function, and present a new method to test based on its exponential ... I. INTRODUCTION. The convergence and divergence of infinite integral plays a significant ...
(IJACSA) International Journal of Advanced Computer Science and Applications, Vol. 2, No. 11, 2011
A New Test Method on the Convergence and Divergence for Infinite Integral Guocheng Li Linyi University at Feixian Feixian, Shandong, P.R.China
Abstract—The way to distinguish convergence or divergence of an infinite integral on non-negative continuous function is the important and difficult question in the mathematical teaching all the time. Using the comparison of integrands to judge some exceptional infinite integrals are hard or even useless. In this paper, we establish the exponential integrating factor of negative function, and present a new method to test based on its exponential integrating factor. These conclusions are convenient and valid additions of the previously known results. Keywords-Iinfinite integral; Convergence and divergence.
I.
Exponential
integrating
INTRODUCTION
It is well-known that the convergence and divergence of infinite integral for the different integrand which its limit is zero is different. Base on the geometric meaning of infinite integral, it equal to the size of the area which is surrounded by the integrand (image) and X axis (not closed). The Convergence and divergence mainly determined by the proximity degree of integrand tends to the x-axis. In this paper, we discuss the test method for the convergence and divergence
a
f ( x)dx
f ( x)
of infinite is the nonnegative continuous function which is defined in the interval [a, ), a 0
the problem by ln f ( x) / ln x [3], it can't intuitive describe the proximity degree of integrand tends to the x-axis, so the exponential integrating factor of negative function is created and we obtain a new test method for the convergence and divergence of infinite integral. This method is more simple and feasible than ever before.
CHARACTERIZATIONS ON CONVERGENCE OF INFINITE INTEGRAL
Definition 2.1 Assume that f x is non-negative differentiable function, we call the formula
R( x) f ' ( x) / f ( x) is the function of the exponential integrating factor of f ( x) , abbreviated as factor function. From definition 2.1, we suppose the function
factor;
The convergence and divergence of infinite integral plays a significant role in mathematical analysis, and has been received much attention of many researchers. Many effective methods have been proposed such as Ordinary Comparison Test, Limiting Test, Dirichlet Test and Abel Test [1]. The basic ideas of these methods are to find another comparison infinite integral that its convergence or divergence is certain. But it is difficult or impossible to find the comparison infinite. So these arouse great interest of many scholars to research ([2-3]).
integral
II.
x f ( x) f ( x0 ) exp R(t )dt , x0 a is completely x0 determined by its factor function R x When the
independent variable tend to be infinite, we can use the size of factor function to compare the different functions reduce speed of which tend to the x-axis. For lim f ( x) 0 is the x
necessary condition of convergence for infinite integrals
a
f ( x)dx [1]), so we can easy to push that R( x) 0 is
the necessary conditions of convergence for
a
f ( x)dx
f ( x) c 0 we have f x and g x are x g ( x) t equivalent. We denoted by f ~ g The nature of the f lim
equivalent functions is incredibly similar, so what about the infinite integration convergence of the equivalent integrand? We have the following conclusion. Theorem 2.1 Assume that Non-negative differentiable functions f x and g x are equivalent, then its convergence and divergence is incredibly similar.
Proof: By lim x
x
a x a
f (t )dt g (t )dt
lim x
f ( x) c0 g ( x)
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(IJACSA) International Journal of Advanced Computer Science and Applications, Vol. 2, No. 11, 2011
There exist constant m, M 0 such that
(ii) If lim xR( x) 1, then x
m f ( x) / g ( x) M , x a. Obviously, we have
m
g ( x)dx
a
1 M
a
a
f ( x)dx
a
f ( x)dx M
a
g ( x)dx
Proof: (i) k 0, we have
g ( x)dx;
1 f ( x)dx m a
From the definition of convergence of the infinite integrals, the desired result follows. Assume that f x g x are non-
Theorem 2.2
Comparison Test, (ii) Let
f ( x) / g ( x) is finally monotone decreasing, then
R( x)
x0 , x x0 , R f ( x) Rg ( x)
convergent, If g ( x)dx is divergent, then (ii) If
a
g ( x)dx is convergent, then
a
a
a
f ( x)dx is f ( x)dx is
a
f ( x)dx is divergent.
f ' ( x) 1 k , then f ( x) x
ln f ( x) ln(cx1 k ), c 0 1 (1 k ) That is f ( x) x i.e., According to the Ordinary c
negative differentiable functions in interval [a, ) and
(i)
a
a
f ( x)dx is convergent.
1 g ( x) , then x
a
1 dx is divergent, for x
1 Rg ( x), from Theorem 2.2, we have that x
f ( x)dx is divergent.
Combining [3], we have that
a
f ( x)dx is possible
divergent.
convergent or divergent if lim xR( x) 1 from some
Proof:
possible examples. In fact, we can continue to study convergence as follow:
x
(i) x0 , x x0 ，
f ( x) ' g ( x) f ' ( x) f ( x) g ' ( x) ) g ( x) g 2 ( x) 1 [ Rg ( x) R f ( x)] 0 f ( x) g ( x) R f ( x) Rg ( x).
(
a
f ( x)dx M
a
lim xR( x) 1 .
x
(i) If lim ln x[ xR( x) 1] 1 then x
(ii) If lim ln x[ xR( x) 1] 1 then x
g ( x)dx Using Ordinary Comparison
Test, we obtain this conclusion. Assume that we know the convergence and divergence of an infinite integral, we can obtain the convergence and divergence of other infinite integrals by comparing the size of the exponential factor. But it's complex. In fact, it can be derived alone by the exponential factor R( x) of integrand.
x
a
f ( x)dx is
a
f ( x)dx is
Proof: (i) We can obtain that R( x) is monotone-nonincreasing function, for all k 0 we have
f ( x) c0 exp xR( x) 1 k c0 exp (1 ) ln x 1 k c0 c0 exp 1 k ln x x
differentiable function in [a, ) (i) If lim xR( x) 1, then
a
divergent.
f x is non-negative
Theorem 2.3 Assume that
convergent,
f ( x) M , then g ( x)
(ii) M 0, let
f x is non-negative differentiable function in interval [a, ) and Theorem 2.4 Assume that
From Ordinary Comparison Test we can obtain that
f ( x)dx is convergent,
a
f ( x)dx
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(IJACSA) International Journal of Advanced Computer Science and Applications, Vol. 2, No. 11, 2011
1 ln x , we have that x ln x 1 1 by dx ln f ( x) ln( x ln x) i.e. f ( x) x ln x x ln x a (ii)Method1:
R( x)
Let
is divergent, combining Ordinary Comparison test, we can obtain that
a
III.
POSSIBLE APPROACH
From Theorem 2.4, we can obtain that the convergence and divergence of
f ( x)dx is determined by the exponential
a
f ( x) .
integrating factor of
f ( x)dx is divergent.
Suppose that
a
Method2: Using the condition, we obtain that
R( x)
1 ln x Rg ( x) x ln x
From theorem2.2, we have that
a
a
f ( x) g ( x)dx are
f ( x) g ( x)
i.e. R fg ( x) R f ( x) Rg ( x) if lim ( x[ R f ( x) Rg ( x)]) 1 then x
f ( x)dx is divergent too.
Using the conclusions above, we can easily think of the following examples. Example 1: Discuss the convergence of
2
1 dx . p x ln q x
a
Theorem 3.1 Assume that
lim x
x
p ln x q p using x ln x
theorem 2.3, we have that the integration is convergent if p 1 and is divergent if p 1 when p 1 , we easy to know x[ R( x) 1] q using theorem 2.4, we get that the integration is convergent when q 1 the integration is divergent when q 1
a
f ( x)dx is convergent, if
the non-negative function g ( x) is monotonous and bounded, then
p cqx c R( x) x
f ( x) g ( x)dx is convergent. In fact, we can construct
some convergent infinite integral for some specific needs.
Solution: For its exponential integrating factor
We easy to know that
Using the exponential integrating factor of
1 1 , and dx is divergent. x ln x x ln x a
g ( x)
f ( x)dx and
convergent, we give the following conditions of the nonnegative function g ( x)
Where
CONSTRUCTING CONVERGENT INFINITE INTEGRAL AND
a
f ( x) g ( x)dx is convergent.
Proof: Combining the conditions and theorem 2.4, we know that lim ln x[ xR( x) 1] 1 f g ( x) is decreasing x
and lower bounded, Then
Rg ( x) 0 and R fg ( x) R f ( x) Rg ( x) hold.
Obviously, lim ( x[ R fg ( x) 1]) 1 f g ( x) is increasing x
and upper bounded, then
x
a
Rg (t )dt is convergent, i.e.
lim x ln xRg ( x) 0 such that
x
Example2: Discuss the convergence of
1
x p exp(qxc )dx
lim ln x[ xR fg ( x) 1] lim ln x[ xR f ( x) 1] 1
x
x
From all above,
Solution: For its exponential integrating factor is
we can obtain that
a
f ( x) g ( x)dx is
convergent.
p cqx c , R( x) x
Theorem 3.2 Suppose
We easy to know that lim xR( x) [ p cqx ] using
a
f ( x)dx is convergent where
f ( x) is continuous and nonnegative in [a, ) .
c
x
theorm2.3, we have
1
x p exp(qxc )dx is convergence if
lim p cqxc 1 i.e c 0 or q 0, p 1 or
(i)If and
1, then x f ( x)dx is convergent.
(ii)If , R, then
a
a
x f ( x)dx is convergent,
x
c>0, q
