Investigate the convergence of an improper integral online with a solution. Definite integral online
Are you here now? =) No, I wasn’t trying to intimidate anyone, it’s just that the topic of improper integrals is a very good illustration of how important it is not to start higher mathematics and other exact sciences. Everything you need to learn the lesson is on the website - in a detailed and accessible form, if you wish...
So, let's start with. Figuratively speaking, an improper integral is an “advanced” definite integral, and in fact there are not so many difficulties with them, and besides, the improper integral has a very good geometric meaning.
What does it mean to evaluate an improper integral?
Calculate improper integral - this means finding the NUMBER(exactly the same as in the definite integral), or prove that it diverges(that is, you end up with infinity instead of a number).
There are two types of improper integrals.
Improper integral with infinite limit(s) of integration
Sometimes such an improper integral is called improper integral of the first kind. IN general view an improper integral with an infinite limit most often looks like this: . How is it different from a definite integral? At the upper limit. It is endless: .
Less common are integrals with an infinite lower limit or with two infinite limits: , and we will look at them later - when you get the hang of it :)
Well, now let’s look at the most popular case. In the vast majority of examples, the integrand function continuous in between, and this one important fact should be checked first! Because if there are gaps, then there are additional nuances. For definiteness, let us assume that even then the typical curved trapezoid will look like this:
Note that it is infinite (not bounded on the right), and improper integral numerically equal to its area. The following options are possible:
1) The first thought that comes to mind: “since the figure is infinite, then 
", in other words, the area is also infinite. It may be so. In this case they say that the improper integral diverges.
2) But. As paradoxical as it may sound, the area of an infinite figure can be equal to... a finite number! For example: . Could this be true? Easily. In the second case, the improper integral converges.
3) About the third option a little later.
In what cases does an improper integral diverge and in what cases does it converge? This depends on the integrand, and we'll look at specific examples very soon.
What happens if an infinite curved trapezoid is located below the axis? In this case, the improper integral 
(diverges) or equal to the final negative number.
Thus, improper integral can be negative.
Important! When you are given ANY improper integral to solve, then, generally speaking, there is no talk about any area and there is no need to build a drawing. I explained the geometric meaning of the improper integral only to make it easier to understand the material.
Since the improper integral is very similar to the definite integral, let us remember the Newton-Leibniz formula: 
. In fact, the formula is also applicable to improper integrals, only it needs to be slightly modified. What is the difference? At the infinite upper limit of integration: . Probably, many guessed that this already smacks of the application of the theory of limits, and the formula will be written like this: 
.
What is the difference from a definite integral? Nothing special! As in the definite integral, you need to be able to find the antiderivative function (indefinite integral), and be able to apply the Newton-Leibniz formula. The only thing that has been added is the calculation of the limit. Whoever has a bad time with them, learn a lesson Function limits. Examples of solutions, because it’s better late than in the army.
Let's look at two classic examples:
Example 1
For clarity, I will draw a drawing, although, I emphasize once again, on practice There is no need to build drawings in this task.
The integrand function is continuous on the half-interval, which means that everything is fine and the improper integral can be calculated by the “standard” method.
Application of our formula 
and the solution to the problem looks like this:
That is, the improper integral diverges, and the area of the shaded curved trapezoid is equal to infinity.
In the example considered, we have the simplest table integral and the same technique for applying the Newton-Leibniz formula as in the definite integral. But this formula will be applied under the sign of the limit. Instead of the usual letter of a “dynamic” variable, the letter “be” appears. This should not confuse or baffle, because any letter is no worse than the standard “X”.
If you do not understand why at , then this is very bad, either you do not understand the simplest limits (and generally do not understand what a limit is), or you do not know what the graph of a logarithmic function looks like. In the second case, attend a lesson Graphs and properties of elementary functions.
When solving improper integrals, it is very important to know what the graphs of basic elementary functions look like!
The finished task should look something like this:
“
! When preparing an example, we always interrupt the solution and indicate what happens to the integrand – is it continuous on the interval of integration or not?. With this we identify the type of improper integral and justify further actions.
Example 2
Calculate the improper integral or establish its divergence.
Let's make the drawing: 
First, we note the following: the integrand is continuous on the half-interval. Hood. We solve using the formula 
:
(1) We take the simplest integral of power function(this special case is in many tables). It is better to immediately move the minus sign beyond the limit sign so that it does not get in the way in further calculations.
(2) Substitute the top and lower limit s according to the Newton-Leibniz formula.
(3) We indicate that at (Gentlemen, this should have been understood a long time ago) and simplify the answer.
Here the area of an infinite curved trapezoid is a finite number! Unbelievable but true.
The finished example should look something like this:
“
The integrand function is continuous on
“
What to do if you come across an integral like - with break point on the integration interval? This means there is a typo in the example. (Most likely), or about an advanced level of training. In the latter case, due to additivity properties, we should consider two improper integrals on intervals and then deal with the sum.
Sometimes, due to a typo or intent, an improper integral may not exist at all, so, for example, if you put the square root of “x” in the denominator of the above integral, then part of the integration interval will not be included in the domain of definition of the integrand at all.
Moreover, the improper integral may not exist even with all the “apparent well-being”. Classic example: . Despite the definiteness and continuity of the cosine, such an improper integral does not exist! Why? It's very simple because:
- does not exist appropriate limit.
And such examples, although rare, do occur in practice! Thus, in addition to convergence and divergence, there is also a third outcome of the solution with a valid answer: “there is no improper integral.”
It should also be noted that the strict definition of an improper integral is given precisely through the limit, and those who wish can familiarize themselves with it in 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 be able to solve the improper integral either.
Which of the table integrals is the integrand similar to? It reminds me of an arctangent: 
. These considerations suggest that it would be nice to have a square in the denominator. This is done by replacement.
Let's replace:
The indefinite integral has been found; in this case, it makes no sense to add a constant.
It is always useful to check the draft, that is, differentiate the result obtained:
The original integrand has been obtained, which means that the indefinite integral has been found correctly.
Now we find the improper integral:
(1) We write the solution in accordance with the formula 
. It is better to immediately move the constant beyond the limit sign so that it does not interfere with further calculations.
(2) We substitute the upper and lower limits in accordance with the Newton-Leibniz formula. Why 
at ? See the arctangent graph in the already recommended article.
(3) We get the final answer. A fact that is useful to know by heart.
Advanced students may not find the indefinite integral separately and not use the replacement method, but rather use the method of substituting the function under the differential sign and solving the improper integral “immediately.” In this case, the solution should look something like this:
“
The integrand is continuous on . 
“
Example 4
Calculate the improper integral or establish its divergence.
! This is a typical example, and similar integrals are found very often. Work it out well! The antiderivative function is found here using the extraction method full square, more details about the method can be found in the lesson Integrating Some Fractions.
Example 5
Calculate the improper integral or establish its divergence.
This integral can be solved in detail, that is, first find the indefinite integral by making a change of variable. Or you can solve it “immediately” - by subsuming the function under the differential sign. Who has any mathematical training?
Complete solutions and answers at the end of the lesson.
Examples of solutions to improper integrals with an infinite lower limit of integration can be found on the page Efficient methods for solving improper integrals. There we also analyzed the case when both limits of integration are infinite.
Improper integrals of unbounded functions
Or improper integrals of the second kind. Improper integrals of the second kind are insidiously “encrypted” under the usual definite integral and look exactly the same: But, unlike the definite integral, the integrand suffers an infinite discontinuity (does not exist): 1) at the point , 2) or at the point , 3) or at both points at once, 4) or even on the integration segment. We will look at the first two cases; for cases 3-4 at the end of the article there is a link to an additional lesson.
Just an example to make it clear: . It seems to be a definite integral. But in fact, this is an improper integral of the second kind; if we substitute the value of the lower limit into the integrand, then our denominator goes to zero, that is, the integrand simply does not exist at this point!
In general, when analyzing an improper integral you always need to substitute both integration limits into the integrand. In this regard, let's check the upper limit: 
. Everything is fine here.
The curvilinear trapezoid for the type of improper integral under consideration fundamentally looks like this:
Here everything is almost the same as in the integral of the first kind.
Our integral is numerically equal to area a shaded curved trapezoid that is not limited at the top. In this case, there can be two options*: the improper integral diverges (the area is infinite) or the improper integral is equal to a finite number (that is, the area of an infinite figure is finite!).
* by default we usually assume that the improper integral exists
All that remains is to modify the Newton-Leibniz formula. It is also modified with the help of a limit, but the limit no longer tends to infinity, but to the value on the right. It is easy to follow from the drawing: along the axis we must approach the breaking point infinitely close on right.
Let's see how this is implemented in practice.
Example 6
Calculate the improper integral or establish its divergence.
The integrand has an infinite discontinuity at a point (don’t forget to check verbally or on a draft that everything is fine with the upper limit!)
First, let's calculate the indefinite integral: 
Replacement: 
If you have any difficulties with replacement, please refer to the lesson Substitution method in indefinite integral.
Let's calculate the improper integral:
(1) What's new here? There is practically nothing in terms of solution technology. The only thing that has changed is the entry under the limit icon: . The addition means that we are striving for the value on the right (which is logical - see the graph). Such a limit in the theory of limits is called one-sided limit. In this case we have right-hand limit.
(2) We substitute the upper and lower limits using the Newton-Leibniz formula.
(3) Let's deal with at . How to determine where an expression is going? Roughly speaking, you just need to substitute the value into it, substitute three quarters and indicate that . Let's comb the answer.
In this case, the improper integral is equal to a negative number. There is no crime in this, just the corresponding curved trapezoid is located under the axis.
And now two examples for independent solutions.
Example 7
Calculate the improper integral or establish its divergence.
Example 8
Calculate the improper integral or establish its divergence.
If the integrand does not exist at the point
An infinite curved trapezoid for such an improper integral fundamentally looks like this.
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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) 
; d) 
.
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) Let's 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 
. Consequently, this integral is an improper integral 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, we speak of improper integrals.
10.1 Improper integrals of the 1st kind
Improper integral Type 1 occurs 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 $+\infty$; we will discuss other options a little 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, we need 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 for $N\rightarrow +\infty$.
Definition. Let there be a finite limit
\[ A=\lim_(N \rightarrow +\infty)I(N)=\lim_(N \rightarrow +\infty)\int _a^(N)f(x)dx. \]
Then we say that the improper integral of the 1st kind (19) is convergent and the value $A$ is assigned to it; the function itself is called integrable on the interval $\left[ a, \, +\infty \right)$. If specified limit does not exist or is equal to $\pm \infty$, then 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 function is known, so
\[ I(N)=\int _0^(N) \frac(dx)(1+x^2)=arctgx|_0^(N)=arctgN. \]
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 $\pi /2$.
Convergent 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 on this interval $f(x)>0$, 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 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)
Let's introduce the function
\[ I(N)=\int _1^(N)\frac(1)(x^k)\,dx. \]
In this case the antiderivative is known, so
\[ I(N)=\int _1^(N)\frac(1)(x^k)\,dx\,=\frac(x^(1-k))(1-k)|_1^N = \frac(N^(1-k))(1-k)-\frac(1)(1-k) \]
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 integral (20) converges for $k>1$, and diverges for $k \leq 1$.
Let us now consider the option when the lower limit of integration is equal to $-\infty$, and the upper one is finite, i.e. let's look at the integrals
\[ I=\int _(-\infty)^af(x)dx. \]
However, this option can be reduced to the previous one if we make a change of variables $x=-s$ and then change the limits of integration in places, 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)
and $f(x)$ 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 and is assigned the value $I=I_1+I_2$ (in accordance with additivity over the interval). If at least one of the integrals $I_1$, $I_2$ diverges, integral (21) is called divergent.
It can be proven that the convergence of integral (21) does not depend on the choice of 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 the corresponding reformulation taking into account the choice integration interval).
10.1.2 Tests 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 $0 a$. Then
1. If the integral \[ \int _a^(+\infty)g(x)dx \] converges, then the integral \[ \int _a^(+\infty)f(x)dx converges. \] 2. If the integral \[ \int _a^(+\infty)f(x)dx \] diverges, then the integral \[ \int _a^(+\infty)g(x)dx diverges. \]
Theorem (second comparison criterion). 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 simultaneously.
Consider the integral
\[ I=\int _1^(+\infty)\frac(1)(x+\sin x)\,dx. \]
The integrand expression 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 comparison criterion, we come to the conclusion that our integral converges or diverges simultaneously with the integral
\[ \int _1^(+\infty)\frac(1)(x)\,dx . \]
As was shown in the previous example, this integral diverges ($k=1$). Consequently, 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. \[ \int _(0)^(+\infty)\frac(xdx)(x^3+1). \]