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Until now, we have talked about improper integrals with one feature associated with the unboundedness of the function at one of the limits of integration or with the unboundedness of this limit itself. Here we will indicate in what sense the others are understood possible options improper integral.

If both limits of integration are singularities of one or another of the above types, then we assume by definition

where c - arbitrary point gap

It is assumed that each of the improper integrals on the right side of relation (12) converges. Otherwise, they say that the integral on the left side of (12) diverges.

Due to Remark 2 and the additivity property of the improper integral, definition (12) is correct in the sense that it actually does not depend on the choice of point with

Example 13.

Example 14. Integral

called the Euler-Poisson integral, and sometimes also the Gawes integral. It obviously converges in the sense indicated above. It will be shown later that it is equal to

Example 15. Integral

diverges, since for any a at least one of the two integrals diverges

Example 16. Integral

converges if each of the integrals converges

The first of these integrals converges if for

at The second integral converges at which can be verified directly by integration by parts, similar to that done in Example 12, or by referring to the Abel-Dirichlet test. Thus, the original integral makes sense when

In the case when the integrand is not limited in the vicinity of one of the internal points and the segment of integration, we assume

requiring that both integrals on the right exist.

Example 17. In the sense of agreement (13)

Example 18. Integral - not defined.

There is also an agreement different from (13) for calculating the integral of a function that is unbounded in the neighborhood internal point and the segment of integration. Namely, they believe

if the limit on the right exists. This limit is called, following Cauchy, an integral in the sense of the principal value and, in order to distinguish definitions (13) and (14), in the second case, the sign of the integral is preceded by initial letters V. R. French words valeur principal (main value). In the English version, the designation is used. (from principal value).

In accordance with this agreement we have

Example 19.

The following definition is also accepted:

Example 20.

Finally, if there are several ( final number) of certain singularities lying inside the interval or coinciding with its ends, then the interval is divided by non-singular points into a finite number of such intervals, each of which has only one singularity, and the integral is calculated as the sum of integrals over the segments of the partition.

It can be verified that the result of such a calculation does not depend on the arbitrariness in the choice of partition.

Example 21. Exact definition the integral logarithm can now be written as

IN the latter case the symbol V.R. refers to the only singularity internal to the interval, located at point 1. Note that in the sense of definition (13) this integral is not convergent.

Effective solution methods
definite and improper integrals

This article contains additional materials on methods for solving definite and improper integrals. The reader is assumed to have intermediate to advanced integration skills. If this is not the case, please start with the basics, intended for dummies: Indefinite integral, examples of solutions.

Where the indefinite integral is, there is nearby and Definite integral, you should also be familiar with the Newton-Leibniz formula firsthand. In addition, be able to solve the simplest problems to calculate the area of ​​a plane figure.

The lesson is intended for those who want to learn how to solve definite and improper integrals faster and more efficiently. First I will look at the features of integration of even and not even function along an interval symmetrical with respect to zero. Then we'll figure it out problem of finding the area of ​​a circle by using definite integral. This problem is also important because it introduces you to a common technique for integrating a definite integral - trigonometric substitution . It has not been reviewed anywhere yet - new material!

The second section is intended for readers familiar with improper integrals. Similarly, let us consider improper integrals of even and odd functions over a symmetric interval. Including more rare types improper integrals that were not included in the main article: when the lower limit tends to “minus infinity”, when both limits tend to infinity, when at both ends of the integration segment the function suffers an infinite discontinuity (this is already an integral of the second kind). And a very rare improper integral - with a discontinuity point on the integration segment.

If you are interested in something specific, here are the links:

• Definite integral of an even function over a symmetrical segment
• Calculation of the area of ​​a circle, trigonometric substitution
• Improper integrals with infinite limits of integration
• Improper integral of the 2nd kind with discontinuities at both ends of the segment
• Improper integrals with discontinuity on the integration interval

Method for solving the definite integral of an even function

Let's consider a definite integral of the form . It is easy to see that the integration segment is symmetrical about zero.

If the integrand function is even, then the integral can be calculate using half of the segment, and double the result: .

Many have guessed why this is so, nevertheless, let’s consider concrete example with drawing:

Example 1

Much has been said about the parity of a function in methodological material Graphs and properties of elementary functions. Let us repeat once: a function is even if the equality holds for it. How to check a function for parity? Need to instead of substitute "x".

IN in this case:
, Means, this function is even.

According to the rule, on a segment symmetric with respect to zero, our integral of an even function can be calculated as follows:

And now geometric interpretation. Yes, we continue to torment the unfortunate parabola...

Any even function, in particular, is symmetrical about the axis:

Definite integral numerically equal to area flat figure, which is shaded green. But, due to the parity of the integrand, and therefore the symmetry of its graph relative to the axis, it is enough to calculate the area of ​​the figure shaded in blue, and double the result. Identical halves!
That is why the action is right

A similar story happens with any even function along a segment symmetrical with respect to zero:

Some will say: “Why is all this necessary, you can calculate the definite integral anyway.” Can. Let's calculate:

But was it convenient to substitute a negative lower limit? Not really. By the way, a non-zero percentage of students will make a mistake in the signs. It is much easier and more pleasant to substitute a zero. I note that this was just a simple demonstration example; in practice, everything can be worse.

In addition, the technique in question is often used when calculating double integrals, triple integrals, where there are already enough calculations.

A short warm-up example for independent decision:

Example 2

Calculate definite integral

Complete solution and the answer at the end of the lesson.

Please note that when you are asked to simply evaluate the definite integral, you do not need to complete the drawing! The illustration for Example 1 is given only to make the rule clear. Just at this moment The following simple problem is devoted to:

Example 3

1) Calculate the definite integral.
2) Calculate the area of ​​a flat figure, limited by lines and the axis on the interval .

These are two different tasks! This has already been discussed in the article How to calculate the area of ​​a flat figure? Let's deal with the first point first:

1) The integrand is even, the integration segment is symmetrical about zero, therefore:

The definite integral turned out to be negative and this happens!

2) Now let's find the area flat figure. Here it is difficult to do without a drawing:

If you have difficulty with naive cosine, please refer to the article Geometric transformations of graphs.

On the segment, the graph of the function is located below the axis, therefore:

Note that no one canceled the parity of the cosine, so we again halved the segment and doubled the integral.

Calculating the area of ​​a circle using a definite integral
Trigonometric substitution

This is very important task, since it will be considered standard integral and a decision making process that will be encountered many times in the future.

But first, a quick reminder on the equation of a circle. An equation of the form specifies a circle with a center at a point of radius . In particular, the equation defines a circle of radius centered at the origin.

Example 4

Calculate the area of ​​a circle bounded by a circle, given by the equation

is a circle with a center at the origin of radius .

Let's make the drawing:

First, let's calculate the area of ​​a circle using the well-known school formula. If the radius of a circle is , then its area is:

In order to calculate the area of ​​a circle using a definite integral, it is necessary to express the “Y” function in explicit form from the equation of the circle:

The upper semicircle is given by the equation
The lower semicircle is given by the equation

Particularly paranoid people, like me, can substitute several points of the circle into these equations and verify the validity of the above statements.

How to calculate the area of ​​a circle? IN in this example the circle is symmetrical relative to the origin, so it is enough to calculate the area of ​​the sector in the 1st quarter (shaded in blue), then multiply the result by 4.

Thus:

The same, but indefinite integral was considered in example 6 of lesson Complex integrals, it was solved by a lengthy and labor-intensive method of reducing the integral to itself. You can go the same way, but for a definite integral there is a convenient and effective method trigonometric substitution:

Let's make a replacement:

Why exactly this replacement will become clear very soon, but for now let’s find the differential:

Let’s find out what the root will turn into, I’ll describe it in great detail:

If during the solution you cannot guess, apply a formula like , then, alas, you will hear from the teacher “come back next time.”

After transforming the root, you can clearly see why the replacement was made, special attention I draw attention to the coefficient for sine - “two”, this coefficient must be selected in such a way that when squaring everything is well taken out of the brackets and from under the root.

It remains to calculate the new limits of integration:
If , then

New lower integration limit:
New upper limit of integration:

Thus:

The area of ​​the sector must be multiplied by 4, therefore, the area of ​​the entire circle is:

Probably, some people have asked why bother with the integral if there is a short school formula? And the trick is that the ability to very accurately calculate the area of ​​a circle appeared only with the development mathematical analysis(although already in ancient times the area of ​​a circle was calculated with decent accuracy).

The analyzed example can be solved in general view, that is, find the area of ​​a circle bounded by a circle of arbitrary radius: . The result is exactly the formula!

It should be noted that another approach could be applied to solving this problem - to calculate the area of ​​the upper semicircle using the integral , and then double the result. But due to the parity of the integrand, the solution is simply reduced to the optimal version:

Once again I emphasize the importance of the trigonometric substitution; it will occur in practice more than once or twice. Therefore, to secure the material a little more difficult task for independent solution:

Example 5

Calculate definite integral

The condition requires calculating a definite integral, so there is no need to complete the drawing. Think carefully about the coefficient in replacement. If you have difficulty with the integral after the replacement, return to the lesson Integrals of trigonometric functions. Be careful! Full solution and answer at the end of the lesson.

Method for solving the definite integral of an odd function
along a segment symmetrical with respect to zero

You'll like it.

Let's consider the same definite integral with a segment of integration symmetrical with respect to zero: .
If the integrand is odd, That .

Why such an integral equal to zero?

Example 6

Calculate definite integral

Let's make the drawing:

Here, at the same time, is the graph of the function, which I have never seen anywhere before; the graph is an inverted cubic parabola.

Let's check our function for even/odd:
, which means that this function is odd, and its graph is symmetrical about the origin. From the symmetry of the graph it follows that the areas shaded in red and blue are equal..

When calculating the definite integral, the area that is shaded in blue is formally negative. And the area that is shaded in red is positive. Since the areas are equal and formally opposite in sign, they cancel each other, therefore.

And once again I emphasize the difference between the tasks:

1) Any definite integral (of course it must exist) – it's still formally an area(even if negative). In particular, therefore, since due to the oddness of the area functions, they cancel each other out. This is illustrated with a specific example.

2) The problem of finding the area is a completely different task. So, if we were asked to find the area of ​​the figure in this example, then it should be calculated as follows:

A few more short examples on the topic of this rule:

And, similarly for any odd function and a segment symmetric with respect to zero.

Should I use it? this method in practice? In fact, the question is not so simple. When you are offered complex example With a large number calculations, then it is possible, and even appropriate, to indicate that such an integral is equal to zero, referring to the oddness of the function and the symmetry of the integration segment about zero. As they say, knowledge is power, and ignorance is labor power.

But when you are offered short example, then the teacher can quite reasonably force you to solve it in detail: take the integral and substitute the limits of integration using the Newton-Leibniz formula. For example, you are asked to calculate the same definite integral. If you immediately write down what you mean and explain in words why it turns out to be zero, it won’t be very good. It is much better to “play dumb” and carry out the full solution:

And you will know in advance that the integral is equal to zero ;-) And this knowledge will 100% allow you to avoid mistakes.

Method for solving an improper integral with an infinite lower limit

The second section of the article is intended for those who have understood the lesson well Improper integrals. Examples of solutions, or at least understood most of it. We will talk about improper integrals first kind with an infinite lower limit: .

Example 7

How does this integral differ from the “ordinary” improper integral with infinite upper limit? There is practically nothing in terms of solution technology. You also need to find the antiderivative (indefinite integral), and you also need to use the limit when calculating the integral. The difference is that it is necessary to direct the lower limit of integration to “minus infinity”: .

From the above, an obvious formula for calculating such an improper integral follows:

In this example, the integrand is continuous on and:
, that is, the improper integral diverges.

Here, the main thing is be careful with signs, and do not forget that . You need to carefully understand what is going where.

Example 8

Calculate the improper integral or establish its divergence.

This is an example for you to solve on your own. Full solution and answer at the end of the lesson.

Method for solving improper integral
with infinite limits of integration

Very interesting case. Improper integral first kind With infinite limits integration has the following form:

How to solve it? This integral must be represented as the sum of two improper integrals:
(everything ingenious is simple) and look at the situation:
Note: Any number can be used instead of zero, but zero is usually the most convenient.

Example 9

Calculate the improper integral or establish its divergence.

The integrand is continuous on the entire number line. We represent the integral as the sum of two integrals:

and deal with them separately:

Thus:
, that is, the improper integral exists and converges.

Now let's turn our attention to the integrand function. She is even.
In improper integrals with infinite limits (and therefore a symmetric integration interval), parity CAN be used. Similar to the definite integral, it is advantageous to halve the interval and double the result:

Why is this possible? The graph of the integrand of an even function is symmetrical about the axis. Consequently, if half the area is finite (the integral converges), then the symmetric half of the area is also finite. If half the area is infinite (the integral diverges), therefore the symmetrical half will also diverge. And don’t forget about the third case: if half does not exist, then the second, and the whole integral, too. For example:
– this limit does not exist, which means that the improper integral does not exist.

Let's move on to an even more interesting case:

Example 10

Investigate the improper integral for convergence.

Pay attention to the task - here the condition no longer states the fact of the existence of the integral.

The integrand is continuous along the entire number line, and in academic style we cut the patient into two parts:

Let's solve the first one:

and the second:

And, despite the fact that both integrals separately diverge – the final integral in general case doesn't exist, because the amount not defined. Why? Because the variable “a” can tend to “minus infinity”, for example, FASTER than the variable “be” to “plus infinity” (or vice versa).

But there is a special special case– when both variables tend to infinity equally. This is expressed by the limit:

and is called convergence of the Cauchy integral . The very value of the limit is called the leading value of the improper integral .

And since the condition required us research, then the following will be literate answer: in the general case, an improper integral does not exist, but Cauchy convergence takes place and the principal value of the integral is equal to zero. The main meaning is usually denoted as follows:

And now Very important point : the integrand is odd, and as you guess correctly, in improper integrals with infinite limits, oddity SHOULD NOT be used!!!

This is the difference from a definite integral. There you can safely write down that, but here you can do the same shouldn't. Why? Because in some cases, as, for example, in the example considered, an automatic error will result, which is not true.

The subtlety is that the integrals of some odd functions and are actually equal to zero! And it is precisely this subtlety that is dedicated next example for an independent decision.