You're still here? =) No, I didn't try to intimidate anyone, just the topic of improper integrals is a very good illustration of how important it is not to run higher mathematics and other exact sciences. To master the lesson, the site has everything - in a detailed and accessible form, there would be a desire….

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, moreover, the improper integral has a very good geometric meaning.

What does it mean to calculate an improper integral?

Calculate improper integral - this means to find the NUMBER(just like in the definite integral), or prove that it diverges(that is, end up with infinity instead of a number).

Improper integrals are of two types.

Improper integral with infinite limit (s) of integration

Sometimes such an improper integral is called improper integral of the first kind... V general view an improper integral with an infinite limit most often looks like this:. What is its difference 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 a taste :)

Well, now let's look at the most popular case. In the overwhelming majority of examples, the integrand function continuous in between, and this important fact should be checked first! For 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 limited to the right), and improper integral numerically equal to its area... In this case, it is possible the following options:

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, the improper integral is said to be diverges.

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

3) About the third option a little later.

In what cases does the improper integral diverge, and in what cases does it converge? It depends on the integrand, and specific examples we will review very soon.

And what will happen if the infinite curved trapezoid is located below the axis? In this case, the improper integral (diverges) or is equal to the final negative number.

Thus, improper integral can be negative.

Important! When ANY improper integral is proposed to you for a solution, then, generally speaking, there is no question of any area and there is no need to build a drawing... I told 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, then we recall 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? In the infinite upper limit of integration:. Probably, many have guessed that this already smacks of application of the theory of limits, and the formula will be written like this: .

What is the difference from a definite integral? Yes, nothing special! As in a definite integral, you need to be able to find the antiderivative function ( indefinite integral), be able to apply the Newton-Leibniz formula. The only thing that has been added is the calculation of the limit. Who's bad with them, learn a lesson Limits of functions. Examples of solutions, for it is better late than in the army.

Let's look at two classic examples:

Example 1

For clarity, I will build a drawing, although, I emphasize again, on practice you do not need to build drawings in this task.

The integrand is continuous on the half-interval, which means that everything is normal 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 curvilinear trapezoid is equal to infinity.

In the example considered, we have the simplest tabular integral and the same technique for applying the Newton-Leibniz formula as in the definite integral. But this formula will be applied under the limit sign. Instead of the usual letter "dynamic" variable, the letter "be" appears. This should not be confusing or perplexing, 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 don’t know what the graph of a logarithmic function looks like. In the second case, visit the lesson Graphs and properties of elementary functions.

When solving improper integrals, it is very important to know how the graphs of the fundamental elementary functions!

The final design of the assignment should look something like this:

“

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

Example 2

Calculate the improper integral or establish its divergence.

Let's execute the drawing:

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

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

(2) Substitute the upper and lower limits according to the Newton-Leibniz formula.

(3) We point out that with (Gentlemen, this has long been understood) and simplify the answer.

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

The final layout of the example should look something like this:

“

The integrand is continuous on

“

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

Sometimes, due to a misprint or intent, an improper integral can not exist at all, so, for example, if the square root of "x" is put 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, an 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 corresponding limit.

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

It should also be noted that a 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 solve the improper integral either.

Which of the tabular integrals does the integrand look like? She reminds the arctangent: ... From these considerations, the thought suggests itself that it would be nice to get a square in the denominator. This is done by replacement.

Let's replace:

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

On a draft, it is always useful to check, that is, to differentiate the result obtained:

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

Now we find the improper integral:

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

(2) Substitute the upper and lower limits according to the Newton-Leibniz formula. Why at ? See the arc tangent graph in the already recommended article.

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

Advanced students may not find the indefinite integral separately, and not use the replacement method, but use the method of bringing the function under the differential sign and solve the improper integral “right away”. 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.

! it typical example, and similar integrals are very common. Work it out well! The antiderivative function is found here by the method of selecting a complete square, for more details on the method, see the lesson Integration of 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 changing the variable. And you can decide "right away" - by bringing the function under the sign of the differential. Who has some kind of mathematical background.

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 Effective Methods for Solving Improper Integrals... In the same paper, the case was analyzed 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 cunningly "encrypted" under the usual definite integral and look exactly the same: But, unlike a definite integral, the integrand suffers an infinite discontinuity (does not exist): 1) at a point, 2) or at a point, 3) or at both points at once, 4) or even on the interval of integration. We will consider 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 vanishes, that is, the integrand simply does not exist at this point!

In general, in the analysis of the improper integral it is always necessary to substitute both limits of integration into the integrand... In this regard, let us check the upper limit: ... Everything is fine here.

The curvilinear trapezoid for the considered kind of improper integral looks like this in principle:

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

Our integral is numerically equal to the area of ​​a shaded curvilinear trapezoid that 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 habitually assume that the improper integral exists

It remains only to modify the Newton-Leibniz formula. It is also modified with the help of the limit, but the limit no longer tends to infinity, but to the value on the right. It is easy to trace the drawing: along the axis we have to approach infinitely close to the break point on right.

Let's see how this is implemented in practice.

Example 6

Calculate the improper integral or establish its divergence.

The integrand suffers an infinite discontinuity at a point (do not forget verbally or on a draft to check if everything is normal with the upper limit!)

First, we calculate the indefinite integral:

Replacement:

For those who have difficulty replacing, refer to the lesson Replacement method in the indefinite integral.

We calculate the improper integral:

(1) What's new here? There is practically nothing in the solution technique. 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) Substitute the upper and lower limits according to the Newton-Leibniz formula.

(3) We deal with at. How to determine where an expression is heading? Roughly speaking, you just need to substitute the value into it, substitute three quarters and indicate that. Combing 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 an independent solution.

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 curvilinear trapezoid for such an improper integral basically looks like this.

The definite integral as the limit of the integral sum

can exist (i.e., have a definite finite value) only if the conditions

If at least one of these conditions is violated, then the definition loses its meaning. Indeed, in the case of an infinite segment, for example [ a; ) it cannot be broken down into NS parts of finite length
, which, moreover, with an increase in the number of segments, would tend to zero. In the case of unbounded at some point with[a; b] the requirement for an arbitrary point selection is violated on partial segments - cannot be selected =with, since the value of the function at this point is undefined. However, for these cases it is possible to generalize the concept of a definite integral by introducing one more passage to the limit. Integrals over infinite intervals and of discontinuous (unbounded) functions are called improper.

Definition.

Let the function
defined on the interval [ a; ) and integrable on any finite segment [ a; b], ie exists
for anyone b > a... Limit of the species
are called improper integral first kind (or an improper integral over an infinite interval) and denote
.

Thus, by definition,
=
.

If the right limit exists and is finite, then the improper integral
are called converging ... If this limit is infinite, or does not exist at all, then the improper integral is said to be diverges .

Similarly, one can introduce the concept of an improper integral of a function
by interval (–; b]:

=
.

And the improper integral of the function
over the interval (–; + ) is defined as the sum of the integrals introduced above:

=
+
,

where a- an arbitrary point. This integral converges if both terms converge, and diverges if at least one of the terms diverges.

From a geometric point of view, the integral
,
, determines the numerical value of the area of ​​an infinite curvilinear trapezoid bounded from above by the graph of the function
, left - straight
, below - by the OX axis. The convergence of the integral means the existence of a finite area of ​​such a trapezoid and its equality to the limit of the area of ​​a curvilinear trapezoid with a movable right wall
.

To the case of an integral with infinite limit, one can generalize and Newton-Leibniz formula:

=
= F ( + ) - F ( a),

where F ( + ) =
... If this limit exists, then the integral converges; otherwise, it diverges.

We have considered a generalization of the notion of a definite integral to the case of an infinite interval.

Let us now consider a generalization for the case of an unbounded function.

Definition

Let the function
defined on the interval [ a; b), is unbounded in some neighborhood of the point b, and is continuous on any segment
, where> 0 (and, therefore, is integrable on this interval, that is,
exists). Limit of the species
called improper integral of the second kind (or an improper integral of an unbounded function) and is denoted
.

Thus, the improper integral of unbounded at the point b functions are by definition

=
.

If the right limit exists and is finite, then the integral is called converging... If there is no finite limit, then the improper integral is called divergent.

Similarly, one can define the improper integral of the function
having an infinite discontinuity at the point a:

=
.

If the function
has an infinite discontinuity at an interior point with
, then the improper integral is defined as follows

=
+
=
+
.

This integral converges if both terms converge, and diverges if at least one term diverges.

From a geometric point of view, the improper integral of an unbounded function also characterizes the area of ​​an unbounded curvilinear trapezoid:

Since the improper integral is derived by passing to the limit from a definite integral, all the properties of a definite integral can be transferred (with appropriate refinements) to improper integrals of the first and second kind.

In many problems leading to improper integrals, it is not necessary to know what this integral is equal to; it is enough to just make sure that it converges or diverges. To do this, use convergence criteria. Convergence criteria for improper integrals:

1) Comparison attribute.

Let for everyone NS

... Then if
converges, then converges and
, and

... If
diverges, then diverges and
.

2) If converges
, then converges and
(the last integral in this case is called absolutely convergent).

The criteria for the convergence and divergence of improper integrals of unbounded functions are similar to those formulated above.

Examples of problem solving.

Example 1.

a)
; b)
; v)

G)
; e)
.

Solution.

a) By definition, we have:

.

b) Similarly

Therefore, this integral converges and is equal to .

c) By definition
=
+
, moreover, a- an arbitrary number. In our case, we put
, then we get:

This integral converges.

Hence, this integral diverges.

e) Consider
... To find the antiderivative of the integrand, it is necessary to apply the method of integration by parts. Then we get:

Since neither
nor
do not exist, then it does not exist and

Consequently, this integral diverges.

Example 2.

Investigate the convergence of the integral depending on the NS.

Solution.

At
we have:

If
, then
and. Consequently, the integral diverges.

If
, then
, a
, then

=,

Consequently, the integral converges.

If
, then

hence, the integral diverges.

Thus,

Example 3.

Calculate the improper integral or establish its divergence:

a)
; b)
; v)
.

Solution.

a) Integral
is an improper integral of the second kind, since the integrand
not limited at the point

... Then, by definition,

.

The integral converges and is equal to .

b) Consider
... Here also the integrand is not bounded at the point
... Therefore, this integral is improper of the second kind and, by definition,

Consequently, the integral diverges.

c) Consider
... Integrand
suffers an infinite gap at two points:
and
, the first of which belongs to the interval of integration
... Consequently, this integral is improper of the second kind. Then, by definition

=

=

.

Consequently, the integral converges and is equal to
.

ThemeNON-PROPER INTEGRALS

In the topic “ Definite integral»The concept of a definite integral was considered for the case of a finite interval
and limited function
(see Theorem 1 of §3). Now we will generalize this concept for the cases of an infinite interval and an unbounded function. The need for such a generalization is shown, for example, by such situations.

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

, where
.

2. Let the body 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 easy to show that the solution to this (differential!) Equation is the function
If we need to calculate the path traveled by the body to a complete stop, i.e. until the moment when
, then we arrive at an 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 any
it is integrable on the interval
, that is, there is an integral
.

Definition 1 ... The finite or infinite limit of this integral for
is called an improper integral of the first kind of the function
by interval
and are denoted by the symbol
... Moreover, if this limit is finite, then the improper integral is called convergent, otherwise (
or does not exist) - divergent.

So by definition

Examples of

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 be
- some antiderivative for the function
(exists on
since
- continuous). Then

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

, where
.

Examples of .

5.
.

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

Now we can find the integral considering that

:

III Properties

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

IV Other definitions

Definition 2 ... If
continuous on
, then

.

Definition 3 ... If
continuous on
, then take by definition

(- arbitrary),

moreover, the improper integral on the left-hand side converges if both integrals on the right-hand side converge.

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

Example 7 .

§2. Convergence criteria for an improper integral of the first kind

Most often, the improper integral cannot be calculated by definition, therefore, the approximate equality is used

(for large ).

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

I Integrals of Positive Functions

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

Theorem 1 ... An improper integral of the first kind of a non-negative function converges if and only if the function
remains limited when increasing .

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

Theorem 2 (1st comparison sign). 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
, then

... Let the integral
converges, then (by Theorem 1) the function
- limited. But then and
is bounded, and hence the integral
also converges. The second part of the theorem is proved similarly.

This criterion is not applicable in the case of divergence of the integral of
or the convergence of the integral of
... This disadvantage 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 improper integrals
and
converge or diverge at the same time.

Proof ... From the condition of the theorem, we obtain the following chain of equally strong statements:

, ,

.

Let, for example,
... Then:

We apply Theorem 2 and property 1) from §1 and obtain the assertion of Theorem 3.

The power function
,
... We invite students to prove themselves that the integral

converges at
and diverges at
.

Examples of . 1.
.

Consider the integrand on the interval
:

,
.

Integral
converges because
... According to the second comparison criterion, the integral
, and by virtue of property 2) in §1 the original integral also converges.

2.
.

Because
then there is
such that for

... For such variable values:

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

,

and hence, starting from some value of the variable, this fraction is less than 1. Therefore,

.

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