Excerpt
2
important examples 4 and 5 [5, pp.300-301], which are treated as properties in proving the
functions elementary and nonelementary. By applying the above properties we get the following
well-known nonelementary functions:
2
x
e dx
,
2
ax
e
dx, (a
0)
,
2
x
e
dx
-
,
x
e
dx
x
-
,
x
e
dx
x
,
sin x
dx
x
,
cos x
dx
x
First Conjecture on Nonelementary Functions: Yadav & Sen [6, 7] have given six standard
forms of indefinite nonintegrable functions out of which form-1 is as follows:
"An indefinite integral of the form
dx
x
f
e
x
f
)
(
'
)
(
, where f(x) is a polynomial function of degree 2,
or a trigonometric (not inverse trigonometric) function, or a hyperbolic (not inverse hyperbolic)
function is always nonintegrable i. e. nonelementary."
Proof: We will prove it taking different possible cases as follows:
Case I: When f(x) is an algebraic function (polynomial) of degree 2:
We have
dx
x
f
e
x
f
)
(
'
)
(
f ( x )
g(x)e
dx
=
, [Taking
1
g(x)
f '(x)
=
]. From strong Liouville theorem
(special case),
f (x )
g(x)e
dx
is elementary if and only if there exists a rational function R(x) such
that
g(x)
R '(x) R(x)f '(x)
=
+
1
R '(x)
R(x)f '(x)
f '(x)
=
+
Let
p(x)
R(x)
q(x)
=
, where g.c.d.(p(x), q(x))=1. Then we have
1
R '(x)
R(x)f '(x)
f '(x)
=
+
2
2
f '(x)q(x)p '(x) f '(x)p(x)q '(x) [f '(x)] p(x)q(x)
[q(x)]
-
+
=
(1.1)
2
f '(x)p(x)q '(x)
f '(x)p '(x) q(x) [f '(x)] p(x)
q(x)
-
+
=
3
Which implies 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 (1.1) we have
2
2
f '(x)kp '(x) [f '(x)] p(x)k
k
+
=
(1.2)
Comparing degrees of x in (1.2) results out in a contradiction. Hence q(x) cannot be a constant.
For q(x) a polynomial of degree less than or equal to the degree of f'(x), we have since q(x)|f'(x),
let us assume that f'(x)=q(x).h(x). Then from (1.1)
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)p(x)q '(x)
h(x)p '(x) 1 q(x)[h(x)] p(x)
q(x)
- +
=
(1.3)
Which implies q(x)|h(x), since q(x) cannot divide p(x) and q`(x). Let h(x)=q(x).(x). Then from
(1.3), we have
3
2
q(x) (x)p '(x) p(x)q '(x) (x) [q(x)] [ (x)] p(x) 1
-
+
=
(1.4)
Comparing the degrees of x in both sides in (1.4) results out in a contradiction. Therefore such
R(x) does not exist, i. e., the given function is nonelementary.
Case II: When f(x) be a trigonometric (not inverse trigonometric) function:
Let us consider them one by one.
1.1 For sine function, we have
f ( x)
sin ( x)
e
dx
e
dx
f '(x)
'(x) cos (x)
=
,
where (x) be any polynomial of degree 1. On putting sin(x)=z, we have
sin ( x)
z
2
2
e
dx
e dz
'(x) cos (x)
[ '(x)] (1 z )
=
-
(1.1.1)
Sub-case I: When (x) is linear, let (x)=x+b. Then from (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)
=
+
-
+
4
where
z
p
e dz
e
e
dp,[Putting1 z
p]
(1 z)
p
-
= -
- =
-
and
z
p
e dz
1 e
dp,[Putting 1 z
p]
(1 z)
e
p
=
+ =
+
Both are nonelementary from example-4 due to Marchisotto et al [5, pp.300]. Therefore the
given function is also nonelementary.
Sub-case II: When (x) is a polynomial of degree 2. Let us consider (x)=x
2
+bx+c. Then we
have from (1.1.1), on putting sin(x)=z
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 theorem, part (b), 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
e
d
U
c log U
4[sin z
k](1 z )
dz
-
=
=
+
+
-
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
. Considering different forms of U
j
like
1
z
z
1
log[(sin z k)e ], e log(sin z k)
-
-
+
+
we find that no such U
j
exist. Hence the given function is nonelementary. Similarly we can prove
it nonelementary for higher degree polynomials (x).
5
1.2 For cosine function, we have
f ( x)
cos (x )
e
dx
e
dx
f '(x)
'(x) sin (x)
=
-
where (x) be any polynomial of degree 1. On putting cos(x)=z, we have
cos (x )
z
z
2
2
2
2
e
dx
e dz
e dz
'(x) sin (x)
[
'(x)] (1 z )
[ '(x)] (1 z )
=
=
-
-
-
-
(1.2.1)
Sub-case I: When (x) is linear, let (x)=x+b. Then from (1.2.1) we have
z
z
2
2
2
e dz
e dz
[ '(x)] (1 z )
(1 z )
=
-
-
which is nonelementary proved in section 1.1 subcase-I.
Sub-case II: When (x) is a polynomial of degree 2. Then from (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 1.1 to prove it nonelementary.
1.3. For tangent function, we have on putting tan(x)=z.
f ( x)
tan ( x)
2
e
dx
e
dx
f '(x)
'(x) sec
(x)
=
z
2
2
e dz
[ '(x)] (1 z )
=
+
(1.3.1)
Sub-case I: When (x) is linear, let (x)=x+b. Then from (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, putting (1 iz)
p
(1 iz)
i
p
-
=
+
=
+
6
By strong Liouville theorem (special case), it is elementary if and only if there exists a rational
function R(x) such that it 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.
Also
z
ip
i
e dz
e
ie
dp, putting (1 iz)
p
(1 iz)
p
-
=
-
=
-
Again by strong Liouville 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 in this case.
Sub-case II: When (x)=x
2
+bx+c. Then from (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 theorem part (b), it is elementary if and only if there exists an identity of the
form containing U
j
, a function of z, y
1
, y
2
, and y
3
z
n
i
i
i
1
2
i 1
i
dU
U '
e
c
4[tan z
k](1 z )
dz
U
-
=
=
+
+
+
Considering different forms of U
j
like e
z
log(tan
-1
z+k), log[e
z
(tan
-1
z+k)], etc. 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 polynomials (x).
7
1.4. For cotangent function, we have on putting cot(x)=z
f ( x)
cot ( x)
2
e
dx
e
dx
f '(x)
'(x) cos ec
(x)
=
-
z
2
2
e dz
[ '(x)] (1 z )
=
+
(1.4.1)
Sub-case-I: For (x)=x+b, we have from (1.4.1)
z
z
2
2
2
e dz
e dz
[ '(x)] (1 z )
(1 z )
=
+
+
Which is nonelementary, proved in section 1.3, sub-case-I.
Sub-case-II: For (x)=x
2
+bx+c, we have from (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 1.3 to prove it nonelementary.
1.5. For cosecant function, we have on putting cosec(x)=z
f ( x)
cosec ( x)
e
dx
e
dx
f '(x)
'(x) cos ec (x) cot (x)
=
-
z
2
2
2
e dz
[ '(x)] z (z
1)
=
-
(1.5.1)
Sub-case I: When (x) is linear, let (x)=x+b. Then from (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 1.1, sub-case-I and the second integral
z
2 z
2
e dz
z e dz
z
-
=
is also nonelementary from example-5 due to Marchisotto et al [5, pp.301].
8
Sub-case II: When (x)=x
2
+bx+c, then from (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
-
-
= =
=
=
=
=
-
-
By strong Liouville theorem part (b), this is elementary if and only if there exists an identity of
the form containg U
j
, a function of z, y
1
, y
2
, and y
3
as follows
z
n
i
i
i
1
2
2
i 1
i
dU
U '
e
c
4[cos ec z
k]z (z
1)
dz
U
-
=
=
+
+
-
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, which satisfy the above identity. Hence the given function is nonelementary.
Similarly we can prove it for higher degree polynomial (x).
1.6. For secant function, we have on putting sec(x)=z
f ( x)
sec ( x)
e
dx
e
dx
f '(x)
'(x) sec (x) tan (x)
=
z
2
2
2
e dz
[ '(x)] z (z
1)
=
-
(1.6.1)
Sub-case-I: For (x)=x+b, we have from (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 1.5, sub-case-I.
Sub-case-II: For (x)=x
2
+bx+c, we have from (1.6.1)
9
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
-
-
=
=
-
+
-
z
2
1
F[z, e , z
1, sec z]dz
-
=
-
It can be proved nonelementary by the similar procedure as has been done in section 1.5.
Case III: When f(x) be a hyperbolic (not inverse hyperbolic) function. Let us consider them
one by one.
1.7. For sine hyperbolic function, we have on putting sinh(x)=z
f ( x)
sinh (x )
e
dx
e
dx
f '(x)
'(x) cosh (x)
=
z
2
2
e dz
[ '(x)] (1 z )
=
+
(1.7.1)
Sub-case I: When (x) is linear, let (x)=x+b. Then from (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 1.3, sub-case-I.
Sub-case II: When (x)=x
2
+bx+c. Then from (1.7.1), we have
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
=
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 theorem part (b), it is elementary if and only if there exists an identity
of the form
10
z
n
0
i
i
1
2
i 1
i
dU
U '
e
c
4(sinh z
k)(1 z )
dz
U
-
=
=
+
+
+
.
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 polynomials.
1.8. For cosine hyperbolic function, we have on putting cosh(x)=z
f ( x)
cosh (x )
e
dx
e
dx
f '(x)
'(x)sinh (x)
=
z
2
2
e dz
[ '(x)] (z
1)
=
-
(1.8.1)
Sub-case-I: For (x)=x+b, we have from (1.8.1)
z
z
2
2
2
e dz
e dz
[ '(x)] (z
1)
(z
1)
=
-
-
Which is nonelementary proved in section 1.1, sub-case-I.
Sub-case-II: For (x)=x
2
+bx+c, we have from (1.8.1)
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 strong Liouville theorem part (b). Similarly we can
prove it for higher degree polynomial (x).
1.9. For tangent hyperbolic function, we have on putting tanh(x)=z
f ( x)
tanh (x )
2
e
dx
e
dx
f '(x)
'(x) sec h
(x)
=
z
2
2
e dz
[ '(x)] (1 z )
=
-
(1.9.1)
Sub-case-I: For (x)=x+b, we have from (1.9.1)
11
z
z
2
2
2
e dz
e dz
[ '(x)] (1 z )
(1 z )
=
-
-
Which is nonelementary, proved in section 1.1, sub-case-I.
Sub-case-II: For (x)=x
2
+bx+c, we have from (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
-
=
-
It can now be proved nonelementary by strong Liouville theorem part (b). Similarly we can
prove it for higher degree polynomial (x).
1.10. For cotangent hyperbolic function, we have on putting coth(x)=z
f (x )
cot h (x )
2
e
dx
e
dx
f '(x)
'(x) cos ech
(x)
=
-
z
2
2
e dz
[ '(x)] (z
1)
=
-
(1.10.1)
Sub-case-I: For (x)=x+b, we have from (1.10.1)
z
2
2
e dz
[ '(x)] (z
1)
-
z
2
e dz
(z
1)
=
-
Which is nonelementary proved in section 1.1, sub-case-I.
Sub-case-II: For (x)=x
2
+bx+c, we have from (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 strong Liouville theorem part (b). Similarly we can
prove it for higher degree polynomial (x).
1.11. For cosecant hyperbolic function, we have on putting cosech(x)=z
12
f ( x )
cosec h (x )
e
dx
e
dx
f '(x)
'(x) cos ech (x) coth (x)
=
-
z
2
2
2
e dz
[ '(x)] z (z
1)
=
+
(1.11.1)
Sub-case-I: For (x)=x+b, we have from (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 1.5, sub-case-I and section 1.7, sub-case-I
respectively.
Sub-case-II: For (x)=x
2
+bx+c, we have from (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 theorem part (b). Similarly we can
prove it for higher degree polynomial (x).
1.12. For secant hyperbolic function, we have on putting sech(x)=z
f (x )
sec h ( x)
e
dx
e
dx
f '(x)
'(x) sec h (x) tanh (x)
=
-
z
2
2
2
e dz
[ '(x)] z (1 z )
=
-
(1.12.1)
Sub-case-I: For (x)=x+b, we have from (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 1.5, sub-case-I and section 1.1, sub-case-I
respectively.
Sub-case-II: For (x)=x
2
+bx+c, we have from (1.12.1)
13
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 theorem part (b). Similarly we can
prove it for higher degree polynomial (x).
Let us consider some examples on this standard form of nonelementary functions:
Example 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 for second integral, putting ax
2
=z we have
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 due to Marchisotto et al [5, pp.301].
Example 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
=
z
2
e dz
(1 z )
=
-
On putting sinx=z. Which is nonelementary, proved in section 1.1, sub-case-I.
Example 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)
-
=
-
-
z
2
e dz
(1 z )
=
-
On putting cosx=z. Which is nonelementary proved in section 1.1, sub-case-I.
14
Example 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)
=
-
-
+
(A)
We have on putting 1-iz = p in the first integral of (A)
z
ip
ip
ip
i
2
2
e dz
e dp
e dp
e dp
ie
(iz)(1 iz)
(1 p)
p
p
-
=
+
+
-
-
(B)
where the second and third integrals are nonelementary from example-5 due to Marchisotto et al
[5, pp.301]. Now putting 1-p=X in the first integral of (B) we have
ip
iX
i
e dp
e
dX
e
(1 p)
X
-
= -
-
which is also nonelementary from example-5 due to Marchisotto et al [5, pp.301]. Therefore the
first integral of (A) is nonelementary. Similarly we can prove that the second integral of (A) is
also nonelementary. Therefore the given function is nonelementary.
Example 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 )
=
+
Which is nonelementary proved in section 1.3, sub-case-I.
Example 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 1.3, sub-case-I.
15
Example 7: Show that the integral
dx
x
x
e
x
tan
.
sec
sec
is nonelementary.
Proof: We have on putting secx=z
sec x
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
=
-
-
Which are nonelementary, proved in section 1.5, sub-case-I.
Example 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 1.5, sub-case-I.
Example 9: Show that the integral
dx
x
e
x
2
sin
2
sin
is nonelementary.
Proof: We have on putting sin
2
x=z,
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 due to Marchisotto et al [5, pp.301].
Now since
z
z
z
2
ze
1
e dz
1
e dz
I
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 due to Marchisotto et al [5, pp.301]. Hence the given function is nonelementary.
16
Example 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
-
-
= =
=
=
=
=
-
-
Applying strong Liouville theorem, part(b), it is elementary if and only if there exists an identity
of the form, containing U
i
a function of z, y
1
, y
2
, and y
3
as
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
we find that no such U
j
exist. Hence the given function is
nonelementary.
Acknowledgement
The conjecture discussed in the paper is the first standard form of indefinite nonintegrable
functions discussed in chapter two in the doctorate thesis of first author cited in reference [7],
which was submitted in the University Department of Mathematics, Vinoba Bhave University,
Hazaribag, Jharkhand, in 2012.
References
1. Hardy G. H., The Integration of Functions of a Single Variable, 2
nd
Ed., Cambridge
University Press, London, Reprint 1928, 1916
2. Ritt J. F., Integration in Finite Terms: Liouville's Theory of Elementary Methods, Columbia
University Press, New York, 1948
17
3. Risch R. H., The Problem of Integration in Finite Terms, Transactions of the American
Mathematical Society, 139, 167-189, 1969
4. Rosenlicht M., Integration in Finite Terms, The American Mathematical Monthly, 79:9,
963-972, 1972
5. Marchisotto E. A. & Zakeri G. A., An Invitation to Integration in Finite Terms, The
College Mathematics Journal, Mathematical Association of America, 25:4, 295- 308, 1994
6. Yadav D. K. & Sen D. K., Revised paper on Indefinite Nonintegrable Functions, Acta
Ciencia Indica, 34:3, 1383-1384, 2008
7. Yadav D. K., A Study of Indefinite Nonintegrable Functions, Ph. D. Thesis, Vinoba Bhave
University, Hazaribag, Jharkhand, 2012, Online:
GRIN Verlag, Munich, Germany,
ISBN: 9783668312784, www.grin.com/ebook/341510/
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