Improper integrals of the first kind. In essence, this is the same definite integral, but in cases where the integrals have infinite upper or lower limits of integration, or both limits of integration are infinite.

Improper integrals of the second kind. In essence, this is the same definite integral, but in cases where the integral is taken from unbounded functions, the integrand at a finite number of points does not have a finite segment of integration, turning to infinity.

For comparison. When introducing the concept of a definite integral, it was assumed that the function f(x) is continuous on the interval [ a, b], and the integration segment is finite, that is, it is limited by numbers, and not by infinity. Some tasks lead to the need to abandon these restrictions. This is how improper integrals appear.

Geometric meaning of the improper integral It turns out quite simply. In the case when the graph of a function y = f(x) is above the axis Ox, the definite integral expresses the area of ​​a curvilinear trapezoid bounded by a curve y = f(x) , x-axis and ordinates x = a , x = b. In turn, the improper integral expresses the area of ​​an unlimited (infinite) curvilinear trapezoid enclosed between the lines y = f(x) (in the picture below - red), x = a and the abscissa axis.

Improper integrals are defined similarly for other infinite intervals:

The area of ​​an infinite curved trapezoid can be a finite number, in which case the improper integral is called convergent. The area can also be infinity, and in this case the improper integral is called divergent.

Using the limit of an integral instead of the improper integral itself. In order to evaluate the improper integral, you need to use the limit of the definite integral. If this limit exists and is finite (not equal to infinity), then the improper integral is called convergent, and otherwise - divergent. What a variable tends to under the limit sign depends on whether we are dealing with an improper integral of the first kind or of the second kind. Let's find out about this now.

Improper integrals of the first kind - with infinite limits and their convergence

Improper integrals with infinite upper limit

So, writing an improper integral differs from the usual definite integral in that the upper limit of integration is infinite.

Definition. An improper integral with an infinite upper limit of integration of a continuous function f(x) in the interval from a before ∞ the limit of the integral of this function with the upper limit of integration is called b and the lower limit of integration a provided that the upper limit of integration grows without limit, i.e.

.

If this limit exists and is equal to some number rather than infinity, then an improper integral is called convergent, and the number to which the limit is equal is taken as its value. Otherwise an improper integral is called divergent and no meaning is attributed to it.

Example 1. Calculate improper integral(if it converges).

Solution. Based on the definition of the improper integral, we find

Since the limit exists and is equal to 1, then this improper integral converges and is equal to 1.

In the following example, the integrand is almost the same as in example 1, only the degree x is not two, but the letter alpha, and the task is to study the improper integral for convergence. That is, the question remains to be answered: at what values ​​of alpha does this improper integral converge, and at what values ​​does it diverge?

Example 2. Examine the improper integral for convergence(the lower limit of integration is greater than zero).

Solution. Let us first assume that , then

In the resulting expression, we move to the limit at:

It is easy to see that the limit on the right side exists and is equal to zero when , that is , and does not exist when , that is .

In the first case, that is, when . If , then and doesn't exist.

The conclusion of our study is as follows: this improper integral converges at and diverges at .

Applying the Newton-Leibniz formula to the type of improper integral being studied , you can derive the following formula, which is very similar to it:

.

This is a generalized Newton-Leibniz formula.

Example 3. Calculate improper integral(if it converges).

The limit of this integral exists:

The second integral, making up the sum expressing the original integral:

The limit of this integral also exists:

.

We find the sum of two integrals, which is also the value of the original improper integral with two infinite limits:

Improper integrals of the second kind - from unbounded functions and their convergence

Let the function f(x) given on the segment from a before b and is unlimited on it. Suppose that the function goes to infinity at the point b , while at all other points of the segment it is continuous.

Definition. An improper integral of a function f(x) on the segment from a before b the limit of the integral of this function with the upper limit of integration is called c , if when striving c To b the function increases without limit, and at the point x = b function not defined, i.e.

.

If this limit exists, then the improper integral of the second kind is called convergent, otherwise it is called divergent.

Using the Newton-Leibniz formula, we derive.

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. Examine 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.Subject

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

Let
Integrals of positive functions
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 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
(1st comparison sign). Let the functions
. Let's denote:
, That

. Because
. 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
(1st comparison sign). Let the functions
(2nd sign of comparison). Let the functions
continuous and non-negative on
. Then if
at
(1st comparison sign). Let the functions
, then the improper integrals

diverges. converge or diverge simultaneously.

, ,

.

. From the conditions of the theorem we obtain the following chain of equivalent statements:
and satisfy the inequality

Let, for example,

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
.

Examples . 1.
.

and diverges at
:

,
.

Consider the integrand on the interval
Integral
converges, because
. Based on the 2nd comparison criterion, the integral also converges

2.
.

, and due to property 2) from §1, the original integral also converges.
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.

.

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

Lecture 24. IMPROPER INTEGRALS

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.

If the integrand has a discontinuity of the second kind on the (finite) interval of integration, we speak of an improper integral of the second kind.

10.2.1 Definition and basic properties

Let us denote the integration interval by $\left[ a, \, b \right ]$; both of these numbers are assumed to be finite below. If there is only 1 discontinuity, it can be located either at point $a$, or at point $b$, or inside the interval $(a,\,b)$. Let us first consider the case when there is a discontinuity of the second kind at point $a$, and at other points the integrand function is continuous. So we are discussing the integral

\begin(equation) I=\int _a^b f(x)\,dx, (22) \label(intr2) \end(equation)

and $f(x) \rightarrow \infty $ when $x \rightarrow a+0$. As before, the first thing to do is to give meaning to this expression. To do this, consider the integral

\[ I(\epsilon)=\int _(a+\epsilon)^b f(x)\,dx. \]

Definition. Let there be a finite limit

\[ A=\lim _(\epsilon \rightarrow +0)I(\epsilon)=\lim _(\epsilon \rightarrow +0)\int _(a+\epsilon)^b f(x)\,dx. \]

Then the improper integral of the second kind (22) is said to converge and the value $A$ is assigned to it; the function $f(x)$ itself is said to be integrable on the interval $\left[ a, \, b\right]$.

Consider the integral

\[ I=\int ^1_0\frac(dx)(\sqrt(x)). \]

The integrand function $1/\sqrt(x)$ at $x \rightarrow +0$ has an infinite limit, so at the point $x=0$ it has a discontinuity of the second kind. Let's put

\[ I(\epsilon)=\int ^1_(\epsilon )\frac(dx)(\sqrt(x))\,. \]

In this case, the antiderivative is known,

\[ I(\epsilon)=\int ^1_(\epsilon )\frac(dx)(\sqrt(x))=2\sqrt(x)|^1_(\epsilon )=2(1-\sqrt( \epsilon ))\rightarrow 2\]

at $\epsilon \rightarrow +0$. Thus, the original integral is a convergent improper integral of the second kind, and it is equal to 2.

Let us consider the option when there is a discontinuity of the second kind in the integrand function at the upper limit of the integration interval. This case can be reduced to the previous one by making the change of variable $x=-t$ and then rearranging the limits of integration.

Let us consider the option when the integrand function has a discontinuity of the second kind inside the integration interval, at point $c \in (a,\,b)$. In this case, the original integral

\begin(equation) I=\int _a^bf(x)\,dx (23) \label(intr3) \end(equation)

presented as a sum

\[ I=I_1+I_2, \quad I_1=\int _a^cf(x)\,dx +\int _c^df(x)\,dx. \]

Definition. If both integrals $I_1, \, I_2$ converge, then the improper integral (23) is called convergent and is assigned a value equal to the sum of the integrals $I_1, \, I_2$, the function $f(x)$ is called integrable on the interval $\left [a, \, b\right]$. If at least one of the integrals $I_1,\, I_2$ is divergent, the improper integral (23) is called divergent.

Convergent improper integrals of the 2nd kind have all the standard properties of ordinary definite integrals.

1. If $f(x)$, $g(x)$ are integrable on the interval $\left[ a, \,b \right ]$, then their sum $f(x)+g(x)$ is also integrable on this interval, and \[ \int _a^(b)\left(f(x)+g(x)\right)dx=\int _a^(b)f(x)dx+\int _a^(b)g (x)dx. \] 2. If $f(x)$ is integrable on the interval $\left[ a, \, b \right ]$, then for any constant $C$ the function $C\cdot f(x)$ is also integrable on this interval , and \[ \int _a^(b)C\cdot f(x)dx=C \cdot \int _a^(b)f(x)dx. \] 3. If $f(x)$ is integrable on the interval $\left[ a, \, b \right ]$, and on this interval $f(x)>0$, then \[ \int _a^(b ) f(x)dx\,>\,0. \] 4. If $f(x)$ is integrable on the interval $\left[ a, \, b \right ]$, then for any $c\in (a, \,b)$ the integrals \[ \int _a^ (c) f(x)dx, \quad \int _c^(b) f(x)dx \] also converge, and \[ \int _a^(b)f(x)dx=\int _a^(c ) f(x)dx+\int _c^(b) f(x)dx \] (additivity of the integral over the interval).

Consider the integral

\begin(equation) I=\int _0^(1)\frac(1)(x^k)\,dx. (24) \label(mod2) \end(equation)

If $k>0$, the integrand tends to $\infty$ as $x \rightarrow +0$, so the integral is improper of the second kind. Let's introduce the function

\[ I(\epsilon)=\int _(\epsilon)^(1)\frac(1)(x^k)\,dx. \]

In this case, the antiderivative is known, so

\[ I(\epsilon)=\int _(\epsilon)^(1)\frac(1)(x^k)\,dx\,=\frac(x^(1-k))(1-k )|_(\epsilon)^1= \frac(1)(1-k)-\frac(\epsilon ^(1-k))(1-k). \]

for $k \neq 1$,

\[ I(\epsilon)=\int _(\epsilon)^(1)\frac(1)(x)\,dx\,=lnx|_(\epsilon)^1= -ln \epsilon. \]

for $k = 1$. Considering the behavior at $\epsilon \rightarrow +0$, we come to the conclusion that integral (20) converges at $k

10.2.2 Tests for the convergence of improper integrals of the 2nd kind

Theorem (the first sign of comparison). Let $f(x)$, $g(x)$ be continuous for $x\in (a,\,b)$, and $0 1. If the integral \[ \int _a^(b)g(x)dx \] converges, then the integral \[ \int _a^(b)f(x)dx converges. \] 2. If the integral \[ \int _a^(b)f(x)dx \] diverges, then the integral \[ \int _a^(b)g(x)dx diverges. \]

Theorem (second comparison criterion). Let $f(x)$, $g(x)$ be continuous and positive for $x\in (a,\,b)$, and let there be a finite limit

\[ \theta = \lim_(x \rightarrow a+0) \frac(f(x))(g(x)), \quad \theta \neq 0, \, +\infty. \]

Then the integrals

\[ \int _a^(b)f(x)dx, \quad \int _a^(b)g(x)dx \]

, then the improper integrals

Consider the integral

\[ I=\int _0^(1)\frac(1)(x+\sin x)\,dx. \]

The integrand is a positive function on the integration interval, the integrand tends to $\infty$ as $x \rightarrow +0$, so our integral is an improper integral of the second kind. Further, for $x \rightarrow +0$ we have: if $g(x)=1/x$, then

\[ \lim _(x \rightarrow +0)\frac(f(x))(g(x))=\lim _(x \rightarrow +0)\frac(x)(x+\sin x)=\ frac(1)(2) \neq 0,\, \infty \, . \]

Applying the second comparison criterion, we come to the conclusion that our integral converges or diverges simultaneously with the integral

\[ \int _0^(+1)\frac(1)(x)\,dx . \]

As was shown in the previous example, this integral diverges ($k=1$). Consequently, the original integral also diverges.

Calculate the improper integral or establish its convergence (divergence).

1. \[ \int _(0)^(1)\frac(dx)(x^3-5x^2)\,. \] 2. \[ \int _(3)^(7)\frac(x\,dx)((x-5)^2)\,. \] 3. \[ \int _(0)^(1)\frac(x\,dx)(\sqrt(1-x^2))\,. \] 4. \[ \int _(0)^(1)\frac(x^3\,dx)(1-x^5)\,. \] 5. \[ \int _(-3)^(2)\frac(dx)((x+3)^2)\,. \] 6. \[ \int _(1)^(2)\frac(x^2\,dx)((x-1)\sqrt(x-1))\,. \] 7. \[ \int _(0)^(1)\frac(dx)(\sqrt(x+x^2))\,. \] 8. \[ \int _(0)^(1/4)\frac(dx)(\sqrt(x-x^2))\,. \] 9. \[ \int _(1)^(2)\frac(dx)(xlnx)\,. \] 10. \[ \int _(1)^(2)\frac(x^3\,dx)(\sqrt(4-x^2))\,. \] 11. \[ \int _(0)^(\pi /4)\frac(dx)(\sin ^4x)\,. \]