Convergence and divergence of internal integrals. Invalid integrals
Examples of the study of improper integrals for convergence
Example 1.
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Thus, this integral converges at a\u003e 1 and dispel at a £ 1.
Example 2.
Explore convergence. Calculate the integral by definition: 
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Thus, this integral converges when a<1 и расходится при a³1.
Example 3.
Explore convergence 
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<0) при x стремящемся к 0, поэтому разобьем исходный интеграл на два
.
The convergence of the first integral I1 is investigating using an equivalent function: (t. N\u003e 0), and the integral converges at m\u003e -1 (example 2). Similarly, for integral I2:
And the integral converges at m + n<-1 (пример2). Следовательно, исходный интеграл сходится при выполнении одновременно двух условий m>-1 and M + N<-1, и будет расходится при нарушении хотя бы одного из них.
Example 4. Explore convergence.
The integrated function can be infinitely large (if M<0) при x стремящемся к 0, поэтому разобьем исходный интеграл на два:
Since arctgx »x at x®0, then the integral I1 is equivalent to the integral, which converges at m + 1\u003e -1 i.e. at m\u003e -2 (example1).
For the integrative function in the incompatible integral of the first kind of I2, we will select equivalent:
T. K. ARCTGX »P / 2 with X® ¥. Consequently, according to the second sign of comparison, the integral I2 will be converged at M + N<-1, и расходится в противном случае.
Combining conditions for the convergence of integrals I1 and I2 We obtain the conditions for the convergence of the original integral: M\u003e -2 and M + N<-1 одновременно.
Comment. In examples 2-4, 2 signs of comparison were used, which provides the necessary and sufficient conditions for convergence, which allows, setting convergence at a certain condition to the parameter values, not to prove the integral divergence in violation of the convergence conditions.
Example 5. Explore convergence.
This integral contains a special point 0, in which the integrand function can turn into infinity at P<0, поэтому снова разобьем исходный интеграл на два:
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The I1 integral is an incompatible integral of the second genus, and the integrand function is equivalent to X®0 XP (E-X ®1 at x®0), i.e. i1 converges at p\u003e -1 (Example 1).
The I2 integral is an incompatible integral of the first kind. Choose a function that is equivalent to the integrand function, so that it does not contain an indicative function, it fails. Therefore, to use a sign of comparison 2, as in previous examples, it is impossible. Apply the first sign of comparison, for which we use the following well-known fact:
With a\u003e 0 and any p. From this, and the fact that the XPE-AX function is continuous, it follows that this function is limited, that is, there is such a constant M\u003e 0 that Xpe-AX< M. Возьмем, например, a=1/2, и оценим интеграл I2 сверху:
That is, the I2 integral converges at any p.
Thus, the original integral converges at p\u003e -1.
Example 6. Explore convergence.
We will replace the variable: t \u003d lnx, and get
The splitting of the integral of two was produced in analogously to example 5. The integral I1 is completely equivalent to the integral I1 from Example 5 and, therefore, converges when Q<1.
Consider the integral I2. Provided 1-p<0 этот интеграл полностью эквивалентен интегралу I2 в примере 5 (доказательство сходимости аналогично, а условие 1-p<0 нужно для выполнения 
and a \u003d (1-p) / 2.).
So, I2 converges at p\u003e 1. However, on this study of the convergence of this integral is not completed, since the used sign of convergence gives only sufficient conditions for convergence. Therefore, it is necessary to study convergence at 1-p £ 0.
Consider the case p \u003d 1. Then the I2 integral is equivalent, which converges at Q\u003e 1 (we note that in this case the integral I1 is diverged) and otherwise dispersed.
At P.<1 оценим интеграл I2 и покажем его расходимость. Для этого вспомним, что 
At 1-p\u003e 0, and, therefore, starting from some A\u003e 1. T.-
Q.E.(1-
P.)
T. ³ M \u003d Const\u003e 0. Then for the I2 integral is valid
,
Where the integral in the right part dispels, which proves the divergence of the integral I2.
Summing the results obtained, we obtain that the source integral converges when Q<1 и p>1, otherwise the integral is diverged.
Example 6. Explore the absolute and conditional convergence.
Severe the original integral of two:
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Convergence. Integral I1 Equivalent 
, i.e. converges at p<2 (пример 1) , причем абсолютно, так как подынтегральная функция положительна на отрезке интегрирования.
The integral I2 converges on the sign of the Dirichlet-Abel at p\u003e 0 t. The first sin (x) is limited, and the function 1 / XP monotonously tends to zero with X-X tend to infinity.
We show that at p £ 0 integral diverges. We use the criterion of Cauchy for this, or rather by denial
.
Take the following values \u200b\u200bas R1I R2: R1 \u003d 2PK and R2 \u003d 2PK + P / 2, then
, with p\u003e 0.
Thus, the integral converges at 0 Absolute convergence The absolute convergence of the integral I1 is already established, consider the absolute convergence of I2. We estimate the integral from above: To prove divergence at p £ 1, we estimate the integral from the bottom We break the last integral from the difference of functions on the difference in integrals If both integrals converge, the integral of the difference converges, if one of the integrals diverges, and the other converges - then the integral is separated from the difference. In the case of divergence of both integrals, the convergence of the integral of the difference is subject to further study. We are interested in the second of the cases described. Divergent (example 1) at p<1. сходится по признаку Дирихле-Абеля при 1>p\u003e 0 (see convergence), therefore the integral is estimated at the bottom with a divergent integral, i.e. it is dispersed. The case of P³1 does not interest us, since these values \u200b\u200bof the parameter integral diverges. Thus, the original integral converges absolutely at 0 Theorem 12.11 (sign of comparison of internal integrals). Let the functions f (x) and g (x) are continuous on the interval [A, "\u003e) and satisfy the condition of 0 Fix)? (X). Then from the convergence of the integral it follows the convergence of integral conversely, the integral divergence (12.64) should include integral divergence (12.63). Evidence. We introduce notation: Function P (K) is inconsistent; In fact, if and I 2, then J. fix) DX\u003e 0, and then Take the sequence of values \u200b\u200b(/? ") -\u003e"\u003e; Then the corresponding sequence of function values (F (R n)) is monotonous and unremarkable. Let the integral (12.63) converges, then the sequence (67 ( R. IT)) limited; But then limited and consistency (F. (/? ")), So, by virtue of Theorem 7.13, it converges. Consequently, there is a limit F (R) for R. - + "\u003e, i.e. The integral (12.64) converges. Now we will prove the second part of the theorem; Let the integral (12.64) dispel. If you assume that the integral (12.63) converges, then on the proven above integral (12.64) should also converge, which contradicts the condition. Theorem is proved. ? Comment. A similar sign of comparison is also fair for improper integrals of the second kind. If functions / (x) and g. (x) Continuous on semi-interval [A\u003e b) and for all points in some neighborhood of a special point b. Completed conditions 0. (x), then from the convergence of the integral JG (X) DX follows the bridge of the integral j / (x) dx, and from the divergence of the integral J / (X) DX - the bridge of the integral JG (X) DX. Consider examples on the study of convergence of internal integrals. Example 27. t. ^ -. X 3 (1 + e l) Decision. Compare the integrated function in this integral with the function DG. Obviously, about - - - h. g * (1 + 0 x j county J-JDX converges; Therefore, due to a sign of comparison converges and Dan- 1 H. ny integral. Example 28. I-. Decision. Comparing the integrand function of this integral with a function of 1 / x, we see that (1 + in x) / x\u003e 1 / x on the interval 1
it is, therefore, this integral is also based on a sign of comparison. In conclusion, we give without proof the criterion of Cauchy convergence of the incomprehensible integral of the first kind. 12.10.4. Absolute and conditional convergence of internal integrals Definition 5. Incompheated EntheGl J / (X) DX is called absolutely convergentIf the integral j | / (x) is converged | DX. Definition 6. Invisible integral J / (x) DX is called conditionally sitting wearingIf it converges, and the integral j | / (x) | DX diverges. Note that from the absolute convergence of the integral and its convergence due to the estimate of the 3 specific integral and the Cauchy criterion. Theorem 12.13 (a sign of Dirichlet - Abel *). Let the function / (x) are continuous and has limited primitive F. (x) In the interval [A, "\u003e), and the function G (x) has a continuous derivative on this gap, does not increase and strives for zero at x -\u003e © about. Then imaginary integral converges. Evidence. Apply integration in parts to the integral J / (x) G (x) DX on an arbitrary cut R R " from [ but, °°). We have: Theorem 12.12. For convergence of the immunity integral (12.64), it is necessary and enough to find such a number for any E\u003e 0 BUT \u003e 0, what for any R " and /? "big than BUT, Inequality is performed: By condition theorem F (X) Limited, i.e. | F (x) | K. The function G (x) does not increase and tends to zero at x - ""\u003e, it means. g (x) \u003e 0, a g "(x) Abel Niels Henrik (1802-1829) - Norwegian mathematician. Since under the condition of the theorem G (x) - "0 at x -\u003e © °, for an arbitrary number E\u003e 0 can be found A \u003e. such that R "L Inequality will be performed g (R ") Substituting this in assessment (12.68), we get: what corresponds to the Curious criterion of the integral convergence (12.66). Theorem is proved. ? Consider the examples of using the feature of Dirichle - Abel convergence of internal integrals. Example 29. F ^^ DX, A\u003e 0. Decision. Put / (x) \u003d sin x g (x) \u003d L / X "; it is easy to make sure that all the conditions of the theorem are made, i.e. This integral converges. When A\u003e 1 this integral ral converges absolutely. Really, | sin x / xp 1 / d l, integral J (L / X E) DX converges, i.e. According to a comparison (Theorem 12.11), this integral is converged. Example 30. JSIN X 2 DX - Fresnel integral, Decision. Imagine this integral in the form of the amount: Since SIN X 2 is a continuous function on the segment (0, 1J, the first integral in (12.69) exists. To determine the convergence of the incompatible integral in the right-hand side (12.69), we put / (x) \u003d x SIN 2, g. (x) \u003d 1 / x. Then for the function / (x) primitive F (X) = -COSX 2 /! It is limited to the interval | 1, "\u003e), and # (x) is positive, tends to zero at x -" °° and has a continuous derivative on (1, © o). It means on the basis of Dirichlet - Abel, the second integral in (12.69) converges, i.e. Fresnel integral also converges. As you know, finding the integral may be a rather complicated task. It would be a great disappointment to make a calculation of an incompatible integral and detect at the end of the way that it dispels. Therefore, methods allow, without serious computing, in one type of functions, make a conclusion about convergence or divergence of an incomplete integral. The first and second comparison theorems that will be discussed below are largely helping to explore incomplete integrals for convergence. Let F (x)? 0. Then functions are monotonously increasing from variables T or-D (as we take d\u003e 0, it seeks to zero on the left). If, with an increase in the arguments of the function F 1 (T) and F 2 (-d) remain limited from above, this means that the corresponding incomprehensible integrals converge. This is based on the first comparison theorem for integrals from non-negative functions. Suppose for the function f (x) and g (x) with x? A conditions: Then from the convergence of the integral follows the convergence of the integral, and the divergence of integral should be Since 0? F (x)? G (x) and functions are continuous, then By condition, the integral converges, i.e. It has the final magnitude. Consequently, the integral converges as well. Let the integral now diverge. Suppose that the integral converges, but then the integral must be converged, which contradicts the condition. Our assumption is incorrect, the integral diverges. The comparison theorem for improper integrals of the 2nd kind. Suppose for functions f (x) and g (x) on the gap, increasingly increases with x\u003e +0. For her at x\u003e +0 inequality<. Несобственный интеграл есть эталонный интеграл 2-го рода, который при p=<1 сходится; следовательно, по 1-й теореме сравнения для несобственных интегралов 2-го рода интеграл сходится также. The comparison theorem for improper integrals of the 1st genus. Suppose for the function f (x) and g (x) on the interval, and the intercom segment is the final, that is, the numbers are limited, and not infinity. Some tasks lead to the need to abandon these restrictions. So indispensable integrals appear. Geometrical meaning of an incompatible integral It turns out quite simple. In the case when the schedule function y. = f.(x.)
located above axis OX. The defined integral expresses the area of \u200b\u200bthe curvilinear trapezion, limited curve y. = f.(x.)
, Axis of abscissa and orders x. = a.
, x. = b. . In turn, inappropriate integral expresses the area of \u200b\u200bunlimited (infinite) curvilinear trapezium, concluded between the lines y. = f.(x.)
(in the figure below - red), x. = a. and an abscissa axis. In the same way, incompatible integrals are determined and for other infinite intervals: The area of \u200b\u200ban infinite curvilinear trapezium can be a finite number and in this case the immutable integral is called convergent. The area may be infinity and in this case the immutable integral is called divergent. Use the integral limit instead of the most incompatible integral. In order to calculate the incompatible integral, you need to use the limit of a specific integral. If this limit exists and is finite (not equal to infinity), then a uncomparable integral is called converging, and otherwise - divergent. What a variable is striving for a sign of a limit, depends on whether we have the case with an incompatible integral of the first kind or second kind. We will find out about it now. So, the record of the immunity integral is different from the usual definite integral in the fact that the upper integration limit is infinite. Definition. Invalid integral with an endless upper limit of integration from continuous function f.(x.)
At the interval OT a.
before ∞
called the limit of the integral of this function with the upper integration limit b.
and lower integration limit a.
provided that the upper limit of integration is indefinitely growing. If this limit exists and is equal to some number, not infinity, then incoming integral is called convergent, and the number to which the limit is equal to its value. Otherwise involved integral is called divergent And he does not attribute any meaning. Example 1. Calculate the incompatible integral (if it converges). Decision. Based on the definition of an incompatible integral, we find Since the limit exists and is equal to 1, then involved integral converge and equal to 1. In the following example, the integrand function is almost as in Example 1, only the degree of ICA is not a twice, but the letter of alpha, but the task is to study an incomplete integral for convergence. That is, to answer the question: under what values \u200b\u200bof the alpha, this incoming integral converges, and at what diverge? Example 2. To explore the convergence of the immobility integral (Lower integration limit is greater than zero). Decision. Suppose first that, then In the resulting expression, we turn to the limit when: It is easy to see that the limit in the right part exists and is zero, when, that is, there is no, when, that is. In the first case, that is, when there is a place. If, then The withdrawal of our study is as follows: this involved integral converge at I. diverge at. Applying to the submitted type of internal integral formula Newton-Leibnia This is the generalized formula of Newton Labitsa. Example 3. Calculate the incompatible integral (if it converges). The limit of this integral exists: The second integral constituting the amount expressing the original integral: The limit of this integral also exists: We find the sum of two integrals, which is and the value of the initial incompatible integral with two infinite limits: Let the function f.(x.)
set on the segment from a.
before b.
And unlimited on it. Suppose that the function addresses infinity at the point b.
, while in all other points of the segment it is continuous. Definition. Incompatible integral function f.(x.)
On the cut from a.
before b.
called the limit of the integral of this function with the upper integration limit c.
If with a pursuit c.
to b.
The function increases indefinitely, and at the point x. = b.
The function is not defined. If this limit exists, the incoming integral of the second kind is called converging, otherwise divergent. Using the Newton-Labender formula, we derive.
, i.e. the integral converges at p\u003e 1.
.
.
Incombaty integrals of the first kind - with infinite limits and their convergence
Unobual integrals with an endless upper limit
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And there is no.
, You can withdraw the following formula very similar to it:
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.Unobual integrals of the second kind - from unlimited functions and their convergence
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