The sign of the equivalence of the integral. How to designate an unclassified integral and z'yasuvati yogo zbіzhnist. Appointment and main power

Nevlasni integrals of the first kind. As a matter of fact, the very same song integration, but in the opposite cases, if they have integrated an inexhaustible upper or a lower interintegration, or an insult between an inextricable integration.

Uneven integrals of a different kind. As a matter of fact, this is the same singable integral, but in reverse, if the integral is taken from non-integrated functions, the integrand function is in the last number the point of the final integrity cannot be integrated, instead of inconsistency.

For povnyannya. With the introduction of the understanding of the sing integral, it was conveyed that the function f(x) without interruption to the wind [ a, b]. Deyakі zavdannya prizvodit to nebhіdnostі vіdmovitysa vіd tsikh obmezhen. This is how unsatisfactory integrals are blamed.

Geometric zmіst of a non-contiguous integral z'yasovuєtsya dosit easily. At times, if the schedule of the function y = f(x) know more than Ox, the singing integral turns the area of ​​the curvilinear trapezium, the circumscribed curve y = f(x) , abscissa and ordinates x = a , x = b. At its edge, the unsheltered integral bends over the area of ​​the uncircumscribed (unskinned) curvilinear trapezium, laid between the lines y = f(x) (a little lower - a red color), x = a i vіsyu abscissa.

An analogous rank is assigned to the unmatched integrals of other non-limiting intervals:

The area of ​​\u200b\u200ban infinite curvilinear trapezium can be the last number and in which way the inconsistent integral is called similar. The area can be and inconsistency and in some way the inconsistent integral is called rozbіzhny.

Vykoristanya mezhі і іntegrаl zamіst іnevlіsnіshії іntegrаl. In order to calculate the ambiguous integral, it is necessary to draw between sing integral. As if the boundary between the original and the end (not the same inconsistencies), then the unclassified integral is called similar, and in a different way - razbіzhny. How pragne is changed under the sign of the boundary, to fall in line with what can be right with the unsettled integral of the first genus and another genus. We know about it at once.

Inconsistent integrals of the first kind - with indistinct boundaries and similarity

Uneven integrals with an unskinned upper boundary

Also, the record of the non-contiguous integral is considered to be the same as the sig- nificant integral, since the upper boundary of the integration is not limited.

Appointment. An indistinct integral with an inexhaustible upper limit of integration into uninterrupted function f(x) for a break a before ∞ is called the boundary of the integral of the function and the upper boundary of integration b that lower boundary integration a remember that the upper boundary of integration is growing unbounded, then.

.

If it is between the true and the real number, and not inconsistency, then undefined integral is called similar and the number, which is closer to the boundary, is taken as its meaning. In a different direction unclassified integral is called razbіzhny and it is not assigned the same meaning.

Example 1. Calculate the undefined integral(how the wines converge).

Solution. On the basis of the designation of the non-consecutive integral, we know

So, as between the current and the old 1, then the Danes non-contiguous integral converge i dorivnyu 1.

In the offensive butt, the pidintegral function may be the same as in the butt 1, only the steps of the x are not a two, but the letter alpha, but the task of the last integral integral is zbіzhnist. In order to depend on the power: for some values ​​of the alpha, does the inconsistent integral converge, and for some ones it diverges?

Example 2(The lower boundary of integration is greater than zero).

Solution. Let's start back, sho, todi

Let's move on to the boundary in the opposite direction at:

It doesn’t matter what the boundary between the right part is true and equal to zero, if, then, and if not, if, then.

In the first mood, that may be the place. Yakscho something i don't know.

Visnovok of our next offensive: Denmark non-contiguous integral converge at i disperse at .

The Newton-Leibniz formula , you can enter the footsteps and go to her formula:

.

The formula of Newton-Leibniz is zagalnen.

Example 3. Calculate the undefined integral(how the wines converge).

Between tsgogo іsnuє:

Another integral, which makes the sum, which turns the external integral:

The inter-integral is also used:

.

We know the sum of two integrals, which is the value of the outer non-continuous integral with two infinite boundaries:

Inconsistent integrals of a different kind - in the form of nonexistent functions and their similarity

Come on function f(x) assigned to the vіdіzku vіd a before b and is not surrounded by new. It is acceptable that the function is transformed into inconsistency at the point b , at that hour, at all other points, the wind is without interruption.

Appointment. The incomprehensible integral of the function f(x) on the vіdіzku vіd a before b is called the boundary of the integral of the function and the upper boundary of integration c , as with exercise c before b the function is growing exponentially, but dots x = b function not assigned, then.

.

For example, if there is a boundary, then an unclassified integral of another kind is called similar, otherwise - razbіzhny.

Vikoristovuyuchi Newton-Leibniz formula, apparently.

As a pіdintegral function can be on the (terminal) interval of integration, having explored another genus, one can speak of an undefined integral of another genus.

10.2.1 Appointment and basic authority

Significantly interval of integration $ \ left [ a, \, b \ right ] $; If there is more than 1 expansion, the vin can be either at the point $a$, or at the point $b$, or in the middle of the interval $(a,\,b)$. Let's take a look at the reverse, if we look at another kind of є at the point $a$, and at other points the integrand is uninterrupted. Otzhe, we discuss the integral

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

moreover, $f(x) \rightarrow \infty $ if $x \rightarrow a+0$. As before, we should next give a sense to that viraz. For whom we can look at the integral

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

Appointment. Nehay isnuє kіnceva boundary

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

Then we say that an undefined integral of a different kind (22) converges, and to it we assign the value $ A $, the function $ f (x) $ itself is called integrated over the interval $ \ left [a, \, b \ right] $.

Let's look at a variant, if we look at another kind of integrand function - the upper interinterval of integration. This turn can be brought to the front by replacing the change $x=-t$ and then rearranging the interintegration.

Let's look at a variant if we look at another kind of integrand function in the middle of the integration interval, at the point $ c \in (a, \, b) $. Whose mind has an outward integral

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

serve at the sight of sumi

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

Appointment. If the two integrals $I_1, \, I_2$ converge, then the non-consecutive integral (23) is called similar and the value assigned to it is equal to the sum of the integrals $I_1, \, I_2$, and the function $f(x)$ is called integrated over the interval $\ left[a,\,b\right]$. If we want one of the integrals $I_1,\, I_2$ to be disjoint, the undefined integral (23) is called disproportionate.

Converging unclassified integrals of the 2nd kind can be used as standard features of traditional singable integrals.

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

Let's look at the integral

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

Like $k>0$, the integrand function is right $\infty$ for $x \rightarrow +0$, and the integral is of a different kind. Introducing the function

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

In this moment, the first view is at home, so

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

for $k \neq 1$,

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

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

10.2.2 Signs of morbidity of obscure integrals of the 2nd kind

Theorem (the first sign of the difference). Let $f(x)$, $g(x)$ be uninterrupted for $x\in (a,\,b)$, moreover, $0 1. This is the integral \[ \int _a^(b)g(x)dx \] converge, then the th integral \[ \int _a^(b)f(x)dx. \] 2. If the integral \[ \int _a^(b)f(x)dx \] diverge, then the і integral \[ \int _a^(b)g(x)dx diverge. \]

Theorem (another sign of difference). Let $f(x)$, $g(x)$ be uninterrupted and positive for $x\in (a,\,b)$, moreover, there is a finite boundary

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

Total integrals

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

converge or disperse at the same time.

Apply the following inconsistent integrals to work

butt 1
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In this way, the given integral converges for a>1 and diverges for a1.

butt 2 Dosliditi on zbіzhnist. We calculate the integral for the appointments:
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In this way, the danish integral converges for a<1 и расходится при a³1.

butt 3 Follow the profitability .

<0) при x стремящемся к 0, поэтому разобьем исходный интеграл на два

.

The convergence of the first integral I1 can be followed by an additional equivalent function: (because n>0), and the integral converges when m>-1 (appendix 2). Similarly, for the integral I2:

And the integral converges after 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.

butt 4 Dosliditi on zbіzhnist.

The integrand function can be infinitely great (like m<0) при x стремящемся к 0, поэтому разобьем исходный интеграл на два:

Oskіlki arctgx »x at x®0, then the integral I1 is equivalent to the integral , which converges at m + 1> -1 then at m> -2 (applied 1).

For the integrand function in the non-wired integral of the first genus I2, it is sub-optimally equivalent to:

sparks arctgx » p/2 at x® ¥. Later, after another sign, the integration of I2 converges for m + n<-1, и расходится в противном случае.

One by one, calculate the efficiency of the integrals I1 and I2, we take away the efficiency of the output integral: m>-2 and m+n<-1 одновременно.

Respect. In the butts of 2-4 vicorists, there are 2 signs of difference, which is necessary for the safety of the necessary sufficient intelligence, which allows, having set the efficiency for the deuce of the mental value of the parameters, not to bring the integration of the integral when the efficiency of the mind is damaged.

butt 5 Dosliditi on zbіzhnist.

The danish integral to mark the singular point 0, in which the integrand function can be transformed into inconsistency at p<0, поэтому снова разобьем исходный интеграл на два:

.

The integral I1 is an incomprehensible integral of a different kind, and the pintegral function is equivalent at x®0 to the function xp (e-x ®1 at x®0), then I1 converges at p>-1 (appendix 1).

Integral I2 is an incomprehensible integral of the first kind. Choose a function that is equivalent to the integrand function, such that it does not take away the display function, do not go into it. To that, it is not possible to win the mark of equal 2, like in the front butts. It is necessary to persue the sign of the difference, it is necessary to win such a fact:

For a>0, it's like p. Because the xpe-ax function is uninterrupted, weep, because the function is limited, so there is such a constant M>0, so xpe-ax< M. Возьмем, например, a=1/2, и оценим интеграл I2 сверху:

That is, the integral I2 converges for whatever p.

In this way, the final integral converges at -1.

butt 6 Dosliditi on zbіzhnist.

Let's change the change: t = lnx i take

Breaking the integral into two is viroblen similarly to the butt 5. The integral I1 is exactly equivalent to the integral I1 from the butt 5 i, then, converge at q<1.

Let's look at the integral I2. Wash your mind 1-p<0 этот интеграл полностью эквивалентен интегралу I2 в примере 5 (доказательство сходимости аналогично, а условие 1-p<0 нужно для выполнения i a = (1-p) / 2.).

Also, I2 converges for p>1. Prote, to whom the investigation of the efficiency of this integral is not completed, the shards of victories of the sign of efficiency give more than sufficient intelligence. Therefore, it is necessary to increase the efficiency at 1-p£0.

Let's take a look at p=1. Then the integral I2 is equivalent, which converges for q>1 (depending on which way the integral I1 diverges) and diverges in a different direction.

For p<1 оценим интеграл I2 и покажем его расходимость. Для этого вспомним, что With 1-p>0, i, then, pochinayuchi z deyakogo A>1 vikonano T- QE(1- P) T³M=const>0. Then, for the integral I2, the estimate

,

The deintegral at the right part diverge, so as to bring the divergence of the integral I2.

Subsuming the results, we assume that the resulting integral converges at q<1 и p>1, otherwise the integral diverges.

butt 6 Follow the absolute mindfulness.

Rozіb'єmo vihіdny Іntegrl for two:

.

Life. Integral I1 equivalent , i.e. converge for p<2 (пример 1) , причем абсолютно, так как подынтегральная функция положительна на отрезке интегрирования.

The integral I2 converges after the Dirichlet-Abel sign at p>0, because the primary sin(x) is delimited, and the function 1/xp is monotonically reduced to zero when x is reduced to inconsistency.

Let us show that the integral diverges for p£0. We are speeding up for which criteria of Kosh, or rather, for the following lists

.

Let's take R1і R2 as the current values: R1=2pk і R2=2pk+p/2, then

for p>0.

In this way, the integral converges at 0

Absolute comfort The absolute efficiency of the integral I1 is already installed, let's look at the absolute efficiency of I2. Let's estimate the integral to the beast:

, i.e., the integral converges for p>1.

To prove the versatility for p £ 1, we estimate the integral from below

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Rozіb'єmo remaining іntegrіl vіd vіd іnіtіnі functions іn іnііtіvаі іntegrаіv

.

If the two integrals converge, those integrals will differ, if one of the integrals diverge, and the other one converges, then the retail integral will diverge. In different razbіzhnosti both іntegrіlіv zbіzhnіst іntegrіl vіd rіznitі pіdlаgaє podslіdzhennyu. We are called by another of the descriptions of the vipadkiv.

Separate (butt 1) ​​at p<1. сходится по признаку Дирихле-Абеля при 1>p>0 (div. convergence), so the integral is evaluated from below by the expansion integral, so that it diverges.

The fall of p³1 does not squash us, because for these values ​​of the parameter the integral diverges.

In this way, the final integral converges absolutely at 0

As you can see, the value of the integral can be achieved by folding tasks. It would be great to be occupied with the calculation of a non-violent integral and for example, to show a way, how to diverge. For this reason, methods that allow, without serious calculation, one type of function to grow visnovoks about the feasibility or rozbіzhnist of a non-linear integral. First of all, the other theorems of equivalence, as they will be looked at below, meaningfully help to add unmatched integrations to the economy.

Let f(x)?0. Same functions

є monotonically increasing in the form of changing t abo-d (scalings taken d> 0, -d pragne zero zliva). If at the same time the arguments of the functions F 1 (t) and F 2 (-d) are overfilled with the boundary of the beast, it means that the inconsistent integrals converge. On which basis the first consistency theorem for integrals in the form of invisible functions is based.

Let for the function f(x) and g(x) with x?

• 1) 0?f(x)?g(x);
• 2) The functions f(x) and g(x) are interruptless.

The same zbіzhnosti іntegrаl vplє zbіzhnіst іntegrа, that yakshchoz zіzbіzhnosti іntegrаl

Skіlki 0?f(x)?g(x) and functions without interruption, then

For the mind integral to converge, tobto. maє kіntsev value. Also, the integral converges in the same way.

Now let the integral diverge. It is acceptable that the integral converge, but if the integral converges, then the mind can supersede. Our allowance is wrong, the integral diverges.

Porіvnyannya theorem for ambiguous integrals of the 2nd kind.

Let the functions f(x) and g(x) grow indefinitely for x>+0. For it, for x>+0, the 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.

Porіvnyannya theorem for ambiguous integrals of the 1st kind.

Let for the function f(x) and g(x) to diverge the integral.

Otzhe, on dilyantsі іntegral also diverge.

In this order, the Danish integral diverges along the whole line [-1, 1]. Significantly, if we began to calculate the integral, without exaggerating respect for the analysis of the integral function at the point x = 0, then we would take away the wrong result. True,

, which is impossible.

Also, on the completion of a non-consecutive integral as a disparate function, it is necessary to "split" the first splice of integrals and add them.