case I: When (x)=x+b. Then from (2.1.7.1) we have
z
z
2
2
2
e dz
e dz
[ '(x)] (1 z )
(1 z )
which is nonelementary, proved in section 2.1.3: sub-case-I.
Sub-case II: When (x)=x
2
+bx+c, then we have from (2.1.7.1)
z
z
2
2
2
2
e dz
e dz
[ '(x)] (1 z )
[2x b] (1 z )
z
2
1
2
e dz
b
4c
, where k
4(sinh z k)(1 z )
4
z
2
1
F[z, e , 1 z ,sinh z]dz
1
2
3
F[z, y , y , y ]dz
20
z
3
1
2
1
2
2
2
2
dy
dy
dy
z
z
1
1
e
y ,
,
dz
dz
y
dz
y
1 z
1 z
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
z
n
0
i
i
1
2
i 1
i
dU
U '
e
c
4(sinh z k)(1 z )
dz
U
Again from strong Liouville's theorem (special case), each U
j
must contain e
z
.
Considering different possible forms of U
j
like e
z
log[sinh
-1
z+k], log[e
z
(sinh
-1
z+k)], etc.
we find that no such U
j
exist. Hence the given function is nonelementary. Similarly we
can prove it nonelementary for higher degree polynomial (x).
We can prove it nonelementary by taking
1
2
sinh z log(z
z
1)
2.1.8 For cosine hyperbolic function, we have
f (x )
cosh (x)
e
dx
e
dx
f '(x)
'(x) sinh (x)
On putting cosh(x)=z, we get
cosh (x )
z
2
2
e
dx
e dz
'(x)sinh (x)
[ '(x)] (z
1)
(2.1.8.1)
Sub-case-I: When (x)=x+b, we have from (2.1.8.1)
z
z
2
2
2
e dz
e dz
[ '(x)] (z
1)
(z
1)
Which is nonelementary, proved in section 2.1.1, sub-case-I.
Sub-case-II: When (x)=x
2
+bx+c, we have from (2.1.8.1)
21
z
z
2
2
2
1
2
e dz
1
e dz
b
4c
,k
[ '(x)] (z
1)
4 (cosh z k)(z
1)
4
z
2
1
F[z, e , z
1, cosh z]dz
It can now be proved nonelementary by the similar procedures as has been applied in
section 2.1.7, sub-case-II. Similarly we can prove it nonelementary for higher degree
polynomial (x).
We can prove it nonelementary by taking
1
2
cosh z log(z
z
1)
2.1.9 For tangent hyperbolic function, we have
f (x )
tanh (x)
2
e
dx
e
dx
f '(x)
'(x)sec h (x)
On putting tanh(x)=z, we have
tanh (x)
z
2
2
2
e
dx
e dz
'(x)sec h (x)
[ '(x)] (1 z )
(2.1.9.1)
Sub-case-I: When (x)=x+b, we have from (2.1.9.1)
z
z
2
2
2
e dz
e dz
[ '(x)] (1 z )
(1 z )
Which is nonelementary, proved in section 2.1.1, sub-case-I.
Sub-case-II: When (x)=x
2
+bx+c, we have from (2.1.9.1)
z
z
2
2
2
1
2
e dz
1
e dz
b
4c
,k
[ '(x)] (1 z )
4 (tanh z k)(1 z )
4
z
2
1
F[z, e , (1 z ), tanh z]dz
22
It can now be proved nonelementary by the similar procedures as has been applied in
section 2.1.3: sub-case-II. Similarly we can prove it nonelementary for higher degree
polynomial (x).
We can prove it nonelementary by taking
1
1
1 z
tanh z
log
2
1 z
2.1.10 For cotangent hyperbolic function, we have
f (x )
cot h(x)
2
e
dx
e
dx
f '(x)
'(x) cos ech (x)
On putting coth(x)=z, we get
cot h(x )
z
2
2
2
e
dx
e dz
'(x) cos ech (x)
[ '(x)] (z
1)
(2.1.10.1)
Sub-case-I: When (x)=x+b, we have from (2.1.10.1)
z
2
2
e dz
[ '(x)] (z
1)
z
2
e dz
(z
1)
Which is nonelementary, proved in section 2.1.1, sub-case-I.
Sub-case-II: When (x)=x
2
+bx+c, we have from (2.1.10.1)
z
z
2
2
2
1
2
e dz
1
e dz
b
4c
,k
[ '(x)] (z
1)
4 (coth z k)(z
1)
4
z
2
1
F[z, e , (z
1), coth z]dz
It can now be proved nonelementary by the similar procedures as has been applied in
section 2.1.3 sub-case-II. Similarly we can prove it nonelementary for higher degree
polynomial (x).
We can prove it nonelementary by taking
23
1
1
z 1
coth z
log
2
z 1
2.1.11 For cosecant hyperbolic function, we have
f (x)
cos ec h (x)
e
dx
e
dx
f '(x)
'(x) cos ech (x)coth (x)
On putting cosech(x)=z, we get
cosec h (x)
z
2 2
2
e
dx
e dz
'(x)cos ech (x)coth (x)
[ '(x)] z (z
1)
(2.1.11.1)
Sub-case-I: When (x)=x+b, we have from (2.1.11.1)
z
z
2 2
2
2
2
e dz
e dz
[ '(x)] z (z
1)
z (z
1)
z
z
2
2
e dz
e dz
z
z
1
Both are nonelementary, proved in section 2.1.5, sub-case-I and section 2.1.7, sub-case-
I respectively. So (2.1.11.1) is nonelementary.
Sub-case-II: When (x)=x
2
+bx+c, we have from (2.1.11.1)
z
z
2
2 2
2
1
2
2
e dz
1
e dz
b
4c
, k
[ '(x)] z (z
1)
4 (cos ech z k)z (z
1)
4
z
2
1
F[z, e , z
1, cos ec h z]dz
It can now be proved nonelementary by strong Liouville's theorem by the similar
procedures as has been applied in section 2.1.5 sub-case-II. Similarly we can prove it
nonelementary for higher degree polynomial (x).
We can prove it nonelementary by taking
2
1
1
1 z
cos ec h z log
z
24
2.1.12 For secant hyperbolic function, we have
f (x )
sec h (x)
e
dx
e
dx
f '(x)
'(x) sec h (x) tanh (x)
On putting sech(x)=z, we get
sec h (x)
z
2 2
2
e
dx
e dz
'(x) sec h (x) tanh (x)
[ '(x)] z (1 z )
(2.1.12.1)
Sub-case-I: When (x)=x+b, we have from (2.1.12.1)
z
z
2 2
2
2
2
e dz
e dz
[ '(x)] z (1 z )
z (1 z )
z
z
2
2
e dz
e dz
z
1 z
Both are nonelementary, proved in section 2.1.5, sub-case-I and section 2.1.1, sub-case-
I respectively. So (2.1.12.1) is nonelementary.
Sub-case-II: When (x)=x
2
+bx+c, we have from (2.1.12.1)
z
z
2 2
2
1
2
2
e dz
1
e dz
[ '(x)] z (1 z )
4 (sec h z k)z (1 z )
z
2
1
F[z, e , 1 z ,s ec h z]dz
It can now be proved nonelementary by strong Liouville's theorem by the similar
procedures as has been applied in section 2.1.5: sub-case-II. Similarly we can prove it
nonelementary for higher degree polynomial (x).
We can prove it nonelementary by taking
2
1
1
1 z
s ec h z log
z
25
Examples on Conjecture-1
Example 2.1.1: Show that the integral
2
ax
b
e
dx, a 0
x
is nonelementary.
Proof: We have
2
2
ax
b
ax
b
e
e
e
dx
dx
dx
x
x
x
2
ax
b
2
2axe
e log x
dx
2ax
Now putting ax
2
=z we get
2
ax
z
1 z
2
2axe
1 e
1
dx
dz
z e dz
2ax
2 z
2
which is nonelementary from example-5 (page-6). Hence the given integral is
nonelementary.
Example 2.1.2: Show that the integral
dx
x
e
x
cos
sin
is nonelementary.
Proof: We have
sin x
sin x
2
e
e
cos x
dx
dx
cos x
cos x
On putting sinx=z, we get
sin x
z
2
2
e
cos x
e dz
dx
cos x
(1 z )
which is nonelementary, proved in section 2.1.1, sub-case-I. Hence the given integral is
nonelementary.
Example 2.1.3: Show that the integral
dx
x
e
x
sin
cos
is nonelementary.
Proof: We have
cos x
cos x
2
e
e
( sin x)
dx
dx
sin x
( sin x)
26
On putting cosx=z, we have
cos x
z
2
2
e
( sin x)
e dz
dx
( sin x)
(1 z )
Which is nonelementary, proved in section 2.1.1, sub-case-I. Hence the given integral is
nonelementary.
Example 2.1.4: Show that the integral
dx
x
e
x
2
tan
sec
is nonelementary.
Proof: We have, on putting tanx=z
tan x
z
2
2 2
e
e
dx
dz
sec x
(1 z )
z
z
2
2
1
e dz
1
e dz
4 (iz)(1 iz)
4 (iz)(1 iz)
(2.1.4.A)
On putting 1-iz = p in the first integral of (2.1.4.A)
z
ip
ip
ip
i
2
2
e dz
e dp
e dp
e dp
ie
(iz)(1 iz)
(1 p)
p
p
(2.1.4.B)
where the second and third integrals are nonelementary from example-5 (page-6). Now
putting 1-p=X in the first integral of (2.1.4.B) we have
ip
iX
i
e dp
e dX
e
(1 p)
X
which is nonelementary from example-5 (page-6). Hence the first integral of (2.1.4.A) is
nonelementary. Similarly we can prove that the second integral of (2.1.4.A) is also
nonelementary. Therefore the given integral is nonelementary.
Example 2.1.5: Show that the integral
dx
x
e
x
cosh
sinh
is nonelementary.
Proof: We have on putting sinhx=z
sinh x
z
2
e
e
dx
dz
cosh x
(1 z )
27
Which is nonelementary, proved in section 2.1.3, sub-case-I. Hence the given integral is
nonelementary.
Example 2.1.6: Show that the integral
cot x
2
e
dx
cos ec x
is nonelementary.
Proof: We have on putting z=cotx
cot x
z
2
2
e
e
dx
dz
cos ec x
(1 z )
Which is nonelementary, proved in section 2.1.3, sub-case-I. Hence the given integral is
nonelementary.
Example 2.1.7: Show that the integral
dx
x
x
e
x
tan
.
sec
sec
is nonelementary.
Proof: We have on putting secx=z
secx
z
2
2
e
e
dx
dz
sec x.tan x
z (z
1)
z
z
2
2
e
e
dz
dz
(z
1)
z
Both are nonelementary, proved in section 2.1.5, sub-case-I. Therefore the given
integral is nonelementary.
Example 2.1.8: Show that the integral
dx
x
ecx
e
ecx
cot
.
cos
cos
is nonelementary.
Proof: We have on putting cosecx=z,
cosecx
z
2
2
e
e
dx
dz
cos ecx.cot x
z (z
1)
z
z
2
2
e
e
dz
dz
(z
1)
z
Which is nonelementary, proved in section 2.1.5, sub-case-I. Hence the given integral is
nonelementary.
Example 2.1.9: Show that the integral
dx
x
e
x
2
sin
2
sin
is nonelementary.
Proof: We have on putting sin
2
x=z,
28
2
sin x
z
2
e
1
e dz
dx
sin 2x
4 z(1 z )
z
z
2
1
ze dz
e dz
4
(1 z )
z
Where
z
e
dz
z
is nonelementary from example-5(page-6).
Now since
z
z
z
2
ze
1
e dz
1
e dz
dz
(1 z )
2 (1 z) 2 (1 z)
where,
z
p
e dz
e
e
dp
(1 z)
p
, on putting 1-z=p, which is nonelementary and
z
p
e dz
1 e
dp
(1 z)
e p
on putting 1+z=p, which is also nonelementary, from example-5
(page-6). Hence the given integral is nonelementary.
Example 2.1.10: Show that the integral
dx
x
x
e
x
2
sin
cos
.
2
2
is nonelementary.
Proof: We have on putting sinx
2
=z,
2
sin x
z
2
2
1
e
e dz
dx
2x.cos x
4(1 z )sin z
z
2
1
F z, e , 1 z ,sin z dz
1
2
3
F z, y , y , y dz
z
3
1
2
1
2
2
2
2
dy
dy
dy
z
z
1
1
e
y ,
,
dz
dz
y
dz
y
1 z
1 z
By strong Liouville's theorem, it is elementary if and only if there exists an identity
containing U
j
a function of z, y
1
, y
2
, and y
3
of the form
z
n
o
i
i
2
1
i 1
i
dU
U '
e
C
dz
U
(1 z ) sin z
Taking different possible forms of U
j
like
z
1
z
2
1
e log sin z,e 1 z log sin z
etc., we find that no such U
j
exist. Hence the given integral is nonelementary.
29
2.2 Conjecture-2
An indefinite integral of the form
dx
x
g
e
x
f
)
(
)
(
, where f(x) and g(x) are polynomial
functions in x of degree greater than or equal to 1, is always nonelementary.
Proof: Applying strong Liouville's theorem (special case), we find that the integral
f (x)
e
dx
g(x)
is elementary if and only if there exists a rational function R(x) which satisfies an
identity of the form
1
R '(x) R(x)f '(x)
g(x)
(2.2.1)
Let
p(x)
R(x)
q(x)
, where g.c.d.(p(x), q(x))=1, then from (2.2.1), we have
2
g(x)q(x)p '(x) g(x)p(x)q '(x) g(x)p(x)q(x)f '(x) [q(x)]
g(x)p(x)q '(x)
g(x)p '(x) q(x) g(x)p(x)f '(x)
q(x)
(2.2.1.1)
Which implies that q(x) divides g(x), since gcd(p(x), q(x))=1 and q(x) cannot divide
q'(x) . Let g(x)=q(x).r(x), then from (2.2.1.1), we have
q(x)r(x)p '(x) q(x) r(x)p(x)q(x)f '(x) p(x)q '(x)r(x)
r(x)p(x)q '(x)
r(x)p '(x) 1 r(x)p(x)f '(x)
q(x)
(2.2.1.2)
Which implies that q(x) divides r(x). Let r(x)=q(x).(x), then we have from (2.2.1.2),
q(x) (x)p '(x) 1 r(x) (x)p(x)f '(x) p(x)q '(x) (x)
30
q(x) (x)p '(x) q(x) (x)p(x)f '(x) p(x)q '(x) (x) 1
(x)[q(x)p '(x) q(x)p(x)f '(x) p(x)q '(x)] 1
(2.2.1.3)
Comparing the degrees of x in both sides, we get that the degree of x in left hand side is
2, whereas the degree of x in right hand side is 0, which results out in a contradiction.
So, such R(x) i. e. such identity given by (2.2.1) does not exist. Therefore, the integral
of this form is nonelementary.
Examples on Conjecture-2
Example 2.2.1: Show that the integral
dx
x
e
x
is nonelementary.
Proof: It is nonelementary proved in section 2.1.1. We can also deduce this from
example-5 (page-6).
Example 2.2.2: Show that the integral
dx
x
e
x
x
)
1
(
2
is nonelementary.
Proof: Applying strong Liouville's theorem (special case), we find that the integral
dx
x
e
x
x
)
1
(
2
is elementary if and only if there exists a rational function R(x), which satisfies an
identity of the form
1
R '(x) (2x 1)R(x)
(x 1)
(2.2.2.1)
Let
p(x)
R(x)
q(x)
where gcd(p(x), q(x))=1.
31
Then from (2.2.2.1) we have
2
(x 1)q(x)p '(x) (x 1)p(x)q '(x) (2x 1)(x 1)p(x)q(x) [q(x)]
(x 1)p(x)q '(x)
(x 1)p '(x) q(x) (2x 1)(x 1)p(x)
q(x)
(2.2.2.2)
which implies that q(x) divides (x+1) i.e., either q(x) = k a constant or q(x)=k(x+1).
When q(x) = k, we have from (2.2.2.2)
(x 1)p '(x) k (2x 1)(x 1)p(x) 0
(x 1)p'(x) (2x 1)(x 1)p(x) k
Comparing the degrees of x in both sides, we get a contradiction. Therefore q(x) cannot
be a constant.
When q(x) = k(x+1), we have from (2.2.2.2)
(x 1)p '(x) k(x 1) (2x 1)(x 1)p(x) p(x)
Again comparing the degrees of x in both sides, we get a contradiction. Therefore we
find that q(x) cannot be equal to k(x+1). Finally we conclude that no such R(x) exist.
Hence the given integral is nonelementary.
Example 2.2.3: Show that the integral
dx
x
x
e
x
x
x
1
2
2
2
2
3
is nonelementary.
Proof: Applying strong Liouville's theorem (special case), we find that the integral
dx
x
x
e
x
x
x
1
2
2
2
2
3
is elementary if and only if there exists a rational function R(x), which satisfies an
identity of the form
2
2
1
R '(x) (3x
4x 1)R(x)
(x
2x 1)
(2.2.3.1)
32
Let
p(x)
R(x)
q(x)
Where gcd(p(x), q(x))=1. Then from (2.2.3.1) we have
2
2
q(x)
q(x)p '(x) p(x)q '(x) (3x
4x 1)p(x)q(x)
(x 1)
(2.2.3.2)
Which implies that q(x)=(x+1)r(x) for some polynomial r(x). Putting it in (2.2.3.2), we
get
2
2
(x 1)r(x)p '(x) p(x)r(x) (x 1)p(x)r '(x)
(3x
4x 1)(x 1)p(x)r(x)
r(x)
(2.2.3.3)
2
(x 1)p(x)r '(x)
(x 1)p '(x) p(x) r(x) (3x
4x 1)(x 1)p(x)
r(x)
But r(x) cannot divide p(x) as q(x) cannot divide p(x), which implies that r(x) divides
r'(x) or (x+1). When r(x) divides r'(x), it means that r(x) is a constant. Then we have
from (2.2.3.3)
2
2
k(x 1)p '(x) kp(x) (3x
4x 1)(x 1)p(x)k k
2
(x 1)p '(x) p(x) (3x
4x 1)(x 1)p(x) k
Comparing the degrees of x in both sides now results out in a contradiction. Therefore
r(x) cannot be a constant. When r(x) divides (x+1), let r(x)=k(x+1). Then from (2.2.3.3)
we have
2
2
2
2
2
k(x 1) p '(x) (3x
4x 1)(x 1) p(x)k k (x 1)
2
p '(x) (3x
4x 1)p(x) k
comparing the degrees of x in both sides again results out in a contradiction. Therefore
r(x) cannot be equal to k(x+1). Finally we conclude that no such R(x) exist. Hence the
given integral is nonelementary.
33
2.3 Conjecture-3
An indefinite integral of the form
dx
x
g
x
f
)
(
)
(
; where f(x) is a trigonometric (not inverse
trigonometric) function, or a hyperbolic (not inverse hyperbolic) function, and g(x) is a
polynomial of degree greater than or equal to 1, is always nonelementary.
Proof: We shall prove it in two different cases:
Case-I: When f(x) is a trigonometric (not inverse trigonometric) function and (x)
be a polynomial of degree 1. Then
2.3.1 For f(x)=sin(x), we have
f (x)
sin (x)
dx
dx
g(x)
g(x)
i (x)
i (x )
1 e
1 e
dx
dx
2i g(x)
2i g(x)
(2.3.1.1)
By strong Liouville's theorem,
i (x )
e
dx
g(x)
is elementary if and only if there exists a
rational function R(x) which satisfies an identity of the form
1
R '(x) i '(x)R(x)
g(x)
1
R '(x)
g(x)
and R(x)=0. But when R(x)=0, R'(x) cannot be
1
g(x)
. Thus no such
R(x) exist i.e., it is nonelementary. Similarly we can prove that the integral
i (x )
e
dx
g(x)
is
nonelementary. Therefore the given integral (2.3.1.1) is nonelementary.
2.3.2 For f(x)=cos(x), we have
f (x)
cos (x)
dx
dx
g(x)
g(x)
i (x)
i (x )
1 e
1 e
dx
dx
2 g(x)
2 g(x)
(2.3.2.1)
34
Both the integrals of (2.3.2.1) are nonelementary, proved in section 2.3.1. Therefore the
given integral is nonelementary.
2.3.3 For f(x)=tan(x), we have
f (x)
tan (x)
sec (x) tan (x) '(x)
dx
dx
dx
g(x)
g(x)
'(x)g(x) sec (x)
On putting sex(x)=x, we get
tan (x)
dz
dx
g(x)
zg(x) '(x)
(2.3.3.1)
Sub-case I: When (x)=x+b, then from (2.3.3.1) we have
1
dz
dz
zg(x) '(x)
zg(sec z b)
2
2
1
z
1dz
z z
1g(sec z b)
(2.3.3.2)
For g(x)=x+a, we have from (2.3.3.2)
2
2
1
tan (x)
z
1dz
dx
, k a b
g(x)
z z
1(sec z k)
2
1
F z, z
1, sec z dz
1
2
F z, y , y dz
1
2
2
2
1
1
dy
dy
z
z
1
1
,
dz
y
dz
zy
z
1
z z
1
Applying strong Liouville's theorem we find that it is elementary if and only if there
exists an identity of the form
'
n
'
i
0
i
1
i 1
i
U
1
U
C
(sec z k)z
U
Considering different possible forms of U
j
, like
35
1
log(sec z k)
we find that no such U
j
exist. Hence the given function is nonelementary. Similar
arguments can be given for higher degree polynomial g(x).
Sub-case-II: When (x)=x
2
+bx+c, then from (2.3.3.1) we have
tan (x)
dx
g(x)
2
2
2
1
1
z
1dz
b
4c
b
,k
, B
4
2
2z z
1 sec z kg
sec z k B
(2.3.3.3)
For g(x)=x+a, (2.3.3.3) becomes
2
2
1
1
z
1dz
, A a B
2z z
1 sec z k
sec z k A
2
1
1
F z, z
1,sec z, sec z k dz
1
2
3
F z, y , y , y dz
3
1
2
2
2
1
1
3 1
dy
dy
dy
z
z
1
1
1
,
,
dz
y
dz
zy
dz
2zy y
z
1
z z
1
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
'
n
i
0
i
1
1
i 1
i
U
1
U
C
U
2z sec z k
sec z k A
Considering different possible forms of U
j
like
1
1
log( sec z k A), sec z k
We find that no such U
j
exist. Therefore the given function is nonelementary. Similarly
we can prove it nonelementary for higher degree polynomial g(x).
36
2.3.4 For f(x)=cot(x), we have
f (x)
cot (x)
dx
dx
g(x)
g(x)
On putting cosec(x)=z, we get
cos ec (x) cot (x) '(x)
dz
dx
g(x) cos ec (x) '(x)
g(x) '(x)z
(2.3.4.1)
Sub-case-I: When (x)=x+b, then we have from (2.3.4.1)
1
dz
dz
g(x) '(x)z
zg(cos ec z b)
For g(x)=x+a, it becomes
1
dz
, k a b
z(cos ec z k)
2
2
1
z
1dz
z z
1(cos ec z k)
2
1
F[z, z
1, cos ec z]dz
which can now be proved nonelementary by the similar procedures as has been applied
in section 2.3.3. Similarly it can be proved nonelementary for higher degree polynomial
g(x).
Sub-case-II: When (x)=x
2
+bx+c, then we have from (2.3.4.1) for g(x)=x+a,
1
1
dz
dz
g(x) '(x)z
2( cos ec z k B)z cos ec z k
2
1
2
1
z
1dz
2( cos ec z k B)z z
1 cos ec z k
2
1
1
F[z, z
1, cos ec z, cos ec z k ]dz
37
which can now be proved nonelementary by the same procedures as has been applied in
section 2.3.3. Similarly it can be proved nonelementary for higher degree polynomial
g(x).
2.3.5 For f(x)=sec(x), we have
i (x )
i (x)
f (x)
sec (x)
2dx
dx
dx
g(x)
g(x)
g(x) e
e
i (x )
i2 (x )
2e
dx
g(x) e
1
On putting e
i(x)
=z, it becomes
2
2
dz
i g(x) '(x)(z
1)
(2.3.5.1)
Sub-case-I: When (x)=x+b, then from (2.3.5.1) we have
2
2
dz
i g(x) '(x)(z
1)
2
2
dz
i g( i log z b)(z
1)
(2.3.5.2)
For g(x)=x+a, it becomes
2
2
dz
i ( i log z a b)(z
1)
2
dz
2
{log z i(a b)}(z
1)
2
dz
2
,i(a b) k
(log z k)(z
1)
F z,log z dz
1
F z, y dz
1
dy
1
dz
z
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
'
n
'
i
0
i
2
i 1
i
U
1
U
C
(log z k)(z
1)
U
38
Considering different possible forms of U
j
like log(logz+k), log[(logz+k)(z
2
+1)], etc. we
find that no such U
j
exist. Therefore the given function is nonelementary. Similarly we
can prove it nonelementary for higher degree polynomial g(x).
Sub-case-II: When (x)=x
2
+bx+c, then from (2.3.5.1) we have
2
2
dz
i g(x) '(x)(z
1)
2
2
dz
b
4c
, k
4
g(x)i(z
1) log z k
For g(x)=x+a, it becomes
2
dz
b
, M a
2
i( log z k M)(z
1) log z k
F[z, log z, log z k ]dz
1
2
F[z, y , y ]dz
1
2
2
dy
dy
1
1
1
,
dz
z dz
2zy
2z log z k
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
'
n
i
0
i
2
i 1
i
U
1
U
C
U
i( log z k M)(z
1) log z k
Considering different possible forms of U
j
, like
log( log z k M)
etc., we find that no such U
j
exist. Therefore the given function is nonelementary.
Similarly we can prove it nonelementary for higher degree polynomial g(x).
2.3.6 For f(x)=cosec(x), we have
i (x )
i (x)
f (x)
cosec (x)
2idx
dx
dx
g(x)
g(x)
g(x) e
e
39
On putting e
i(x)
=z, it becomes
2
dz
2
g(x) '(x)(z
1)
(2.3.6.1)
Sub-case-I: When (x)=x+b, then we have from (2.3.6.1)
2
2
2dz
2dz
g(x) '(x)(z
1)
g(x)(z
1)
For g(x)=x+a, it becomes
1
2
2dz
, k b a
(cos ec z k)(z
1)
1
2
2
2zdz
(cos ec z k)z z
1 z
1
2
1
F[z, z
1, cos ec z]dz
1
2
F[z, y , y ]dz
By strong Liouville's theorem it is elementary if and only if there exists an identity of
the form
'
n
i
0
i
1
2
i 1
i
U
2
U
C
(cos ec z k)(z
1)
U
But no such U
j
exist. Therefore the given function is nonelementary. Similarly we can
prove it nonelementary for higher degree polynomial g(x).
Sub-case-II: When (x)=x
2
+bx+c, then we have from (2.3.6.1)
2
2
2
1
2dz
dz
b
4c
, k
g(x) '(x)(z
1)
4
g(x)(z
1) cos ec z k
For g(x)=x+a, it becomes
2
1
1
dz
b
, M a
2
(z
1)( cos ec z k M) cos ec z k
40
2
1
1
F[z, z
1, cos ec z, cos ec z k ]dz
1
2
3
F[z, y , y , y ]dz
By strong Liouville's theorem it is elementary if and only if there exists an identity of
the form
'
n
i
0
i
2
1
1
i 1
i
U
1
U
C
U
(z
1)( cos ec z k M) cos ec z k
But no such U
j
exist. Therefore the given function is nonelementary. Similarly we can
prove it nonelementary for higher degree polynomial g(x). We can extend this proof for
higher degree polynomials (x) and g(x).
Case-II: When f(x) is a hyperbolic (not inverse hyperbolic) function and (x) be an
arbitrary polynomial of degree 1. Then
2.3.7 For f(x)=sinh(x), we have
(x )
(x)
f (x)
1 e
1 e
dx
dx
dx
g(x)
2 g(x)
2 g(x)
(2.3.7.1)
By strong Liouville's theorem, the first integral in (2.3.7.1),
(x )
e
dx
g(x)
is elementary if
and only if there exists a rational function R(x) satisfying an identity of the form
1
R '(x)
'(x)R(x)
g(x)
(2.3.7.2)
Let
p(x)
R(x)
; where gcd(p(x), q(x)) 1
q(x)
Then from (2.3.7.2) we get
2
1
q(x)p '(x) p(x)q '(x)
'(x)p(x)q(x)
g(x)
q(x)
41
p(x)g(x)q '(x) q(x)[g(x)p '(x) q(x)
'(x)p(x)g(x)]
p(x)g(x)q '(x)
[g(x)p '(x) q(x)
'(x)p(x)g(x)]
q(x)
(2.3.7.3)
Since g.c.d.(p(x), q(x))=1, we have from (2.3.7.3) either q(x) divides q'(x) or q(x)
divides g(x).
Sub-case-I: When q(x)|q'(x), it means q(x) is a constant. Then without loss of
generality we may take R(x)=p(x). Comparing the degrees of x on the two sides of
(2.3.7.3) now results out in a contradiction, since the right hand side has degree 0 in x,
but the left hand side has degree 1. Hence q(x) cannot divide q'(x).
Sub-case-II: When q(x)|g(x), let g(x)=(x)q(x), i.e.,
g(x)
q(x)
(x)
; where degree of
(x)degree of g(x). Then from (2.3.7.3), we have
p(x) (x)q(x)q '(x)
(x)q(x)p '(x) q(x)
'(x)p(x) (x)q(x)
q(x)
p(x) (x)q '(x)
(x)p '(x) 1
'(x)p(x) (x)
q(x)
(2.3.7.4)
But q(x) does not divide p(x) and q'(x), then it must divide (x). If q(x)|(x), let
(x)=q(x)r(x). Then from (2.3.7.4) we have
q(x)r(x)p '(x) 1
'(x)p(x)q(x)r(x) p(x)r(x)q '(x)
r(x)[q(x)p '(x)
'(x)p(x)q(x) p(x)q '(x)] 1
Which again results out in a contradiction by comparing the degrees of x in both sides
i.e., such R(x) does not exist. Hence this integral is nonelementary.
Similarly we can prove that the second integral in (2.3.7.1)
(x )
e
dx
g(x)
is
nonelementary. Therefore the given integral is nonelementary.
42
2.3.8 For f(x)=cosh(x), we have
(x)
(x)
f (x)
1 e
1 e
dx
dx
dx
g(x)
2 g(x)
2 g(x)
Both are nonelementary proved in section 2.3.7. Therefore the given integral is also
nonelementary.
2.3.9 For f(x)=tanh(x), we have
f (x)
tanh (x)
dx
dx
g(x)
g(x)
sec h (x).tanh (x). '(x)dx
g(x) '(x)sec h (x)
On putting sech(x)=z, it becomes
dz
g(x) '(x)z
(2.3.9.1)
Sub-case-I: When (x)=x+b, then from (2.3.9.1) we have
dz
dz
g(x) '(x)z
zg(x)
For g(x)=x+a, it becomes
1
dz
, M a b
z(sec h z M)
2
2
1
1 z dz
z 1 z (sec h z M)
2
1
F[z, 1 z ,sec h z]dz
1
2
F[z, y , y ]dz
1
2
2
2
1
1
dy
dy
z
z
1
1
,
dz
y
dz
zy
1 z
z 1 z
By strong Liouville's theorem it is elementary if and only if there exists an identity of
the form
43
'
n
i
0
i
1
i 1
i
U
1
U
C
z(sec h z M)
U
But no such U
j
exist, which satisfies this identity. Therefore the given function is
nonelementary. Similarly we can prove it nonelementary for higher degree polynomial
g(x).
Sub-case-II: When (x)=x
2
+bx+c, then from (2.3.9.1) we have
2
1
dz
dz
b
4c
, k
g(x) '(x)z
4
2zg(x) sec h z k
For g(x)=x+a, it becomes
1
1
dz
b
, B a
2
2z( sec h z k B) sec h z k
2
2
1
1
1 z dz
2z 1 z ( sec h z k B) sec h z k
2
1
1
F[z, 1 z ,sec h z, sec h z k ]dz
1
2
3
F[z, y , y , y ]dz
3
1
2
1
1
1 3
dy
dy
dy
z
1
1
,
,
dz
y
dz
zy
dz
2zy y
By strong Liouville's theorem it is elementary if and only if there exists an identity of
the form
'
n
i
0
i
1
1
i 1
i
U
1
U
C
U
2z sec h z k ( sec h z k B)
But no such U
j
exist, which satisfies this identity. Therefore the given function is
nonelementary. Similarly we can prove it nonelementary for higher degree polynomial
g(x).
44
2.3.10 For f(x)=coth(x), we have
f (x)
coth (x)
dx
dx
g(x)
g(x)
cosec h (x).coth (x). '(x)dx
g(x) '(x)co sec h (x)
On putting cosech(x)=z, it becomes
dz
g(x) '(x)z
(2.3.10.1)
Sub-case-I: When (x)=x+b, then from (2.3.10.1) we have
dz
dz
g(x) '(x)z
zg(x)
For g(x)=x+a, it becomes
1
dz
, B a b
z(cos ech z B)
2
2
1
z
1dz
z z
1(cos ech z B)
2
1
F[z, z
1,cos ech z]dz
1
2
F[z, y , y ]dz
1
2
2
2
1
1
dy
dy
z
z
1
1
,
dz
y
dz
zy
z
1
z z
1
By strong Liouville's theorem it is elementary if and only if there exists an identity of
the form
'
n
i
0
i
1
i 1
i
U
1
U
C
z(cos ech z B)
U
Considering different possible forms of U
j
, like
1
1
1
log[z(cos ech z B)],
log(cos ech z B)
z
45
etc., we find that no such U
j
exist, which satisfies this identity. Therefore the given
function is nonelementary. Similarly we can prove it nonelementary for higher degree
polynomial g(x).
Sub-case-II: When (x)=x
2
+bx+c, then from (2.3.10.1) we have
2
1
dz
dz
b
4c
, k
g(x) '(x)z
4
2zg(x) cos ech z k
For g(x)=x+a, it becomes
1
1
dz
b
, B a
2
2z cos ech z k ( cos ech z k B)
2
1
1
F[z, z
1, cos ech z, cos ech z k ]dz
which can now be proved nonelementary by strong Liouville's theorem as has been
proved in sub-case-I above. Similarly we can prove it nonelementary for higher degree
polynomial g(x). We can extend this proof for higher degree polynomials (x) and g(x).
2.3.11 For f(x)=sech(x), we have
(x )
(x )
f (x)
sec h (x)
2dx
dx
dx
g(x)
g(x)
g(x) e
e
On putting e
(x)
=z, it becomes
2
dz
2
g(x) '(x)(z
1)
(2.3.11.1)
Sub-case-I: When (x)=x+b, then from (2.3.11.1) we have
2
2
2dz
2dz
g(x) '(x)(z
1)
g(x)(z
1)
46
For g(x)=x+a, it becomes
2
2dz
,B a b
(log z B)(z
1)
1
F[z, log z]dz
F[z, y ]dz,
1
dy
1
dz
z
By strong Liouville's theorem it is elementary if and only if there exists an identity of
the form
'
n
i
0
i
2
i 1
i
U
2
U
C
(z
1)(log z B)
U
But no such U
j
exist, which satisfies this identity. Therefore the given function is
nonelementary. Similarly we can prove it nonelementary for higher degree polynomial
g(x).
Sub-case-II: When (x)=x
2
+bx+c, then from (2.3.11.1) we have
2
2
2
2dz
2dz
b
4c
, k
g(x) '(x)(z
1)
4
g(x)(z
1) log z k
For g(x)=x+a, it becomes
2
2dz
b
, B a
2
(z
1) log z k ( log z k B)
1
2
F[z, log z, log z k ]dz
F[z, y , y ]dz
Which is nonelementary, proved in section 2.3.5: sub-case-II. Similarly we can prove it
nonelementary for higher degree polynomial g(x).
In a similar process, we can extend the proof for higher degree polynomials (x) and
g(x).
47
2.3.12 For f(x)=cosech(x), we have
(x )
(x)
f (x)
cosec h (x)
2dx
dx
dx
g(x)
g(x)
g(x) e
e
On putting e
(x)
=z, it becomes
2
dz
2
g(x) '(x)(z
1)
(2.3.12.1)
Sub-case-I: When (x)=x+b, then from (2.3.12.1) we have
2
2
2dz
2dz
g(x) '(x)(z
1)
g(x)(z
1)
For g(x)=x+a, it becomes
2
2dz
, B a b
(log z B)(z
1)
1
F[z, log z]dz
F[z, y ]dz,
which can now be proved nonelementary by the similar procedures as has been applied
in sections 2.3.5 and 2.3.11. Similarly it can be proved nonelementary for higher degree
polynomial g(x).
Sub-case-II: When (x)=x
2
+bx+c, then from (2.3.12.1) we have
2
2
2
2dz
dz
b
4c
, k
g(x) '(x)(z
1)
4
g(x)(z
1) log z k
For g(x)=x+a, it becomes
2
2dz
b
, B a
2
(z
1) log z k ( log z k B)
1
2
F[z, log z, log z k ]dz
F[z, y , y ]dz
48
Which can now be proved nonelementary by the similar procedures as has been applied
in section 2.3.5, sub-case-II. Similarly we can prove it nonelementary for higher degree
polynomial g(x). We can extend this proof for higher degree polynomials (x) and g(x).
Examples on Conjecture-3
Example 2.3.1: Show that the integral
dx
x
x
sin
is nonelementary.
Proof: It is nonelementary from example 12 (Marchisotto et al, 41, pp.302, 1994).
Alternate Proof: We have using Euler's identity
ix
sin x
e
dx img
dx
x
x
Taking g(x) = ix, f(x) = 1/x, and applying strong Liouville's theorem (special case), we
find that it is elementary if and only if there exists a rational function R(x) which
satisfies the identity
1
R '(x) iR(x)
x
1
R '(x)
R(x) 0
x
which is impossible. Hence the given function is nonelementary.
Example 2.3.2: Show that the integral
dx
x
x
cosh
is nonelementary.
Proof: We have
x
x
cosh x
1 e
1 e
dx
dx
dx
x
2 x
2
x
Both are well proved nonelementary functions follow from examples-5 (page-6) as well
as proved in section 2.1.1. Therefore the given integral is nonelementary.
49
Example 2.3.3: Show that the integral
2
sinh x
dx,a 0
(ax
b)
is nonelementary.
Proof: We have
x
x
2
2
2
sinh x
1
e
1
e
dx
dx
dx
(ax
b)
2 (ax
b)
2 (ax
b)
Where
x
x
x
2
2
e
1
e
e
b
dx
dx
dx ;k
(ax
b)
2iak
(x ik)
(x ik)
a
Both are nonelementary, proved in section 2.1.3, sub-case-I.
Similarly
x
x
x
2
2
e
1
e
e
b
dx
dx
dx ;k
(ax
b)
2iak
(x ik)
(x ik)
a
Both are nonelementary, proved in section 2.1.3, sub-case-I. Therefore the given
function is nonelementary.
Example 2.3.4: Show that the integral
tan x
dx
x
is nonelementary.
Proof: We have
tan x
sec x tan x
dx
dx
x
x sec x
On putting secx=z, it becomes
2
1
2
1
dz
z
1dz
z sec z
z z
1sec z
2
1
F z, z
1, sec z dz
1
2
F z, y , y dz
1
2
2
1
1
dy
dy
z
z
1
,
dz
y
dz
zy
z
1
50
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
'
n
'
i
0
i
1
i 1
i
U
1
U
C
z sec z
U
But no such U
j
exist. Hence the given function is nonelementary.
Example 2.3.5: Show that the integral
dx
b
x
ax
x
)
(
tan
2
3
is nonelementary.
Proof: We have
3
2
3
2
tan x
sec x tan x
dx
dx
(ax
x
b)
(ax
x
b)sec x
On putting secx=z, it becomes
3
2
dz
(ax
x
b)z
1
3
1
2
dz
z a(sec z)
(sec z)
b
2
1
F[z, z
1, sec z]dz
1
2
F[z, y , y ]dz
1
2
2
1
1
dy
dy
z
z
1
,
dz
y
dz
zy
z
1
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
'
n
'
i
0
i
1
3
1
i 1
i
U
1
U
C
z[a(sec z)
(secc z) b]
U
Giving different values of a and b, we can find that no such U
j
exist. Hence the given
function is nonelementary.
51
2.4 Conjecture-4
An indefinite integral of the form
dx
e
x
f
)
(
, where f(x) is a trigonometric (not inverse
trigonometric) function, or a hyperbolic (not inverse hyperbolic) function, or a
polynomial of degree greater than or equal to 2, is always nonelementary.
Proof: We shall prove it in three different cases:
Case I: When f(x) be a polynomial of degree 2.
Then by strong Liouville's theorem
dx
e
x
f
)
(
is elementary if and only if there exists a
rational function R(x) which satisfies an identity of the form
R '(x) f '(x)R(x) 1
(2.4.1)
Let
p(x)
R(x)
; where gcd(p(x),q(x)) 1.
q(x)
Then from (2.4.1) we have
2
q(x)p '(x) p(x)q '(x) f '(x)p(x)q(x) [q(x)]
2
q(x)p '(x) f '(x)p(x)q(x) [q(x)]
p(x)q '(x)
p(x)q '(x)
p '(x) f '(x)p(x) q(x)
q(x)
But q(x) cannot divide p(x) as g.c.d.(p(x), q(x))=1. Therefore q(x) divides q'(x), which
means that q(x) is a constant.
Without loss of generality, we can take R(x)=p(x) i. e., R'(x)=p'(x). Then from (2.4.1),
we have
p'(x)+f'(x)p(x)=1 (2.4.2)
52
where degree of p(x)1 as p(x) cannot be a constant. Comparing the degrees of x in
both sides of (2.4.2) now results out in a contradiction, since the left hand side has
degree 1 of x, but the right hand side has degree 0. Hence no such R(x) exist, so the
given function is nonelementary.
Case II: When f(x) be a trigonometric (not inverse trigonometric) function and
(x) be a polynomial of degree 1. Then
2.4.1 For f(x)=sin(x), we have
f (x )
sin (x )
e
dx
e
dx
On putting sin(x)=z, we get
sin ( x )
e
dx
z
2
e dz
'(x) 1 z
(2.4.1.1)
Sub-case-I: For (x)=x+b, we have from (2.4.1.1)
z
z
2
2
e dz
e dz
'(x) 1 z
1 z
z
2
F[z, e , 1 z ]dz
By strong Liouville's theorem (special case), it is elementary if and only if there exists
an identity of the form
'
z
n
i
0
i
2
i 1
i
U
e
U
C
U
1 z
But no such U
j
exist. Hence the given function is nonelementary.
Sub-case-II: For (x)=x
2
+bx+c, we have from (2.4.1.1)
z
z
2
2
e dz
e dz
'(x) 1 z
(2x b) 1 z
z
2
1
2
1
e dz
b
4c
; k
2
4
sin z k 1 z
z
2
1
F[z, e , 1 z ,sin z]dz
53
Which is nonelementary by strong Liouville's theorem, proved in section 2.1.1, sub-
case-II. Similarly we can prove it nonelementary for higher degree polynomial (x).
2.4.2 For f(x)=cos(x), we have
f (x )
cos (x )
e
dx
e
dx
On putting cos(x)=z, we get
z
cos (x )
2
e dz
e
dx
'(x) 1 z
(2.4.2.1)
Sub-case-I: For (x)=x+b, the integral (2.4.2.1) becomes
z
z
z
2
2
2
e dz
e dz
F[z,e , 1 z ]dz
'(x) 1 z
1 z
(2.4.2.2)
Sub-case-II: For (x)=x
2
+bx+c, the integral (2.4.2.1) becomes
z
z
2
2
1
2
e dz
e dz
b
4c
, k
4
'(x) 1 z
2 cos z k 1 z
z
2
1
F[z, e , 1 z , cos z]dz
(2.4.2.3)
By applying strong Liouville's theorem and the similar procedures as has been applied
in sections 2.1.1, 2.1.2 and 2.4.1, we can prove that both (2.4.2.2) and (2.4.2.3) are
nonelementary.
2.4.3 For f(x)=tan(x), we have
f (x )
tan ( x )
e
dx
e
dx
On putting tan(x)=z, we get
54
z
tan (x)
2
e dz
e
dx
'(x)(1 z )
(2.4.3.1)
Sub-case-I: For (x)=x+b, we have from (2.4.3.1)
z
2
e dz
'(x)(1 z )
z
2
e dz
(1 z )
Which is nonelementary proved in section 2.1.3, subcase-I.
Sub-case-II: For (x)=x
2
+bx+c, we have from (2.4.3.1)
z
2
e dz
'(x)(1 z )
z
2
1
2
e dz
b
4c
;k
4
2 tan z k (1 z )
z
2
1
F[z, e , (1 z ), tan z]dz
Which can be proved nonelementary by the similar procedure applied in section 2.1.3,
sub-case-II.
2.4.4 For f(x)=cot(x), we have
f (x )
cot (x )
e
dx
e
dx
On putting cot(x)=z, we get
z
cot (x )
2
e dz
e
dx
'(x)(1 z )
(2.4.4.1)
Sub-case-I: For (x)=x+b, the integral (2.4.4.1) becomes
z
z
2
2
e dz
e dz
'(x)(1 z )
(1 z )
Which is nonelementary, proved in section 2.1.3, sub-case-I.
Sub-case-II: For (x)=x
2
+bx+c, the integral (2.4.4.1) becomes
55
z
z
2
2
2
1
e dz
e dz
b
4c
, k
'(x)(1 z )
4
2(1 z ) cot z k
z
2
1
F[z, e , (1 z ), cot z]dz
which can now be proved nonelementary by strong Liouville's theorem and the similar
procedures as has been applied in section 2.1.3, sub-case-II.
2.4.5 For f(x)=cosec(x), we have
f ( x )
cos ec ( x )
e
dx
e
dx
On putting cosec(x)=z, we get
z
cosec (x)
2
e dz
e
dx
'(x)z z
1
(2.4.5.1)
Sub-case-I: For (x)=x+b, we have (2.4.5.1)
z
z
2
2
e dz
e dz
'(x)z z
1
z z
1
z
2
F[z, e , z
1]dz
1
2
F[z, y , y ]dz
z
1
2
1
2
2
dy
dy
z
z
e
y ,
dz
dz
y
z
1
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
'
z
n
'
i
0
i
2
i 1
i
U
e
U
C
U
z z
1
By strong Liouville's theorem (special case), each term containing U
j
's must contain e
z
.
By considering different forms of U
j
like
z
2
z
2
e log z z
1 , e
z
1
56
etc. we find that no such U
j
exist. Hence the given function is nonelementary.
Sub-case-II: For (x)=x
2
+bx+c, we have from (2.4.5.1)
z
2
e dz
'(x)z z
1
z
z
2
1
2
e dz
1
e dz
2
(2x b)z z
1
z cos ec z k z
1
z
2
1
F[z, e , z
1, cos ec z]dz
1
2
3
F[z, y ,y , y ]dz
z
1
2
1
2
2
dy
dy
z
z
e
y ,
dz
dz
y
z
1
which is nonelementary by strong Liouville's theorem and the similar procedure as has
been applied in section 2.1.5, sub-case-II. Similarly we can prove it nonelementary for
higher degree polynomial (x).
2.4.6 For f(x)=sec(x), we have
f (x )
sec ( x )
e
dx
e
dx
On putting sec(x)=z, we get
z
sec (x)
2
e dz
e
dx
'(x)z z
1
(2.4.6.1)
Sub-case-I: For (x)=x+b, the integral (2.4.6.1) becomes
z
z
2
2
e dz
e dz
'(x)z z
1
z z
1
Which is nonelementary proved in section 2.4.5, sub-case-I.
Sub-case-II: For (x)=x
2
+bx+c, the integral (2.4.6.1) becomes
z
z
2
2
1
2
e dz
e dz
b
4c
, k
4
'(x)z z
1
2z sec z k z
1
57
z
2
1
F[z, e , z
1,sec z]dz
which can now be proved nonelementary by strong Liouville's theorem and the similar
procedures as has been applied in section 2.4.5, sub-case-II.
Case III: When f(x) be a hyperbolic (not inverse hyperbolic) function and (x) is a
polynomial of degree 1. Then
2.4.7 For f(x)=sinh(x), we have
f ( x)
sinh ( x )
e
dx
e
dx
On putting sinh(x)=z, we get
z
sinh (x )
2
e dz
e
dx
'(x) 1 z
(2.4.7.1)
Sub-case-I: For (x)=x+b, we have from (2.4.7.1)
z
z
2
2
e dz
e dz
'(x) 1 z
1 z
z
2
F[z, e , 1 z ]dz
1
2
F[z, y , y ]dz
z
1
2
1
2
2
dy
dy
z
z
e
y ,
dz
dz
y
1 z
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
z
n
0
i
i
2
i 1
e
d
U
C log U
dz
1 z
where each U
j
's must contain e
z
(by strong Liouville's theorem, special case).
Considering different possible forms of U
j
like
z
2
z
2
e 1 z , e log 1 z
58
etc. we find that no such U
j
exist i.e., the given function cannot be integrable in this
case.
Sub-case-II: For (x)=x
2
+bx+c, we have from (2.4.7.1)
z
z
2
1
2
e dz
e dz
'(x) 1 z
2 sinh z k 1 z
,
2
b
4c
k
4
z
2
1
F[z, e , 1 z ,sinh z]dz
1
2
3
F[z, y , y , y ]dz
z
3
1
2
1
2
2
2
2
dy
dy
dy
z
z
1
1
e
y ,
,
dz
dz
y
dz
y
1 z
1 z
By strong Liouville's theorem it is elementary if and only if there exists an identity of
the form
z
n
0
i
i
1
2
i 1
e
d
U
C log U
dz
sinh z k 1 z
By strong Liouville's theorem (special case), each U
j
must contain e
z
. Considering
different possible forms of U
j
like
z
1
e sin z k
we find that no such U
j
exist. Hence the given function is nonelementary. Similarly it
can be proved nonelementary for higher degree polynomial (x).
2.4.8 For f(x)=cosh(x), we have
f (x)
cosh (x )
e
dx
e
dx
On putting cosh(x)=z, we get
z
cosh (x )
2
e dz
e
dx
'(x) z
1
(2.4.8.1)
59
Sub-case-I: For (x)=x+b, the integral (2.4.8.1) becomes
z
z
2
2
e dz
e dz
'(x) z
1
z
1
z
2
F[z, e , z
1]dz
which can be proved nonelementary by strong Liouville's theorem and the similar
procedure as has been applied in section 2.4.2, sub-case-I.
Sub-case-II: For (x)=x
2
+bx+c, the integral (2.4.8.1) becomes
z
z
2
2
1
2
e dz
e dz
b
4c
,k
4
'(x) z
1
2 cosh z k z
1
z
2
1
F[z, e , z
1, cosh z]dz
which can now be proved nonelementary by strong Liouville's theorem and the similar
procedures as has been applied in section 2.4.7, sub-case-II.
2.4.9 For f(x)=tanh(x), we have
f ( x)
tanh ( x )
e
dx
e
dx
On putting tanh(x)=z, we get
z
tanh (x)
2
e dz
e
dx
'(x)(1 z )
(2.4.9.1)
Sub-case-I: For (x)=x+b, we have from (2.4.9.1)
z
z
2
2
e dz
e dz
'(x)(1 z )
(1 z )
Which is nonelementary, proved in section 2.1.1, sub-case-I.
Sub-case-II: For (x)=x
2
+bx+c, we have from (2.4.9.1)
z
z
2
1
2
e dz
e dz
'(x)(1 z )
2 tanh z k (1 z )
, where
2
b
4c
k
4
60
z
2
1
F[z, e ,1 z , tanh z]dz
1
2
3
F[z, y , y , y ]dz
z
3
1
2
1
2
2
dy
dy
dy
1
1
e
y ,
2z,
dz
dz
dz
1 z
y
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
z
n
0
i
i
1
2
i 1
e
d
U
C log U
dz
tanh z k (1 z )
Again by strong Liouville's theorem (special case), each U
j
must contain e
z
.
Considering different possible form of U
j
like
z
1
e
tanh z k
we find that no such U
j
exist, which can satisfy the above identity. Hence the given
function is nonelementary. Similarly it can be proved nonelementary for higher degree
polynomial (x).
2.4.10 For f(x)=coth(x), we have
f ( x)
coth ( x )
e
dx
e
dx
On putting coth(x)=z, we get
z
coth (x)
2
e dz
e
dx
'(x)(1 z )
(2.4.10.1)
Sub-case-I: For (x)=x+b, the integral (2.4.10.1) becomes
z
z
2
2
e dz
e dz
'(x)(1 z )
(1 z )
which is nonelementary proved in section 2.1.1, sub-case-I.
Sub-case-II: For (x)=x
2
+bx+c, the integral (2.4.10.1) becomes
61
z
z
2
2
1
2
e dz
e dz
b
4c
,k
'(x)(1 z )
4
2 coth z k (1 z )
z
2
1
F[z, e , (1 z ), coth z]dz
which can now be proved nonelementary by strong Liouville's theorem and the similar
procedures as has been applied in section 2.4.9: sub-case-II.
2.4.11 For f(x)=cosech(x), we have
f (x )
cos ech ( x)
e
dx
e
dx
On putting cosech(x)=z, we get
z
cosech (x )
2
e dz
e
dx
'(x)z 1 z
(2.4.11.1)
Sub-case-I: For (x)=x+b, we have from (2.4.11.1)
z
z
2
2
e dz
e dz
'(x)z 1 z
z 1 z
z
2
F[z, e , 1 z ]dz
1
2
F[z, y , y ]dz
z
1
2
1
2
2
dy
dy
z
z
e
y ,
dz
dz
y
1 z
It can now be proved nonelementary by the similar procedures as has been applied in
section 2.4.5, sub-case-I.
Sub-case-II: For (x)=x
2
+bx+c, we have from (2.4.11.1)
z
z
2
2
1
2
e dz
e dz
b
4c
, k
4
'(x)z 1 z
2z cos ech z k 1 z
z
2
1
F[z, e , 1 z , cos ech z]dz
62
It can now be proved nonelementary by the similar procedures as has been applied in
section 2.4.5, sub-case-II.
2.4.12 For f(x)=sech(x), we have
f (x)
sech (x)
e
dx
e
dx
On putting sech(x)=z, we get
z
sech (x)
2
e dz
e
dx
'(x)z 1 z
(2.4.12.1)
Sub-case-I: For (x)=x+b, the integral (2.4.12.1) becomes
z
z
2
2
e dz
e dz
'(x)z 1 z
z 1 z
z
2
F[z, e , 1 z ]dz
which can now be proved nonelementary by the similar procedure as has been given in
section 2.4.5, sub-case-I.
Sub-case-II: For (x)=x
2
+bx+c, the integral (2.4.12.1) becomes
z
z
2
2
1
2
e dz
e dz
b
4c
, k
4
'(x)z 1 z
2z sec h z k 1 z
z
2
1
F[z, e , 1 z ,sech z]dz
which can now be proved nonelementary by strong Liouville's's theorem and the
similar procedures as has been applied in section 2.4.5, sub-case-II.
Examples on Conjecture-4
Example 2.4.1: Show that the integral
dx
e
x
2
is nonelementary.
63
Proof: It is a well proved nonelementary function (example-7, Marchisotto et al, 41,
1994). It follows also from example-4 (page-6).
Alternative Proof: By strong Liouville's theorem (special case) this integral is
elementary if and only if there exists a rational function R(x) such that
1 R '(x) 2xR(x)
(2.4.1.A)
Let
p(x)
R(x)
, where gcd{p(x),q(x)} 1
q(x)
.
Then from (2.4.1.A) we have
p(x)q '(x)
p '(x) q(x) 2xp(x)
q(x)
which implies that q(x) divides q'(x) i. e., q(x) is a constant. Without loss of generality,
we can assume that R(x)=p(x). Then from (2.4.1.A) we have
1 p '(x) 2xp(x)
(2.4.1.B)
where p(x) is a polynomial of degree 1. Comparing the degrees of x on both sides of
(2.4.1.B) now results out in a contradiction. Hence such R(x) does not exist i. e., the
given integral is nonelementary.
Example 2.4.2: Show that the integral
dx
e
x
2
is nonelementary.
Proof: It is a well proved nonelementary function (example-7, Marchisotto et al 41,
1994). It follows also from example-4 (page-6). We can prove it nonelementary by the
similar procedures as has been applied in alternative proof in example 2.4.1.
Example 2.4.3: Show that the integral
dx
e
x
sin
is nonelementary.
Proof: On putting sinx=z, we have
64
z
sin x
2
e dz
e
dx
1 z
Which is nonelementary proved in section 2.4.1 for (x)=x+b in sub-case-I.
Example 2.4.4: Show that the integral
dx
e
x
tan
is nonelementary.
Proof: On putting tanx=z, we get
z
tan x
2
e dz
e
dx
(1 z )
Which is nonelementary, proved in section 2.1.3, sub-case-I.
Example 2.4.5: Show that the integral
dx
e
x
sinh
is nonelementary.
Proof: On putting sinhx=z, we get
z
sinh x
2
e dz
e
dx
1 z
which is nonelementary proved in section 2.4.7 for (x)=x+b, sub-case-I.
Example 2.4.6: Show that the integral
dx
e
x
tanh
is nonelementary.
Proof: On putting tanhx=z, we get
z
tanh x
2
e dz
e
dx
(1 z )
Which is nonelementary, proved in section 2.1.1, sub-case-I.
65
2.5 Conjecture-5
An indefinite integral of the form
dx
x
f
g
)]
(
[
, where f(x) is a polynomial of degree
greater than or equal to 2 and g(x) is a trigonometric (not inverse trigonometric) or
a hyperbolic (not inverse hyperbolic) function is always nonelementary.
Proof: We shall prove it in two different cases:
Case I: When g(x) is a trigonometric (not inverse trigonometric) function. Then
2.5.1 For g(x)=sinx, we have
if (x)
if (x )
e
e
sin f (x)dx
dx
2i
if (x )
if (x)
1
e
dx
e
dx
2i
Both are nonelementary functions proved in section 2.4., case-I.
2.5.2 For g(x)=cosx, we have
if (x)
if (x )
e
e
cos f (x)dx
dx
2
if (x )
if (x)
1
e
dx
e
dx
2
Both are nonelementary functions proved in section 2.4., case-I.
2.5.3 For g(x)=tanx, we have
secf (x).tan f (x).f '(x)
tan f (x)dx
dx
secf (x).f '(x)
On putting secf(x)=z, it becomes
dz
zf '(x)
(2.5.3.1)
For f(x)=x
2
+bx+c, we have from (2.5.3.1)
66
1
dz
dz
zf '(x)
2z sec z k
,
2
b
4c
k
4
2
2
1
1
z
1dz
2 z z 1 sec z k
2
1
F[z, z
1, sec z]dz
1
2
F[z, y , y ]dz
1
2
2
2
1
1
dy
dy
z
z
1
1
,
dz
y
dz
zy
z
1
z z
1
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
'
n
'
i
0
i
1
i 1
i
U
1
U
C
U
2z sec z k
Considering different possible forms of U
j
, like
1
sec z k
, we find that no such U
j
exist. Hence the given function is nonelementary. Similarly it can be proved
nonelementary for higher degree polynomial f(x).
2.5.4 For g(x)=cotx, we have
co sec f (x).co t f (x).f '(x)
cot f (x)dx
dx
co sec f (x).f '(x)
On putting cosecf(x)=z, it becomes
dz
zf '(x)
(2.5.4.1)
For f(x)=x
2
+bx+c, we have from (2.5.4.1)
2
1
dz
dz
b
4c
, k
zf '(x)
4
z cos ec z k
2
2
1
z
1dz
z z
1 cos ec z k
2
1
F[z, z
1, cos ec z]dz
67
which can now be proved nonelementary by the same procedure as has been applied in
section 2.5.3.
2.5.5 For g(x)=cosecx, we have
if (x)
if (x )
dx
2idx
cos ecf (x)dx
sin f (x)
e
e
On putting e
if(x)
=z, it becomes
2
dz
2
f '(x)(z
1)
(2.5.5.1)
For f(x)=x
2
+bx+c, we have from (2.5.5.1)
2
2
2
2dz
2
dz
b
4c
, k
f '(x)(z
1)
4
4i
log z k (z
1)
1
dz
dz
2
i
log z k (z 1)
log z k (z 1)
(2.5.5.2)
Now for
1
dz
I
(z 1) log z k
F[z, log z]dz
1
F[z, y ]dz
,
1
dy
1
dz
z
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
'
n
'
i
0
i
i 1
i
U
1
U
C
U
(z 1) log z k
Considering different possible forms of U
j
, we find that no such U
j
exist. Hence it is
nonelementary.
Similarly
68
2
dz
I
(z 1) log z k
F[z, log z]dz
can be proved nonelementary. Therefore from (2.5.5.2) the given function is
nonelementary. Similarly we can prove it nonelementary for higher degree polynomial
f(x).
2.5.6 For g(x)=secx, we have
if (x)
if (x )
dx
2dx
s ecf (x)dx
cos f (x)
e
e
On putting e
if(x)
=z, it becomes
2
2
dz
i f '(x)(z
1)
(2.5.6.1)
For f(x)=x
2
+bx+c, it becomes
2
1
dz
i
i
log z k (z
1)
1
dz
dz
2i
i
(1 iz) log z k
(1 iz) log z k
Both are nonelementary proved in section 2.5.5. Hence the given function is
nonelementary. Similarly we can prove it nonelementary for higher degree polynomial
f(x).
Case II: When g(x) is a hyperbolic (not inverse hyperbolic) function. Then
2.5.7 For g(x)=sinhx, we have
f (x )
f (x )
e
e
sinh f (x)dx
dx
2
f (x)
f (x )
1
e
dx
e
dx
2
Both are nonelementary functions proved in section 2.4, case-I.
69
2.5.8 For g(x)=coshx, we have
f (x)
f (x)
e
e
cosh f (x)dx
dx
2
f (x)
f (x)
1
e
dx
e
dx
2
Both are nonelementary functions proved in section 2.4, case-I.
2.5.9 For g(x)=tanhx, we have
sec hf (x) tanh f (x)f '(x)
tanh f (x)dx
dx
f '(x) sec hf (x)
On putting sechf(x)=z, we get
dz
tanh f (x)dx
zf '(x)
(2.5.9.1)
For f(x)=x
2
+bx+c, we have from (2.5.9.1)
2
1
dz
dz
b
4c
, k
zf '(x)
4
2z sec h z k
2
2
1
1 z dz
2z 1 z
sec h z k
2
1
F[z, 1 z sec h z]dz
which can now be proved nonelementary by the same procedure as has been applied in
section 2.5.3. Similarly we can prove it nonelementary for higher degree polynomial
f(x).
2.5.10 For g(x)=cothx, we have
cos echf (x) coth f (x)f '(x)
coth f (x)dx
dx
f '(x) cos echf (x)
On putting cosechf(x)=z, it becomes
dz
zf '(x)
(2.5.10.1)
70
For f(x)=x
2
+bx+c, we have from (2.5.10.1)
2
1
dz
dz
b
4c
, k
zf '(x)
4
2z cos ech z k
2
2
1
z
1dz
2z z
1 cos ech z k
2
1
F[z, z
1,cos ech z]dz
which can now be proved nonelementary by the same procedure as has been applied in
section 2.5.3. Similarly we can prove it nonelementary for higher degree polynomial
f(x).
2.5.11 For g(x)=cosechx, we have
f (x )
f (x )
dx
2dx
cos echf (x)dx
sinh f (x)
e
e
On putting e
f(x)
=z, it becomes
2
dz
2
f '(x)(z
1)
(2.5.11.1)
For f(x)=x
2
+bx+c, we have from (2.5.11.1)
2
2dz
f '(x)(z
1)
2
2
dz
b
4c
, k
4
log z k (z
1)
1
dz
dz
2
(z 1) log z k
(z 1) log z k
Both are nonelementary proved in section 2.5.5.
Similarly we can prove it nonelementary for higher degree polynomial f(x).
2.5.12 For g(x)=sechx, we have
71
f (x )
f (x )
dx
2dx
sec hf (x)dx
cosh f (x)
e
e
On putting e
f(x)
=z, it becomes
2
dz
2
f '(x)(z
1)
(2.5.12.1)
For f(x)=x
2
+bx+c, we have from (2.5.12.1)
2
2dz
f '(x)(z
1)
2
2
dz
b
4c
, k
4
(z
1) log z k
1
dz
dz
2
(1 iz) log z k
(1 iz) log z k
Both are nonelementary proved in section 2.5.5. Similarly we can prove it
nonelementary for higher degree polynomial f(x).
Examples on Conjecture-5
Example 2.5.1: Show that the integral
dx
x
)
3
sin(
2
is nonelementary.
Proof: We have
2
sin(x
3)dx
2
2
i(x
3)
i(x
3)
1
e
dx
e
dx
2i
Both are nonelementary proved in section 2.4: case-I. It follows also from example-4
(page-6).
Alternative Proof: By strong Liouville's theorem (special case) the first integral
2
i(x
3)
e
dx
is elementary if and only if there exists a rational function R(x) such that
1 R '(x) i2xR(x)
R '(x) 1 and xR(x) 0
72
But such R(x) cannot exist, which satisfies both conditions. Hence this is
nonelementary. Similarly we can proof that the second integral
2
i(x
3)
e
dx
is
nonelementary. Therefore the given integral is also nonelementary.
Example 2.5.2: Show that the integral
dx
b
x
)
6
cosh(
2
is nonelementary.
Proof: We have
2
2
2
(6x
b)
(6x
b)
1
cosh(6x
b)dx
e
dx
e
dx
2
Both are nonelementary proved in section 2.4., case-I. It follows also from example-4
(page-6) as well as from examples 2.4.1 and 2.4.2.
Example 2.5.3: Show that the integral
2
tan(x
bx c)dx
is nonelementary.
We have
2
tan(x
bx c)dx
2
2
2
(2x b) sec(x
bx c) tan(x
bx c)dx
(2x b)sec(x
bx c)
On putting sec(x
2
+bx+c)=z, it becomes
dz
(2x b)z
2
1
1
dz
b
4c
, k
2
4
z sec z k
2
2
1
1
z
1dz
2 z z 1 sec z k
2
1
[z, z
1,sec z]dz
1
2
[z, y , y ]dz
1
2
2
2
1
2
dy
dy
z
z
1
1
,
dz
y
dz
zy
z
1
z z
1
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
'
n
'
i
0
i
1
i 1
i
U
1
U
C
U
z sec z k
73
Considering different forms of U
j
we find that no such U
j
exist. Hence it is
nonelementary.
Example 2.5.4: Show that the integral
dx
d
cx
bx
x
)
3
tan(
2
3
is nonelementary.
Proof: We have
2
3
2
3
2
3
2
2
3
2
(9x
2bx c) sec(3x
bx
cx d) tan(3x
bx
cx d)
tan(3x
bx
cx d)dx
dx
(9x
2bx c) sec(3x
bx
cx d)
On putting
3
2
sec(3x
bx
cx d) z
, we get
dx
d
cx
bx
x
)
3
tan(
2
3
2
dz
(9x
2bx c)z
2
2
dz
b
b
3x
c
z
3
3
1
dz
1
dz
2
(3x
)z 2
(3x
)z
, where
b
3
and
2
2
b
c
3
1
dz
1
dz
6
(x A)z 6
(x B)z
, where A
, B
3
3
2
2
2
2
1
z
1dz
1
z
1dz
6
6
(x A)z z
1
(x B)z z
1
Since z=sec(3x
3
+bx
2
+cx+d), x is a function of sec
-1
z. Then both (x+A) and (x+B) are
functions of sec
-1
z. Therefore both will be of the form
2
1
F[z, z
1, sec z]dz
1
2
F[z, y , y ]dz
, where
1
2
2
2
1
2
dy
dy
z
z
1
1
,
dz
y
dz
zy
z
1
z z
1
Giving the exact values of b, c, d, we can apply the same procedure as has been applied
in example 2.5.3 to prove it nonelementary.
74
2.6 Conjecture-6
An indefinite integral of the form
dx
x
h
x
g
x
f
)
(
)
(
).
(
, where f(x), h(x) are polynomials in x
(degree of h(x) is greater than the degree of f(x)) and g(x) is a trigonometric (not inverse
trigonometric) or a hyperbolic (not inverse hyperbolic) function is always
nonelementary.
Proof: There arise two different cases:
Case-I: When f(x) divides h(x) exactly, let
f (x)
1
h(x)
(x)
Then, we have
f (x).g(x)
g(x)
dx
dx
h(x)
(x)
which is of the form section 2.3 and its proof of nonelementary have been discussed
there.
Case-II: When f(x) does not divide h(x) exactly. Let us denote
f (x)
H(x)
h(x)
where H(x) a rational function with gcd(f(x), h(x))=1. If it is not so, we can make it by
dividing them by their common factors. Then the given integral can be written as
f (x).g(x)
dx
H(x)g(x)dx
h(x)
(2.6.1)
Now we consider the following sub-cases:
Sub-case-I: When g(x) be a trigonometric (not inverse trigonometric) function and
(x) be a polynomial of degree 1. Then
75
2.6.1 For g(x)=sin(x), we have from (2.6.1)
H(x)g(x)dx
H(x)sin (x)dx
i (x )
i (x)
1
1
H(x)e
dx
H(x)e
dx
2i
2i
(2.6.1.1)
By strong Liouville's theorem (special case)
i (x )
H(x)e
dx
is elementary if and only if
there exists a rational function R(x) which satisfies an identity of the form
H(x) R '(x) i '(x)R(x)
R '(x) H(x)
and R(x)=0 because '(x)0. But when R(x)=0, R'(x) cannot be H(x).
Thus no such R(x) exists. Therefore it is nonelementary. Similarly we can prove that the
second integral in (2.6.1.1) is nonelementary. Therefore the given integral (2.6.1) is
nonelementary for g(x)=sin(x) for any H(x).
2.6.2 For g(x)=cos(x), we have from (2.6.1)
H(x)g(x)dx
H(x) cos (x)dx
i (x)
i (x )
1
1
H(x)e
dx
H(x)e
dx
2i
2i
(2.6.2.1)
which can now be proved nonelementary by strong Liouville's theorem (special case)
and the similar procedures as has been applied in section 2.6.1 and in section 2.3.2.
2.6.3 For g(x)=tan(x), we have from (2.6.1)
H(x)g(x)dx
H(x) tan (x)dx
H(x)sec (x) tan (x) '(x)
dx
sec (x) '(x)
On putting sec(x)=z, it becomes
H(x)dz
z '(x)
(2.6.3.1)
Let us take (x)=x+b, f(x)=x+a, h(x)=x
2
+cx+d. Then from (2.6.3.1) we have
76
2
H(x)dz
(x a)dz
z '(x)
z(x
cx d)
1
1
2
1
(sec z a b)dz
z{(sec z b)
c(sec z b) d}
1
2
2
1
2
1
(sec z a b) z
1dz
z z
1{(sec z b)
c(sec z b) d}
2
1
F[z, z
1, sec z]dz
1
2
F[z, y , y ]dz
1
2
1
2
dy
dy
z
1
,
dz
y
dz
zy
By strong Liouville's theorem it is elementary if and only if there exists an identity of
the form
1
1
2
1
(sec z a b)
z{(sec z b)
c(sec z b) d}
'
n
'
i
0
i
i 1
i
U
U
C
U
But no such U
j
exist. Hence it is nonelementary. Similar arguments can be given for
higher degrees polynomials f(x) and h(x).
2.6.4 For g(x)=cot(x), we have from (2.6.1)
H(x)g(x)dx
H(x) cot (x)dx
H(x) cos ec (x) cot (x) '(x)
dx
cos ec (x) '(x)
On putting cosec(x)=z, it becomes
H(x)dz
z '(x)
(2.6.4.1)
Let us take (x)=x+b, f(x)=x+a, h(x)=x
2
+cx+d. Then from (2.6.4.1) we have
H(x)dz
z '(x)
1
1
2
1
(cos ec z a b)dz
z{(cos ec z b)
c(cos ec z b) d}
2
1
F[z, z
1, cos ec z]dz
1
2
F[z, y , y ]dz
77
which can now be proved nonelementary by the similar procedures as has been applied
in sections 2.6.3 and 2.3.4. Similarly it can be proved nonelementary for higher degree
polynomials f(x) and h(x).
2.6.5 For g(x)=sec(x), we have from (2.6.1)
H(x)g(x)dx
H(x) sec (x)dx
i (x )
i 2 (x )
2H(x)e
dx
{e
1}
i (x)
i 2 (x )
2H(x)i '(x)e
dx
i '(x){e
1}
On putting e
i(x)
=z, it become
2
2H(x)dz
i '(x)(z
1)
(2.6.5.1)
Let us take (x)=x+b, f(x)=x+a, h(x)=x
2
+cx+d. Then from (2.6.5.1) we have
2
2H(x)dz
i '(x)(z
1)
2
2
2( i log z a b)dz
i{( i log z b)
c( i log z b) d}(z
1)
F[z, log z]dz
1
F[z, y ]dz
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
'
n
i
0
i
2
2
i 1
i
U
2( i log z a b)
U
C
i{( i log z b)
c( i log z b) d}(z
1)
U
Considering different possible forms of U
j
, we find that no such U
j
exist. Therefore the
given function is nonelementary. Similarly it can be proved nonelementary for higher
degree polynomials f(x) and h(x).
2.6.6 For g(x)=cosec(x), we have from (2.6.1)
H(x)g(x)dx
H(x) cos ec (x)dx
i (x )
i 2 (x )
2iH(x)e
dx
{e
1}
i (x)
i2 (x)
2H(x)i '(x)e
dx
'(x){e
1}
78
On putting e
i(x)
=z, it becomes
2
2H(x)dz
'(x)(z
1)
(2.6.6.1)
Let us take (x)=x+b, f(x)=x+a, h(x)=x
2
+cx+d. Then from (2.6.6.1) we have
2
2H(x)dz
'(x)(z
1)
2
2
2( i log z a b)dz
{( i log z b)
c( i log z b) d}(z
1)
F[z, log z]dz
1
F[z, y ]dz
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
'
n
i
0
i
2
2
i 1
i
U
2( i log z a b)
U
C
{( i log z b)
c( i log z b) d}(z
1)
U
Considering different possible forms of U
j
, we find that no such U
j
exist. Therefore the
given function is nonelementary. Similarly it can be proved nonelementary for higher
degree polynomials f(x) and h(x).
Sub-case-II: When g(x) is a hyperbolic (not inverse hyperbolic) function and (x)
be a polynomial of degree 1. Then
2.6.7 For g(x)=sinh(x), we have from (2.6.1)
H(x)g(x)dx
H(x)sinh (x)dx
(x )
(x )
1
1
H(x)e
dx
H(x)e
dx
2
2
(2.6.7.1)
Now by strong Liouville's theorem (special case)
( x )
H(x)e
dx
is elementary if and only
if there exists a rational function R(x) which satisfies an identity of the form
H(x) R '(x)
'(x)R(x)
f (x)
R '(x)
'(x)R(x)
h(x)
Let
p(x)
R(x)
, where gcd{p(x),q(x)} 1
q(x)
79
Then we have
2
f (x)
q(x)p '(x) p(x)q '(x)
'(x)p(x)q(x)
h(x)
{q(x)}
h(x)p(x)q '(x)
h(x)p '(x) f (x)q(x)
'(x)h(x)p(x)
q(x)
Which implies that q(x) | h(x). Let h(x)=q(x).r(x), then
q(x)r(x)p '(x) f (x)q(x)
'(x)p(x)q(x)r(x) r(x)p(x)q '(x)
r(x)p(x)q '(x)
r(x)p '(x) f (x)
'(x)p(x)r(x)
q(x)
Which implies that q(x) | r(x). Let r(x)=q(x).(x), then
q(x) (x)p '(x) f (x)
'(x)p(x)q(x) (x)
(x)p(x)q '(x)
f (x)
q(x)p '(x) p(x)q '(x)
'(x)p(x)q(x)
(x)
Which implies that (x) | f(x). Let f(x)=(x).(x). Then
q(x)p '(x) p(x)q '(x)
'(x)p(x)q(x)
(x)
(2.6.7.2)
Let us take (x)=x+b, f(x)=x+a, h(x)=x
2
+cx+d.
Then from (2.6.7.2) we have, since (x)|f(x)=x+a, which implies that either (x)=k or
k(x+a) and f(x)=x+a=(x).(x) which again implies that either (x)=M or M(x+a),
where both k and M are constants.
But for (x)=M or M(x+a), we get a contradiction from (2.6.7.2) by comparing the
degrees of x in both sides, i. e., such R(x) does not exist. Hence it is nonelementary for
(x)=x+b, f(x)=x+a and h(x)=x
2
+cx+d. In the similar way we can prove that the second
integral
( x )
H(x)e
dx
is nonelementary. Therefore the given function is
nonelementary. Similarly we can prove it nonelementary for higher degree polynomials
(x), f(x) and h(x).
80
2.6.8 For g(x)=cosh(x), we have from (2.6.1)
H(x)g(x)dx
H(x) cosh (x)dx
(x)
(x)
1
1
H(x)e
dx
H(x)e
dx
2
2
which can now be proved nonelementary by strong Liouville's theorem (special case)
and the similar procedures as has been applied in sections 2.6.7 and in 2.3.8.
2.6.9 For g(x)=tanh(x), we have from (2.6.1)
H(x)g(x)dx
H(x) tanh (x)dx
H(x)sec h (x) tanh (x) '(x)
dx
sec h (x) '(x)
On putting sech(x)=z, it becomes
H(x)dz
z '(x)
(2.6.9.1)
Let us take (x)=x+b, f(x)=x+a, h(x)=x
2
+cx+d. Then from (2.6.9.1) we have
2
H(x)dz
(x a)dz
z '(x)
z(x
cx d)
1
1
2
1
(sec h z a b)dz
z{(sec h z b)
c(sec h z b) d}
1
2
2
1
2
1
(sec h z a b) 1 z dz
z 1 z {(sec h z b)
c(sec h z b) d}
2
1
F[z, 1 z ,sec h z]dz
1
2
F[z, y , y ]dz
which can now be proved nonelementary by strong Liouville's theorem and the similar
procedures as has been applied in section 2.3.9.
2.6.10 For g(x)=coth(x), we have from (2.6.1)
H(x)g(x)dx
H(x) coth (x)dx
H(x) cos ech (x) coth (x) '(x)
dx
cos ech (x) '(x)
On putting cosech(x)=z, it becomes
81
H(x)dz
z '(x)
(2.6.10.1)
Let us take (x)=x+b, f(x)=x+a, h(x)=x
2
+cx+d. Then from (2.6.10.1) we have
H(x)dz
z '(x)
1
2
2
1
2
1
(cos ech z a b) z
1dz
z z
1{(cos ech z b)
c(cos ech z b) d}
2
1
F[z, z
1, cos ech z]dz
1
2
F[z, y , y ]dz
which can now be proved nonelementary by the similar procedures as has been applied
in sections 2.6.9 and 2.3.10.
2.6.11 For g(x)=sech(x), we have from (2.6.1)
H(x)g(x)dx
H(x) sec h (x)dx
(x)
2 (x )
2H(x)e
dx
{e
1}
(x )
2 (x)
2H(x) '(x)e
dx
'(x){e
1}
On putting e
(x)
=z, it becomes
2
2H(x)dz
'(x)(z
1)
(2.6.11.1)
Let us take (x)=x+b, f(x)=x+a, h(x)=x
2
+cx+d. Then from (2.6.11.1) we have
2
2H(x)dz
'(x)(z
1)
2
2
2(log z a b)dz
{(log z b)
c(log z b) d}(z
1)
F[z, log z]dz
1
F[z, y ]dz
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
'
n
i
0
i
2
2
i 1
i
U
2(log z a b)
U
C
{(log z b)
c(log z b) d}(z
1)
U
But no such U
j
exist. Therefore the given function is nonelementary. Similarly it can be
proved nonelementary for higher degree polynomials f(x) and h(x).
82
2.6.12 For g(x)=cosech(x), we have from (2.6.1)
H(x)g(x)dx
H(x) cos ech (x)dx
(x )
2 (x )
2H(x) '(x)e
dx
'(x){e
1}
On putting e
(x)
=z, it becomes
2
2H(x)dz
'(x)(z
1)
(2.6.12.1)
Let us take (x)=x+b, f(x)=x+a, h(x)=x
2
+cx+d. Then from (2.6.12.1) we have
2
2H(x)dz
'(x)(z
1)
2
2
2(log z a b)dz
{(log z b)
c(log z b) d}(z
1)
F[z, log z]dz
1
F[z, y ]dz
which can now be proved nonelementary by strong Liouville's theorem and the similar
procedures as has been applied in sections 2.6.11 and 2.3.12.
Examples on Conjecture-6
Example 2.6.1: Show that the integral
dx
x
x
x
x
)
4
3
(
tan
).
2
(
4
2
is nonelementary.
Proof: We have
2
4
(x
2).tan x
dx
(3x
4x)
2
4
(x
2)sec x tan x
dx
(3x
4x) sec x
On putting secx=z, it becomes
2
4
(x
2)
dz
(3x
4x)z
1
2
1
4
1
(sec z)
2
dz
3(sec z)
4(sec z) z
1
2
2
1
4
1
2
(sec z)
2
z
1
dz
3(sec z)
4(sec z) z z
1
2
1
F[z, z
1, sec z]dz
1
2
F[z, y , y ]dz
1
2
2
2
1
2
dy
dy
z
z
1
1
,
dz
y
dz
zy
z
1
z z
1
83
By strong Liouville's theorem, it is elementary if and only if there exists an identity of
the form
1
2
'
n
'
i
0
i
1
4
1
i 1
i
(sec z)
2
U
U
C
U
3(sec z)
4(sec z) z
Considering different forms of U
j
like log{3(sec
-1
z)
4
+4(sec
-1
z)}, we find that no such U
j
exist. Therefore the given function is nonelementary.
Example 2.6.2: Show that the integral
dx
x
x
x
x
x
)
6
2
(
)
2
3
sin(
).
3
2
(
4
2
3
is nonelementary.
Proof: Taking
3
2
4
(2x
3x)
f (x) and 3x
2 g(x)
(2x
6x)
We have
3
2
4
(2x
3x).sin(3x
2)
dx
f (x) sin g(x)dx
(2x
6x)
ig(x )
ig(x)
e
e
f (x)
dx
2i
ig(x)
ig(x)
1
f (x)e
dx
f (x)e
dx
2i
By strong Liouville's theorem (special case),
ig(x)
f (x)e
dx
is elementary if and only if
there exists a rational function R(x) which satisfies an identity of the form
f (x) R '(x) ig '(x)R(x)
(2.6.2.A)
3
4
2x
3x
R '(x) i6xR(x)
2x
6x
6xR(x) 0
R(x) 0
.
But R(x) cannot be zero, i. e., no such R(x) exist, which satisfies the identity (2.6.2.A).
This implies that
ig(x)
f (x)e
dx
is nonelementary. Similarly we can show that the second
integral
ig( x)
f (x)e
dx
is nonelementary. Therefore the given function is nonelementary.
84
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, Advance Research International Publication House, U.P.
88
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,
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