Improper integrals of the first kind: extension of the concept of a definite integral to cases of integrals with infinite upper or lower limits of integration, or both limits of integration are infinite.
Improper integrals of the second kind: extension of the concept of a definite integral to the cases of integrals of unbounded functions; the integrand does not exist at a finite number of points of 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 segment of integration 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 curved 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 from 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 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 integral with infinite integration limit
Sometimes such an improper integral is also called an improper integral of the first kind..gif" width="49" height="19 src=">.
Less common are integrals with an infinite lower limit or with two infinite limits: .
We will consider the most popular case https://pandia.ru/text/80/057/images/image005_1.gif" width="63" height="51"> ? No not always. Integrandhttps://pandia.ru/text/80/057/images/image007_0.gif" width="47" height="23 src=">
Let us depict in the drawing the graph of the integrand function. A typical graph and curved trapezoid for this case looks like this:
Improper integralhttps://pandia.ru/text/80/057/images/image009_0.gif" width="100" height="51">", 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: .. In the second case, the improper integral converges.
What happens if an infinite curved trapezoid is located below the axis?.gif" width="217" height="51 src=">.
: 
.
Example 1
The integrand function https://pandia.ru/text/80/057/images/image017_0.gif" width="43" height="23">, which means everything is fine and the improper integral can be calculated using the “standard” method.
Application of our formula https://pandia.ru/text/80/057/images/image018_0.gif" width="356" height="49">
That is, the improper integral diverges, and the area of the shaded curved trapezoid is equal to infinity.
When solving improper integrals, it is very important to know what the graphs of basic elementary functions look like!
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. Good..gif" width="327" height="53">
(1) We take the simplest integral of power function(this special case is in many tables). It is better to immediately place the minus outside 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 point out that https://pandia.ru/text/80/057/images/image024.gif" width="56" height="19 src="> (Gentlemen, this has long needed to be understood) and simplify answer.
Here the area of an infinite curved trapezoid is a finite number! Unbelievable but true.
Example 3
Calculate the improper integral or establish its divergence.
The integrand is continuous on .
First, let's try to find the antiderivative function (indefinite integral).
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:
It is always useful to perform a check, that is, to differentiate the result obtained:
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..gif" width="56" height="19 src=">? See the arctangent graph in the already repeatedly 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 function is continuous at https://pandia.ru/text/80/057/images/image041.gif" width="337" height="104">
“
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 is found here using the method of isolating a complete square.
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..
Improper integrals of unbounded functions
Sometimes such improper integrals are called 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: ..gif" width="39" height="15 src=">, 2) or at point , 3) or at both points at once, 4) or even on the integration segment. We will consider the first two cases; for cases 3-4 there is a link to an additional lesson at the end of the article.
Just an example to make it clear: https://pandia.ru/text/80/057/images/image048.gif" width="65 height=41" height="41">, 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..jpg" alt="Improper integral, discontinuity point at the lower limit of integration" width="323" height="380">!}
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!).
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 valuehttps://pandia.ru/text/80/057/images/image052.gif" width="28" height="19"> on right.
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: 
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 a one-sided limit. In this case we have a right-handed limit.
(2) We substitute the upper and lower limits using the Newton-Leibniz formula.
(3) Let's understand https://pandia.ru/text/80/057/images/image058.gif" width="69" height="41 src=">. How to determine where the expression should go? Roughly speaking, in you just need to substitute the value, substitute three quarters and indicate that comb the answer.
In this case, the improper integral is equal to a negative number.
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:
Here everything is absolutely the same, except that our limit tends to to valuehttps://pandia.ru/text/80/057/images/image052.gif" width="28" height="19"> we must approach infinitely close to the breaking point left.
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.
.
Let us 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.
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Definite integral as the limit of the integral sum
can exist (i.e. have a certain final value) only if the conditions are met
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 divided into P parts of finite length 
, which, moreover, would tend to zero with an increase in the number of segments. In the case of unlimited at some point With[a;
b] the requirement for arbitrary point selection is violated 
on partial segments – cannot be selected 
=With, since the value of the function at this point is undefined. However, even for these cases it is possible to generalize the concept of a definite integral by introducing another passage to the limit. Integrals over infinite intervals and over discontinuous (unbounded) functions are called not your own.
Definition.
Let the function 
is defined on the interval [ a; ) and is integrable on any finite interval [ a;
b], i.e. exists 
for anyone b
> a. Type limit 
called improper integral
first kind
(or an improper integral over an infinite interval) and denote 
.
Thus, by definition, 
=
.
If the limit on the right exists and is finite, then the improper integral 
called convergent
. If this limit is infinite, or does not exist at all, then they say that the improper integral diverges
.
Similarly, we can introduce the concept of an improper integral of the function 
along the interval (–; b]:
=
.
And the improper integral of the function 
over the interval (–; +) is defined as the sum of the integrals introduced above:
=
+
,
Where A– 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 above by the graph of the function 
, left – straight 
, from 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 an infinite limit, we can generalize Newton-Leibniz formula:
=
=F( +
) – F( a),
where F( +
)
=
. If this limit exists, then the integral converges, otherwise it diverges.
We considered a generalization of the concept 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 
is defined on the interval [ a;
b), is unlimited in some neighborhood of the point b, and is continuous on any interval 
, where>0 (and, therefore, is integrable on this interval, i.e. 
exists). Type limit 
called improper integral of the second kind
(or an improper integral of an unbounded function) and is denoted 
.
Thus, the improper integral of the unbounded at the point b functions exist by definition
=
.
If the limit on the right exists and is finite, then the integral is called convergent. If there is no finite limit, then the improper integral is called divergent.
Similarly, we can define the improper integral of the function 
having an infinite discontinuity at the point A:
=
.
If the function 
has an infinite gap at the 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 curved trapezoid:
Since an 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 that lead to improper integrals, it is not necessary to know what this integral is equal to, it is enough just to verify its convergence or divergence. For this they use signs of convergence. Signs of convergence of improper integrals:
1) Comparison sign.
Let it be for everyone X
. Then if 
converges, then converges 
, and 
. If 
diverges, then diverges and 
.
2) If converges 
, then converges and 
(the last integral in this case is called absolutely convergent).
The signs of 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) Likewise
Therefore, this integral converges and is equal to 
.
c) By definition 
=
+
, and A– arbitrary number. Let's put in our case 
, then we get:
This integral converges.
This means that 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 does not exist and
Therefore, this integral diverges.
Example 2.
Investigate the convergence of the integral 
depending on the P.
Solution.
At 
we have:
If 
, That 
And. Therefore, the integral diverges.
If 
, That 
, A 
, Then
=,
Therefore, the integral converges.
If 
, That
therefore, 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 a point 
. Then, by definition,
.
The integral converges and is equal to 
.
b) Consider 
. Here also the integrand is not limited at the point 
. Therefore, this integral is improper of the second kind and, by definition,
Therefore, the integral diverges.
c) Consider 
. Integrand 
suffers an infinite gap at two points: 
And 
, the first of which belongs to the integration interval 
. Therefore, this integral is an improper integral of the second kind. Then, by definition
=
=
.
Therefore, the integral converges and is equal to 
.
