SubjectIMPROPER INTEGRALS

In the topic “Definite Integral” the concept of a definite integral for the case of a finite interval was considered
and limited function
(see Theorem 1 from §3). Now let's generalize this concept to the cases of an infinite interval and an unbounded function. The need for such a generalization is demonstrated, for example, by the following situations.

1. If, using the formula for arc length, try to calculate the length of a quarter circle
,
, then we arrive at the integral of the unbounded function:

, Where
.

2. Let the body have mass
moves by inertia in a medium with a resistance force
, Where
- body speed. Using Newton's second law (
, Where
acceleration), we get the equation:
, Where
. It is not difficult to show that the solution to this (differential!) equation is the function
If we need to calculate the path traveled by the body before it comes to a complete stop, i.e. until the moment when
, then we arrive at the integral over an infinite interval:

§1. Improper integrals of the 1st kind

I Definition

Let the function
defined and continuous on the interval
. Then for anyone
it is integrable on the interval
, that is, there is an integral
.

Definition 1 . The finite or infinite limit of this integral at
is called an improper integral of the 1st kind of the function
along the interval
and is designated by the symbol
. Moreover, if the specified limit is finite, then the improper integral is called convergent, otherwise (
or does not exist) – divergent.

So, by definition

Examples

2.
.

3.
- does not exist.

The improper integral from Example 1 converges; in Examples 2 and 3 the integrals diverge.

II Newton–Leibniz formula for an improper integral of the first kind

Let
- some antiderivative for the function
(exists on
, because
- continuous). Then

From here it is clear that the convergence of the improper integral (1) is equivalent to the existence of a finite limit
. If this limit is defined
, then we can write the Newton-Leibniz formula for integral (1):

, Where
.

Examples .

5.
.

6. More complex example:
. First, let's find the antiderivative:

Now we can find the integral , given that

:

III Properties

Let us present a number of properties of the improper integral (1), which follow from the general properties of limits and the definite integral:

IV Other definitions

Definition 2 . If
continuous on
, That

.

Definition 3 . If
continuous on
, then we accept by definition

(– arbitrary),

Moreover, the improper integral on the left side converges if only both integrals on the right side converge.

For these integrals, as well as for integral (1), one can write the corresponding Newton–Leibniz formulas.

Example 7 .

§2. Tests for the convergence of an improper integral of the 1st kind

Most often, it is impossible to calculate an improper integral by definition, so they use the approximate equality

(for large ).

However, this relation makes sense only for convergent integrals. It is necessary to have methods for clarifying the behavior of the integral bypassing the definition.

I Integrals of positive functions

Let
on
. Then the definite integral
as a function of the upper limit it is an increasing function (this follows from the general properties of the definite integral).

Theorem 1 . An improper integral of the first kind of a nonnegative function converges if and only if the function
remains limited with increasing .

This theorem is a consequence of the general properties of monotone functions. The theorem has almost no practical meaning, but it allows us to obtain the so-called signs of convergence.

Theorem 2 (1st sign of comparison). Let the functions
And
continuous for
and satisfy the inequality
. Then:

1) if the integral
converges, then
converges;

2) if the integral
diverges, then
diverges.

Proof . Let's denote:
And
. Because
, That

. Let the integral
converges, then (by Theorem 1) the function
- limited. But then
is limited, and therefore the integral
also converges. The second part of the theorem is proved in a similar way.

This criterion is not applicable if the integral diverges from
or convergence of the integral of
. This drawback is absent in the 2nd comparison feature.

Theorem 3 (2nd sign of comparison). Let the functions
And
continuous and non-negative on
. Then if
at
, then the improper integrals
And
converge or diverge simultaneously.

Proof . From the conditions of the theorem we obtain the following chain of equivalent statements:

, ,

.

Let, for example,
. Then:

Let us apply Theorem 2 and property 1) from §1 and obtain the statement of Theorem 3.

The standard function with which this one is compared is a power function
,
. We invite students to prove for themselves that the integral

converges at
and diverges at
.

Examples . 1.
.

Consider the integrand on the interval
:

,
.

Integral
converges, because
. Based on the 2nd comparison criterion, the integral also converges
, and due to property 2) from §1, the original integral also converges.

2.
.

Because
, then exists
such that when

. For such variable values:

It is known that the logarithmic function grows more slowly than the power function, i.e.

,

which means, starting from a certain value of the variable, this fraction is less than 1. Therefore

.

Integral converges as a reference. By virtue of the 1st comparison criterion, it converges and
. Applying the 2nd criterion, we obtain that the integral
converges. And again property 2) from §1 proves the convergence of the original integral.

Are you here now? =) No, I wasn’t trying to intimidate anyone, it’s just that the topic of improper integrals is a very good illustration of how important it is not to neglect higher mathematics and other exact sciences. Everything you need to learn the lesson is on the website - in a detailed and accessible form, if you wish...

So, let's start with. Figuratively speaking, an improper integral is an “advanced” definite integral, and in fact there are not so many difficulties with them, and besides, the improper integral has a very good geometric meaning.

What does it mean to evaluate an improper integral?

Calculate improper integral - this means finding the NUMBER(exactly the same as in the definite integral), or prove that it diverges(that is, you end up with infinity instead of a number).

There are two types of improper integrals.

Improper integral with infinite limit(s) of integration

Sometimes such an improper integral is called improper integral of the first kind. In general, an improper integral with an infinite limit most often looks like this: . How is it different from a definite integral? At the upper limit. It is endless: .

Less common are integrals with an infinite lower limit or with two infinite limits: , and we will consider them later - when you get the hang of it :)

Well, now let's look at the most popular case. In the vast majority of examples, the integrand function continuous in between, and this one important fact should be checked first! Because if there are gaps, then there are additional nuances. For definiteness, let us assume that even then the typical curved trapezoid will look like this:

Note that it is infinite (not bounded on the right), and improper integral numerically equal to its area. The following options are possible:

1) The first thought that comes to mind: “since the figure is infinite, then ", in other words, the area is also infinite. It may be so. In this case they say that the improper integral diverges.

2) But. As paradoxical as it may sound, the area of ​​an infinite figure can be equal to... a finite number! For example: . Could this be true? Easily. In the second case, the improper integral converges.

3) More on the third option a little later.

In what cases does an improper integral diverge and in what cases does it converge? This depends on the integrand, and we'll look at specific examples very soon.

What happens if an infinite curved trapezoid is located below the axis? In this case, the improper integral (diverges) or is equal to a finite negative number.

Thus, improper integral can be negative.

Important! When you are given ANY improper integral to solve, then, generally speaking, There is no talk about any area and there is no need to build a drawing. I explained the geometric meaning of the improper integral only to make it easier to understand the material.

Since the improper integral is very similar to the definite integral, let us remember the Newton-Leibniz formula: . In fact, the formula is also applicable to improper integrals, only it needs to be slightly modified. What is the difference? At the infinite upper limit of integration: . Probably, many guessed that this already smacks of the application of the theory of limits, and the formula will be written like this: .

What is the difference from a definite integral? Nothing special! As in the definite integral, you need to be able to find the antiderivative function (indefinite integral), and be able to apply the Newton-Leibniz formula. The only thing that has been added is the calculation of the limit. Whoever has a bad time with them, learn a lesson Function limits. Examples of solutions, because it’s better late than in the army.

Let's look at two classic examples:

Example 1

For clarity, I will draw a drawing, although, I emphasize once again, on practice There is no need to build drawings in this task.

The integrand function is continuous on the half-interval, which means that everything is fine and the improper integral can be calculated by the “standard” method.

Application of our formula and the solution to the problem looks like this:

That is, the improper integral diverges, and the area of ​​the shaded curved trapezoid is equal to infinity.

In the example considered, we have the simplest table integral and the same technique for applying the Newton-Leibniz formula as in the definite integral. But this formula will be applied under the sign of the limit. Instead of the usual letter of a “dynamic” variable, the letter “be” appears. This should not confuse or baffle, because any letter is no worse than the standard “X”.

If you do not understand why at , then this is very bad, either you do not understand the simplest limits (and generally do not understand what a limit is), or you do not know what the graph of a logarithmic function looks like. In the second case, attend a lesson Graphs and properties of elementary functions.

When solving improper integrals, it is very important to know what the graphs of basic elementary functions look like!

The finished task should look something like this:

“

! When preparing an example, we always interrupt the solution and indicate what happens to the integrand – is it continuous on the interval of integration or not?. With this we identify the type of improper integral and justify further actions.

Example 2

Calculate the improper integral or establish its divergence.

Let's make the drawing:

First, we note the following: the integrand is continuous on the half-interval. Hood. We solve using the formula :

(1) We take the simplest integral of a power function (this special case is in many tables). It is better to immediately move the minus sign beyond the limit sign so that it does not get in the way in further calculations.

(2) We substitute the upper and lower limits using the Newton-Leibniz formula.

(3) We indicate that at (Gentlemen, this should have been understood a long time ago) and simplify the answer.

Here the area of ​​an infinite curved trapezoid is a finite number! Unbelievable but true.

The finished example should look something like this:

“

The integrand function is continuous on

“

What to do if you come across an integral like - with break point on the integration interval? This means there is a typo in the example. (Most likely), or about an advanced level of training. In the latter case, due to additivity properties, we should consider two improper integrals on intervals and then deal with the sum.

Sometimes, due to a typo or intent, an improper integral may not exist at all, so, for example, if you put the square root of “x” in the denominator of the above integral, then part of the integration interval will not be included in the domain of definition of the integrand at all.

Moreover, the improper integral may not exist even with all the “apparent well-being”. Classic example: . Despite the definiteness and continuity of the cosine, such an improper integral does not exist! Why? It's very simple because:
- does not exist appropriate limit.

And such examples, although rare, do occur in practice! Thus, in addition to convergence and divergence, there is also a third outcome of the solution with a valid answer: “there is no improper integral.”

It should also be noted that the strict definition of an improper integral is given precisely through the limit, and those who wish can familiarize themselves with it in the educational literature. Well, we continue the practical lesson and move on to more meaningful tasks:

Example 3

Calculate the improper integral or establish its divergence.

First, let's try to find the antiderivative function (indefinite integral). If we fail to do this, then naturally we will not be able to solve the improper integral either.

Which of the table integrals is the integrand similar to? It reminds me of an arctangent: . These considerations suggest that it would be nice to have a square in the denominator. This is done by replacement.

Let's replace:

The indefinite integral has been found; in this case, it makes no sense to add a constant.

It is always useful to check the draft, that is, differentiate the result obtained:

The original integrand has been obtained, which means that the indefinite integral has been found correctly.

Now we find the improper integral:

(1) We write the solution in accordance with the formula . It is better to immediately move the constant beyond the limit sign so that it does not interfere with further calculations.

(2) We substitute the upper and lower limits in accordance with the Newton-Leibniz formula. Why at ? See the arctangent graph in the already recommended article.

(3) We get the final answer. A fact that is useful to know by heart.

Advanced students may not find the indefinite integral separately and not use the replacement method, but rather use the method of substituting the function under the differential sign and solving the improper integral “immediately.” In this case, the solution should look something like this:

“

The integrand is continuous on .

“

Example 4

Calculate the improper integral or establish its divergence.

! This is a typical example, and similar integrals are found very often. Work it out well! The antiderivative function here is found using the method of selecting a complete square; more details on the method can be found in the lesson Integrating Some Fractions.

Example 5

Calculate the improper integral or establish its divergence.

This integral can be solved in detail, that is, first find the indefinite integral by making a change of variable. Or you can solve it “immediately” - by subsuming the function under the differential sign. Who has any mathematical training?

Complete solutions and answers at the end of the lesson.

Examples of solutions to improper integrals with an infinite lower limit of integration can be found on the page Efficient methods for solving improper integrals. There we also analyzed the case when both limits of integration are infinite.

Improper integrals of unbounded functions

Or improper integrals of the second kind. Improper integrals of the second kind are insidiously “encrypted” under the usual definite integral and look exactly the same: But, unlike the definite integral, the integrand suffers an infinite discontinuity (does not exist): 1) at the point , 2) or at the point , 3) ​​or at both points at once, 4) or even on the integration segment. We will look at the first two cases; for cases 3-4 at the end of the article there is a link to an additional lesson.

Just an example to make it clear: . It seems to be a definite integral. But in fact, this is an improper integral of the second kind; if we substitute the value of the lower limit into the integrand, then our denominator goes to zero, that is, the integrand simply does not exist at this point!

In general, when analyzing an improper integral you always need to substitute both integration limits into the integrand. In this regard, let's check the upper limit: . Everything is fine here.

The curvilinear trapezoid for the type of improper integral under consideration fundamentally looks like this:

Here everything is almost the same as in the integral of the first kind.

Our integral is numerically equal to the area of ​​the shaded curved trapezoid, which is not bounded from above. In this case, there can be two options*: the improper integral diverges (the area is infinite) or the improper integral is equal to a finite number (that is, the area of ​​an infinite figure is finite!).

* by default we usually assume that the improper integral exists

All that remains is to modify the Newton-Leibniz formula. It is also modified with the help of a limit, but the limit no longer tends to infinity, but to the value on the right. It is easy to follow from the drawing: along the axis we must approach the breaking point infinitely close on right.

Let's see how this is implemented in practice.

Example 6

Calculate the improper integral or establish its divergence.

The integrand has an infinite discontinuity at a point (don’t forget to check verbally or on a draft that everything is fine with the upper limit!)

First, let's calculate the indefinite integral:

Replacement:

If you have any difficulties with replacement, please refer to the lesson Substitution method in indefinite integral.

Let's calculate the improper integral:

(1) What's new here? There is practically nothing in terms of solution technology. The only thing that has changed is the entry under the limit icon: . The addition means that we are striving for the value on the right (which is logical - see the graph). Such a limit in the theory of limits is called one-sided limit. In this case we have right-hand limit.

(2) We substitute the upper and lower limits using the Newton-Leibniz formula.

(3) Let's deal with at . How to determine where an expression is going? Roughly speaking, you just need to substitute the value into it, substitute three quarters and indicate that . Let's comb the answer.

In this case, the improper integral is equal to a negative number. There is no crime in this, just the corresponding curved trapezoid is located under the axis.

And now two examples for independent solutions.

Example 7

Calculate the improper integral or establish its divergence.

Example 8

Calculate the improper integral or establish its divergence.

If the integrand does not exist at the point

An infinite curved trapezoid for such an improper integral fundamentally looks like this.

Improper integrals

Lk5.6(4h)

The concept was introduced under the assumption that:

1) the integration interval is finite (segment [ a;b]),

2) function f(x) is limited to [ a;b].

Such a definite integral is called own(the word “own” is omitted). If any of these conditions are not met, then the definite integral is called not your own. There are improper integrals of the first and second kind.

1. Definition of an improper integral of the first kind

Let us generalize the concept of a definite integral to an infinite interval. Let f(x) is defined on the interval [ a;+¥) and is integrable in each of its finite parts, i.e. . In this case, there is an integral. It is clear that there is a function defined on [ a;+¥). Let's consider. This limit may or may not exist, but regardless of this it is called improper integral of the first kind and is designated .

Definition. If exists and is finite, then the improper integral is called convergent, and the value of this limit is the value of the improper integral. . If does not exist or is equal to ¥, then the improper integral is called divergent.

Similarly defined,

Example 1. Investigate the convergence of the integral , .

D is continuous on [ a;+¥) .

If , then , and Þ the integral converges.

If , then the integral diverges.

So, converges at and ;

diverges at .D

2. Properties of an improper integral of the first kind

Since the improper integral is defined as the limit of the Riemann integral, then all the properties that are preserved during the passage to the limit are transferred to the improper integral, that is, properties 1-8 are satisfied. The mean value theorem makes no sense.

3. Newton–Leibniz formula

Let the function f is continuous on [ a;+¥), F- is antiderivative and exists. Then the Newton–Leibniz formula is valid:

Indeed,

Example 2. D. D

Geometric meaning of an improper integral of the first kind

Let the function f is non-negative and continuous on [ a;+¥) and the improper integral converges. equal to the area of ​​a curved trapezoid with base [ a;b], and is equal to the area with the base [ a;+¥).

4. Improper integrals of nonnegative functions

Theorem 1. Let f(x)³0 on [ a;+¥) and integrable on [ a;b] "b>a. For the convergence of an improper integral, it is necessary and sufficient that the set of integrals be bounded from above, and .

Proof.

Consider the function, a£ b. Because f(x)³0, then F does not decrease Indeed, " b 1 , b 2: a£ b 1 <b 2 due to the fact that , is fulfilled

By definition, an improper integral converges if and only if there is a finite . Because F(b) does not decrease, then this limit exists if and only if the function F(b) is bounded from above, that is, $ M>0: "b>a. Wherein

The divergence of the improper integral means that , that is .

Theorem 2. Let the functions f And g non-negative on [ a;+¥) and integrable on [ a;b] "b>a. Let on [ a;+¥) done

1) from the convergence of integral (2) the convergence of integral (3) follows;

2) from the divergence of the integral (3) the divergence of the integral (2) follows.

Proof.

From (1) " b>a.

1) Let integral (2) converge. By Theorem 1, the set is bounded bounded bounded. By Theorem 1 it converges.

2) Let them disperse. Let us prove that integral (2) diverges. From the opposite. Let us assume that integral (2) converges, but then, by the first part of the theorem, integral (3) converges - a contradiction with the condition.

Theorem 3. Let the functions f And g non-negative on [ a;+¥) and integrable on [ a;b] "b>a. If exists (0£ k£¥), then

1) from the convergence of the integral at k<¥ следует сходимость интеграла ,

2) from the divergence of the integral at k>0 follows the divergence of the integral.

Proof.

1) Let k<¥ и сходится.

Because it converges, it converges, it means it converges. Then, by virtue of (4), converges. From here it converges.

2) Let k>0 and diverges. In this case - a finite number. If we assume the opposite - that the integral converges, then by what was proved in point 1) we will find that it also converges, and this contradicts the condition. Therefore, the assumption made is incorrect and divergent. converges absolutely, then by definition it converges. So it fits. But it fits.

Lecture 24. IMPROPER INTEGRALS

Plan:

1. The concept of an improper integral
2. Improper integrals of the first kind.
3. Improper integrals of the second kind.

1. The concept of an improper integral

Let's consider finding both types of improper integrals.

Let the function be given y=f(x), continuous on the interval [ a;+∞). If there is a finite limit, then it is called improper integral of the first kind and denote .

converges diverges .

Geometric meaning of an improper integral of the first kind is as follows: if converges (provided that f(x)≥0), then it represents the area of ​​an “infinitely long” curved trapezoid (Fig. 24.1).

Similarly, the concept of an improper integral with an infinite lower limit of integration is introduced for a continuous line on the interval ( -∞ ;b] functions: = .

An improper integral with two infinite limits of integration is defined by the formula: = + , where With– arbitrary number.

Let us consider examples of finding improper integrals of the first kind.

Example 24.1.

Solution. To find an improper integral with an infinite upper bound of a continuous function, we use the formula: = . Then = . First, let's calculate the integral of e x:

= = = =∞. We found that the improper integral diverges.

Answer: diverges.

Example 24.2. Calculate the improper integral or establish its divergence: .

Solution. The integrand is continuous on the interval ( -∞ ;- 1]. To find an improper integral of the first kind with an infinite lower bound, we use the formula: = . Then = . Let's calculate the integral contained under the limit sign: = . Let's get rid of the minus sign by swapping the boundaries of integration:

1. We found that the improper integral under consideration converges.

Answer: =1.

1. Improper integrals of the second kind.

Let the function be given y=f(x), continuous on the interval [ a;b). Let b– point of discontinuity of the second kind. If there is a finite limit, then it is called improper integral of the second kind and denote .

Thus, by definition = .

If the found limit is equal to a finite number, then the improper integral is said to be converges . If the specified limit does not exist or is infinite, then the integral is said to be diverges .

Geometric meaning of an improper integral of the second kind, Where b– point of discontinuity of the second kind, f(x)≥0, is as follows: if it converges, then it represents the area of ​​an “infinitely high” curved trapezoid (Fig. 24.2).

Similarly, the concept of an improper integral of the second kind is introduced for a continuous line on the interval ( a;b]functions provided that A– point of discontinuity of the second kind: = .

Example 24.3. Calculate the improper integral of the second kind: .

Solution. The integrand is continuous on the interval (0;1], and x= 0 - point of discontinuity of the second kind (). To calculate the improper integral, we use the formula: = . We get that

= = = = = = ∞. We see that the improper integral of the second kind diverges.

Answer: diverges.

Control questions:

1. What is an improper integral called?
2. What integrals are called improper integrals of the first kind?
3. What is the geometric meaning of an improper integral of the first kind?
4. Which improper integrals are called convergent and which are called divergent?
5. What integrals are called improper integrals of the second kind?
6. What is the geometric meaning of an improper integral of the second kind?

BIBLIOGRAPHY:

1. Abdrakhmanova I.V. Elements of higher mathematics: textbook. manual – M.: Center for Intensive Educational Technologies, 2003. – 186 p.

2. Algebra and the beginnings of analysis (Part 1, Part 2): Textbook for secondary educational institutions / ed. G.N.Yakovleva. – M.: Nauka, 1981.

3. Aleksandrova N.V. Mathematical terms. Directory.- M.: Higher. school, 1978. - 190 p.

4. Valutse I.I., Diligul G.D. Mathematics for technical schools based on secondary schools: Proc. allowance. – M.: Nauka, 1989. – 576 p.

5. Grigoriev V.P., Dubinsky Yu.A. Elements of higher mathematics: Textbook. for students vocational education institutions. - M.: Publishing center "Academy", 2004. – 320 p.

6. Lisichkin V.T., Soloveichik I.L. Mathematics: textbook. manual for technical schools. – M.: Higher. school, 1991. – 480 p.

7. Lukankin G.L., Martynov N.N., Shadrin G.A., Yakovlev G.N. Higher mathematics: textbook. manual for pedagogical students. institutions. – M.: Education, 1988. – 431 p.

8. Written D.T. Lecture notes on higher mathematics: Part 1. – M.:Iris-press, 2006.- 288 p.

9. Filimonova E.V. Mathematics: textbook. allowance for colleges. – Rostov n/d: Phoenix, 2003. – 384 p.

10. Shipachev V.S. Higher mathematics: textbook for universities. – M.: Higher School, 2003. – 479 p.

11. Shipachev V.S. Course of higher mathematics: higher education. – M.: PROYUL M.A. Zakharov, 2002. – 600 p.

12. Encyclopedia for children. T.11. Mathematics / Ch. ed. M.V.Aksenova. - M.: Avanta+, 2000.- 688 p.

1. Definition of an improper integral of the second kind

Let f(x) is set to [ a;b], but is unlimited on it. Let for definiteness f(x) is unbounded in the left neighborhood of the point b: , but in any interval the function is integrable. In this case, the point b called special point.

Definition. Improper integral of the second kind functions f(x) on [ a;b] is called the finite or infinite limit of the integral at

If limit (1) exists and is finite, then the integral is said to converge, and the value of the limit is considered to be the value of the integral. If limit (1) does not exist or is equal to infinity, then the integral is said to diverge.

The integral of the function is defined similarly f(x), unbounded in the right neighborhood of the point A:

Example 1. Examine for convergence.

D 1) : the integral diverges.

So, the integral converges at , diverges at . D

2. Newton-Leibniz formula for an improper integral of the second kind

Let the function f(x) is defined and continuous in the interval [ a;b) and near the point b function is unlimited ( b- singular point of the function f(x)). Then for f(x) in this interval there is an antiderivative F(x) And " h>0 according to the Newton-Leibniz formula we have

It follows that the improper integral (1) exists if and only if there is a finite limit. In this case the function F(x) is continuous on the interval [ a;b]. Then, passing in (2) to the limit at h®0, we obtain the Newton-Leibniz formula for the improper integral of the second kind

So, to calculate improper integrals of the second kind, you can use the Newton-Leibniz formula if the function F(x) is continuous on the interval [ a;b] And F¢(x)=f(x) at all points where f(x) is finite.

Example 2. Calculate.

D X=0 – special point. The antiderivative is continuous on [-1;27], including at the point X=0, therefore, we can apply the Newton-Leibniz formula:

Example 3. Examine for convergence.

D X=0 – special point. The antiderivative has at the point X=0 infinite gap. Therefore, this integral diverges and is equal to ¥.

Note, that if we ignore this and formally apply the Newton-Leibniz formula, we get incorrect result:

3. Improper integrals of the second kind for nonnegative functions

Theorem 1. Let f(x)³0 on [ a;b) and integrable on [ a;b-h] "h>0. For the improper integral (1) to converge, it is necessary and sufficient that the set of integrals ( h>0) is limited from above. Otherwise, integral (1) diverges and is equal to ¥.

For improper integrals of the second kind, as well as for improper integrals of the first kind, comparison theorems 2 and 3 hold. Let us formulate them without proof.

Theorem 2. Let the functions f And g non-negative on [ a;b) and integrable on [ a;b-h] "h>0. Let on [ a;b) done

Then: 1) from the convergence of the integral follows the convergence of the integral;

2) from the divergence of the integral follows the divergence of the integral.

Theorem 3. Let the functions f And g non-negative on [ a;b) and integrable on [ a;b-h] "h>0. If exists (0£ k£¥), then

1) from the convergence of the integral at k<¥ следует сходимость интеграла ,

2) from the divergence of the integral at k>0 follows the divergence of the integral.

Comment. If under the conditions of Theorem 3 0<k< ¥ (a finite number not equal to 0), then the integrals both converge or diverge at the same time.

It is convenient to take power functions as comparison functions: for the interval [ a;b) , and for the interval ( a;b] . The corresponding integrals converge at and diverge at (this is easy to verify by reducing the indicated integrals by linear change of variable to the integral considered in Example 1).

Example 4. Examine for convergence. .

Converges. This means, by Theorem 3, and converges. D

Example 6. Examine for convergence.

D X=0 – singular point of the function f(x)=ln x. Let .

This occurs, among other things, when a<1, когда сходится. Значит, по теореме 3 сходится и данный интеграл. D