You're still here? =) No, I didn't try to intimidate anyone, it's just that the topic of improper integrals is a very good illustration of how important it is not to run higher mathematics and others exact sciences. To master the lesson on the site, everything is there - in detail and accessible form, there would be a desire ....
So, let's start. Figuratively speaking, improper integral is an “advanced” definite integral, and in fact there are not so many difficulties with them, besides, the improper integral has a very good geometric meaning.
What does it mean to calculate an improper integral?
Calculate Improper Integral - it means to find a NUMBER(exactly the same as 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: . How is it different from a definite integral? In 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 analyze the most popular case. In the vast majority of examples, the integrand function continuous in between and this one important fact to check first! For if there are gaps, then there are additional nuances. For definiteness, we assume that even then the typical curvilinear trapezoid will look like this:
Note that it is infinite (not bounded on the right), and improper integral numerically equal to its area. In this case, the following options are possible:
1) The first thought that comes to mind is: “since the figure is infinite, then 
”, in other words, the area is also infinite. So it may be. In this case, we 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 instance: . Could it be? Easy. In the second case, the improper integral converges.
3) About the third option a little later.
When does an improper integral diverge and when does it converge? It depends on the integrand, and concrete examples we will review very soon.
But what happens if an infinite curvilinear trapezoid is located below the axis? In this case, the improper integral 
(diverges) or equal to the final negative number.
In this way, improper integral can be negative.
Important! When ANY improper integral is offered to you to solve, then, generally speaking, there is no talk 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's the Difference? In the infinite upper limit of integration: . Probably, many have guessed that this already smacks of applying the theory of limits, and the formula will be written as follows: 
.
How is it different 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. Solution examples because better late than in the army.
Consider two classic examples:
Example 1
For clarity, I will build a drawing, although, I emphasize once again, on practice it is not necessary to build drawings in this task.
The integrand is continuous on the half-interval , which means that everything is fine and the improper integral can be calculated using the “regular” method.
Application of our formula 
and the solution looks like this:
That is, the improper integral diverges, and the area of the shaded curvilinear trapezoid is equal to infinity.
In the considered example, we have the simplest tabular integral and the same technique for applying the Newton-Leibniz formula as in the definite integral. But this formula is applied under the sign of the limit. Instead of the usual letter of the "dynamic" variable, the letter "be" appears. This should not confuse or confuse, because any letter is no worse than the standard "X".
If you don’t understand why at , then this is very bad, either you don’t understand the simplest limits (and don’t understand what a limit is at all), or you don’t know what the graph looks like logarithmic function. 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 main elementary functions look like!
A clean job design should look something like this:
“
! When designing an example, we always interrupt the solution and indicate what happens to the integrand – is it continuous on the interval of integration or not. By this we identify the type of improper integral and substantiate further actions.
Example 2
Calculate the improper integral or establish its divergence.
Let's make a drawing: 
First, we notice the following: the integrand is continuous on the half-interval . Good. Solving with formula 
:
(1) We take the simplest integral of power function(this particular case is found in many tables). It is better to immediately move the minus beyond the limit sign so that it does not get underfoot in further calculations.
(2) We substitute the upper and lower limits according to the Newton-Leibniz formula.
(3) We indicate that when (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 clean design of the example should look something like this:
“
The integrand is continuous on
“
What to do if you come across an integral like - with breaking point on the interval of integration? This means that there is a typo in the example (Most likely) or an advanced level of education. In the latter case, due to additivity properties, one should consider two improper integrals on the intervals and and then deal with the sum.
Sometimes, due to a typo or the intention of an improper integral, it can not exist at all, so, for example, if we put in the denominator of the above integral Square root from "x", then part of the integration interval will not enter 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, although rare, are found in practice! Thus, in addition to convergence and divergence, there is also a third outcome of the solution with a full answer: "there is no improper integral."
It should also be noted that the strict definition of the improper integral is given precisely through the limit, and those who wish can familiarize themselves with it in the educational literature. Well, we continue 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 table integrals does the integrand look like? It reminds me of the arc tangent: 
. From these considerations, the thought suggests itself that it would be nice to get a square in the denominator. This is done by substitution.
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 perform a check, that is, to differentiate the result:
The original integrand was obtained, which means that the indefinite integral was 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 in further calculations.
(2) We substitute the upper and lower limits in accordance with the Newton-Leibniz formula. Why 
at ? See the arc tangent graph in the already repeatedly 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 summing the function under the differential sign and solve 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 very common. Work it out well! antiderivative function here is the selection method full square, more details about the method can be found in 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 solve it "immediately" - by summing up the function under the sign of the differential. Who has some mathematical background.
Complete solutions and answers at the end of the lesson.
Examples of solutions of improper integrals with an infinite lower limit of integration can be found on the page Efficient Methods for Solving Improper Integrals. The case where both integration limits are infinite is also considered there.
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 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 the denominator vanishes, that is, the integrand simply does not exist at this point!
In general, when analyzing the improper integral it is always necessary to substitute both integration limits into the integrand. In this regard, we also check the upper limit: 
. Everything is good here.
The curvilinear trapezoid for the considered variety of the improper integral fundamentally looks like this:
Here, almost everything is the same as in the integral of the first kind.
Our integral is numerically equal to the area of the shaded curvilinear 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 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 follow along 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 suffers an infinite break at a point (do not forget to check verbally or on a draft if everything is fine with the upper limit!)
First, we calculate the indefinite integral: 
Replacement: 
For those who have difficulty with the replacement, refer to the lesson Replacement method in indefinite integral.
We calculate the improper integral:
(1) What's new here? Practically nothing in terms of technique. The only thing that has changed is the entry under the limit icon: . The addition means that we are aiming for the value on the right (which is logical - see graph). Such a limit in the theory of limits is called unilateral limit. In this case we have right-hand limit.
(2) We substitute the upper and lower limits according to the Newton-Leibniz formula.
(3) Dealing with at . How do you determine where an expression is headed? 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 curvilinear 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.
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 broken down into P parts of finite length 
, which, moreover, would tend to zero with an increase in the number of segments. In the case of unbounded at some point With[a;
b] the requirement of an arbitrary choice of a point is violated 
on partial segments – cannot be selected 
=With, since the value of the function at this point is undefined. However, the notion of a definite integral can also be generalized for these cases by introducing one more passage to the limit. Integrals over infinite intervals and from discontinuous (unbounded) functions are called non-own.
Definition.
Let the function 
defined on the interval [ a; ) and is integrable on any finite interval [ a;
b], i.e. exists 
for anyone b
> a. view limit 
called improper integral
first kind
(or by 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 converging
. If this limit is infinite, or does not exist at all, then the improper integral is said to be diverges
.
Similarly, we 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 is 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 
, from below - 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 moving right wall 
.
For the case of an integral with an infinite limit, one can also generalize 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 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 
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 segment, i.e. 
exists). view limit 
called improper integral of the second kind
(or by the improper integral of an unbounded function) and is denoted 
.
Thus, the improper integral of unbounded at a point b functions are by definition
=
.
If the limit on the right 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 an improper integral of the function 
having an infinite discontinuity at a 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 the definite integral, then all the properties of the definite integral can be transferred (with appropriate refinements) to the 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 make sure that it converges or diverges. For this use signs of convergence. Signs of convergence of improper integrals:
1) Comparison sign.
Let for everyone X
. 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 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 is an arbitrary number. Let's put in our case 
, then we get:
This integral converges.
So 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 do not exist and
Therefore, this integral diverges.
Example 2
Investigate the convergence of the integral 
depending on the P.
Solution.
At 
we have:
If 
, then 
and. Therefore, the integral diverges.
If 
, then 
, a 
, then
=,
Therefore, the integral converges.
If 
, then
hence the integral diverges.
In this way, 
Example 3
Calculate the improper integral or set its divergence:
a) 
; b) 
; v) 
.
Solution.
a) Integral 
is an improper integral of the second kind, because the integrand 
not limited at a point 
. Then, by definition,
.
The integral converges and is equal to 
.
b) Consider 
. Here, too, the integrand is not bounded 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 discontinuity at two points: 
and 
, the first of which belongs to the interval of integration 
. Therefore, this integral is improper of the second kind. Then, by definition
=
=
.
Therefore, the integral converges and is equal to 
.
Definite integral
\[ I=\int_a^bf(x)dx \]
was constructed under the assumption that the numbers $a,\,b$ are finite and $f(x)$ is a continuous function. If one of these assumptions is violated, one speaks of improper integrals.
10.1 Improper integrals of the 1st kind
An improper integral of the first kind arises when at least one of the numbers $a,\,b$ is infinite.
10.1.1 Definition and basic properties
Let us first consider the situation when the lower limit of integration is finite and the upper limit is equal to $+\infty$; other options will be discussed later. For $f(x)$ continuous for all $x$ of interest to us, consider the integral
\begin(equation) I=\int _a^(+\infty)f(x)dx. \quad(19) \label(inf1) \end(equation)
First of all, it is necessary to establish the meaning of this expression. To do this, we introduce the function
\[ I(N)=\int _a^(N)f(x)dx \]
and consider its behavior as $N\rightarrow +\infty$.
Definition. Let there be a limit
\[ A=\lim_(N \rightarrow +\infty)I(N)=\lim_(N \rightarrow +\infty)\int _a^(N)f(x)dx. \]
Then the improper integral of the 1st kind (19) is said to be convergent and the value $A$ is assigned to it, the function itself is called integrable on the interval $\left[ a, \, +\infty \right)$. If the indicated limit does not exist or it is equal to $\pm \infty$, then the integral (19) is said to diverge.
Consider the integral
\[ I=\int _0^(+\infty) \frac(dx)(1+x^2). \]
\[ I(N)=\int _0^(N) \frac(dx)(1+x^2). \]
In this case, the antiderivative of the integrand is known, so that
\[ I(N)=\int _0^(N) \frac(dx)(1+x^2)=arctgx|_0^(N)=arctgN. \]
It is known that $arctg N \rightarrow \pi /2 $ for $N \rightarrow +\infty$. Thus, $I(N)$ has a finite limit, our improper integral converges and is equal to $\pi /2$.
Converging improper integrals of the 1st kind have all the standard properties of ordinary definite integrals.
1. If $f(x)$, $g(x)$ are integrable on the interval $\left[ a, \, +\infty \right)$, then their sum $f(x)+g(x)$ is also is integrable on this interval, and \[ \int _a^(+\infty)\left(f(x)+g(x)\right)dx=\int _a^(+\infty)f(x)dx+\int _a^(+\infty)g(x)dx. \] 2. If $f(x)$ is integrable on the interval $\left[ a, \, +\infty \right)$, then for any constant $C$ the function $C\cdot f(x)$ is also integrable on this interval, and \[ \int _a^(+\infty)C\cdot f(x)dx=C \cdot \int _a^(+\infty)f(x)dx. \] 3. If $f(x)$ is integrable on the interval $\left[ a, \, +\infty \right)$ and $f(x)>0$ on this interval, then \[ \int _a^ (+\infty) f(x)dx\,>\,0. \] 4. If $f(x)$ is integrable on the interval $\left[ a, \, +\infty \right)$, then for any $b>a$ the integral \[ \int _b^(+\infty) f(x)dx \] converges, and \[ \int _a^(+\infty)f(x)dx=\int _a^(b) f(x)dx+\int _b^(+\infty) f( x)dx \] (additivity of the integral over the interval).
The formulas for the change of variable, integration by parts, etc., are also valid. (with natural reservations).
Consider the integral
\begin(equation) I=\int _1^(+\infty)\frac(1)(x^k)\,dx. \quad (20) \label(mod) \end(equation)
We introduce the function
\[ I(N)=\int _1^(N)\frac(1)(x^k)\,dx. \]
In this case, the antiderivative is known, so that
\[ I(N)=\int _1^(N)\frac(1)(x^k)\,dx\,=\frac(x^(1-k))(1-k)|_1^N = \frac(N^(1-k))(1-k)-\frac(1)(1-k) \]
for $k \neq 1$,
\[ I(N)=\int _1^(N)\frac(1)(x)\,dx\,=lnx|_1^N= lnN \]
for $k = 1$. Considering the behavior for $N \rightarrow +\infty$, we come to the conclusion that the integral (20) converges for $k>1$, and diverges for $k \leq 1$.
Let us now consider the case when the lower limit of integration is equal to $-\infty$ and the upper one is finite, i.e. consider the integrals
\[ I=\int _(-\infty)^af(x)dx. \]
However, this variant can be reduced to the previous one if we make the change of variables $x=-s$ and then swap the limits of integration, so that
\[ I=\int _(-a)^(+\infty)g(s)ds, \]
$g(s)=f(-s)$. Let us now consider the case when there are two infinite limits, i.e. integral
\begin(equation) I=\int _(-\infty)^(+\infty)f(x)dx, \quad (21) \label(intr) \end(equation)
where $f(x)$ is continuous for all $x \in \mathbb(R)$. Let's split the interval into two parts: take $c \in \mathbb(R)$, and consider two integrals,
\[ I_1=\int _(-\infty)^(c)f(x)dx, \quad I_2=\int _(c)^(+\infty)f(x)dx. \]
Definition. If both integrals $I_1$, $I_2$ converge, then integral (21) is called convergent, it is assigned the value $I=I_1+I_2$ (according to the interval additivity). If at least one of the integrals $I_1$, $I_2$ diverges, integral (21) is said to be divergent.
It can be proved that the convergence of the integral (21) does not depend on the choice of the point $c$.
Improper integrals of the 1st kind with integration intervals $\left(-\infty, \, c \right]$ or $(-\infty, \, +\infty)$ also have all the standard properties of definite integrals (with a corresponding reformulation that takes into account the choice integration interval).
10.1.2 Criteria for the convergence of improper integrals of the 1st kind
Theorem (the first sign of comparison). Let $f(x)$, $g(x)$ be continuous for $x>a$, and let $0 a$. Then
1. If the integral \[ \int _a^(+\infty)g(x)dx \] converges, then the integral \[ \int _a^(+\infty)f(x)dx converges as well. \] 2. If the integral \[ \int _a^(+\infty)f(x)dx \] diverges, then the integral \[ \int _a^(+\infty)g(x)dx also diverges. \]
Theorem (the second sign of comparison). Let $f(x)$, $g(x)$ be continuous and positive for $x>a$, and let there be a finite limit
\[ \theta = \lim_(x \rightarrow +\infty) \frac(f(x))(g(x)), \quad \theta \neq 0, \, +\infty. \]
Then the integrals
\[ \int _a^(+\infty)f(x)dx, \quad \int _a^(+\infty)g(x)dx \]
converge or diverge at the same time.
Consider the integral
\[ I=\int _1^(+\infty)\frac(1)(x+\sin x)\,dx. \]
The integrand is a positive function on the integration interval. Further, for $x \rightarrow +\infty$ we have:
$\sin x$ is a "small" correction to the denominator. More precisely, if we take $f(x)=1/(x+\sin x)$, \, $g(x)=1/x$, then
\[ \lim _(x \rightarrow +\infty)\frac(f(x))(g(x))=\lim _(x \rightarrow +\infty)\frac(x)(x+\sin x) =1. \]
Applying the second criterion of comparison, we come to the conclusion that our integral converges or diverges simultaneously with the integral
\[ \int _1^(+\infty)\frac(1)(x)\,dx . \]
As shown in the previous example, this integral diverges ($k=1$). Therefore, the original integral diverges.
Calculate the improper integral or establish its convergence (divergence).
1. \[ \int _(0)^(+\infty)e^(-ax)\,dx. \] 2. \[ \int _(0)^(+\infty)xe^(-x^2)\,dx. \] 3. \[ \int _(-\infty)^(+\infty)\frac(2xdx)(x^2+1). \] 4. \[ \int _(0)^(+\infty)\frac(xdx)((x+2)^3). \] 5. \[ \int _(-\infty)^(+\infty)\frac(dx)(x^2+2x+2). \] 6. \[ \int _(1)^(+\infty)\frac(lnx)(x^2)\,dx. \] 7. \[ \int _(1)^(+\infty)\frac(dx)((1+x)\sqrt(x)). \] 8. \[ \int _(0)^(+\infty)e^(-\sqrt(x))\,dx. \] 9. \[ \int _(0)^(+\infty)e^(-ax)\cos x\,dx. \] 10. \[ \int _(0)^(+\infty)\frac(xdx)(x^3+1). \]
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's depict the graph of the integrand in the drawing. A typical graph and a curvilinear 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. So it may be. In this case, we 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 curvilinear trapezoid is located below the axis?.gif" width="217" height="51 src=">.
: 
.
Example 1
The integrand https://pandia.ru/text/80/057/images/image017_0.gif" width="43" height="23">, which means that everything is fine and the improper integral can be calculated using the "regular" 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 curvilinear trapezoid is equal to infinity.
When solving improper integrals, it is very important to know how the graphs of the main elementary functions look like!
Example 2
Calculate the improper integral or establish its divergence.
Let's make a drawing: 
First, we notice the following: the integrand is continuous on the half-interval . Good..gif" width="327" height="53">
(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 beyond the limit sign so that it does not get underfoot in further calculations.
(2) We substitute the upper and lower limits according to the Newton-Leibniz formula.
(3) We indicate that https://pandia.ru/text/80/057/images/image024.gif" width="56" height="19 src="> (Gentlemen, this has long been understood) and simplify answer.
Here, the area of an infinite curvilinear trapezoid is equal to 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 does the integrand look like? It reminds me of the arc tangent: 
. From these considerations, the thought suggests itself that it would be nice to get a square in the denominator. This is done by substitution.
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 in further calculations.
(2) We substitute the upper and lower limits in accordance with the Newton-Leibniz formula..gif" width="56" height="19 src=">?
(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 summing the function under the differential sign and solve the improper integral "immediately". In this case, the solution should look something like this:
“
The integrand 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 very common. Work it out well! The antiderivative function is found here by the full square selection method.
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 solve it "immediately" - by summing the function under the sign of the differential ..
Improper integrals of unbounded functions
Sometimes such improper integrals are called 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: ..gif" width="39" height="15 src=">, 2) or at the 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: https://pandia.ru/text/80/057/images/image048.gif" width="65 height=41" height="41">, then our denominator turns to zero, that is, the integrand simply does not exist at this point!
In general, when analyzing the improper integral it is always necessary to substitute both integration limits into the integrand..jpg" alt="(!LANG:Improper integral, discontinuity point in the lower limit of integration" width="323" height="380">!}
Here, almost everything is the same as in the integral of the first kind.
Our integral is numerically equal to the area of the shaded curvilinear 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!).
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 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 suffers an infinite break at a point (do not forget to check verbally or on a draft if everything is fine with the upper limit!)
First, we calculate the indefinite integral: 
Replacement: 
We calculate the improper integral:
(1) What's new here? Practically nothing in terms of technique. The only thing that has changed is the entry under the limit icon: . The addition means that we are aiming for the value on the right (which is logical - see graph). Such a limit in the theory of limits is called a one-sided limit. In this case, we have a right-hand limit.
(2) We substitute the upper and lower limits according to the Newton-Leibniz formula.
(3) Understanding 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 for it, substitute three quarters and indicate that... We 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 curvilinear trapezoid for such an improper integral fundamentally looks like this:
Everything is exactly the same here, except that the limit tends to to valuehttps://pandia.ru/text/80/057/images/image052.gif" width="28" height="19"> we must get infinitely close to the breaking point left.
Improper integrals of the first kind: dissemination of the concept definite integral for the 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 the finite interval 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 interval 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.
The geometric meaning of the improper integral turns out to be quite simple. When the graph of the 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) , abscissa and ordinates x = a , x = b. In turn, the improper integral expresses the area of an unbounded (infinite) curvilinear trapezoid enclosed between the lines y = f(x) (pictured below in red) x = a and the abscissa axis.
Improper integrals are defined similarly for other infinite intervals:
The area of an infinite curvilinear trapezoid can be a finite number, in which case the improper integral is called convergent. The area can also be infinity, in which case the improper integral is called divergent.
Using the limit of an integral instead of the improper integral itself. In order to calculate 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, otherwise it is divergent. What the variable under the limit sign tends to depends on whether we are dealing with an improper integral of the first kind or of the second kind. Let's find out about it now.
Improper integrals of the first kind - with infinite limits and their convergence
Improper integrals with an infinite upper limit
So, the record of the 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) between a before ∞ is called the limit of the integral of this function with the upper limit of integration b and the lower limit of integration a provided that the upper limit of integration grows indefinitely, i.e.
.
If this limit exists and is equal to some number, and not to infinity, then the improper integral is called convergent, and the number equal to the limit is taken as its value. Otherwise the improper integral is called divergent and no value 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 the given 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 of 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. Investigate the convergence of an improper integral(the lower integration limit is greater than zero).
Solution. Suppose first that , then
In the resulting expression, we pass to the limit at :
It is easy to see that the limit on the right-hand side exists and zero, when , that is , and does not exist when , that is .
In the first case, that is, when . If , then 
and does not exist.
The conclusion of our study is the following: improper integral converges at and diverges at .
Applying to the studied type of improper integral the Newton-Leibniz formula 
, we can derive the following very similar formula:
.
This is the generalized Newton-Leibniz formula.
Example 3. Compute Improper Integral(if it converges).
The limit of this integral exists:
The second integral, which is 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) set on the segment from a before b and unlimited on it. Suppose the function goes to infinity at the point b , while at all other points of the segment it is continuous.
Definition. Improper integral of the function f(x) on the segment from a before b is called the limit of the integral of this function with the upper limit of integration c , if when striving c To b the function increases indefinitely, 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 divergent.
Using the Newton-Leibniz formula, we derive.