If some sequence converges to a finite number a, then write
.
Previously, we introduced infinitely large sequences into consideration. We assumed that they were convergent and denoted their limits with the symbols and . These symbols represent endlessly remote points . They do not belong to the set of real numbers. But the concept of limit allows us to introduce such points and provides a tool for studying their properties using real numbers.

Definition
Point at infinity, or unsigned infinity, is the limit towards which an infinitely large sequence tends.
Point at infinity plus infinity, is the limit to which an infinitely large sequence with positive terms tends.
Point at infinity minus infinity, is the limit to which an infinitely large sequence with negative terms tends.

For anyone real number a the following inequalities hold:
;
.

Using real numbers, we introduced the concept neighborhood of a point at infinity.
The neighborhood of a point is the set.
Finally, the neighborhood of a point is the set.
Here M is an arbitrary, arbitrarily large real number.

Thus, we have expanded the set of real numbers by introducing new elements into it. In this regard, there is following definition:

Extended number line or extended set of real numbers is the set of real numbers complemented by the elements and :
.

First, we will write down the properties that the points and . Next we consider the issue of strict mathematical definition operations for these points and proofs of these properties.

Properties of points at infinity

Sum and difference.
; ;
; ;

Product and quotient.
; ; ;
;
;
; ; .

Relationship with real numbers.
Let a be an arbitrary real number. Then
; ;
; ; ; .
Let a > 0 . Then
; ; .
Let a < 0 . Then
; .

Undefined operations.
; ; ; ;
; ; ;
; ;
.

Proofs of the properties of points at infinity

Defining Mathematical Operations

We have already given definitions for points at infinity. Now we need to define mathematical operations for them. Since we defined these points using sequences, operations with these points should also be defined using sequences.

So, sum of two points
c = a + b,
belonging to the extended set of real numbers,
,
we will call the limit
,
where and are arbitrary sequences having limits
And .

The operations of subtraction, multiplication and division are defined in a similar way. Only, in the case of division, the elements in the denominator of the fraction should not be equal to zero.
Then the difference of two points:
- this is the limit: .
Product of points:
- this is the limit: .
Private:
- this is the limit: .
Here and are arbitrary sequences whose limits are a and b , respectively. IN the latter case, .

Proofs of properties

To prove the properties of points at infinity, we need to use the properties of infinitely large sequences.

Consider the property:
.
To prove it, we must show that
,

In other words, we need to prove that the sum of two sequences that converge to plus infinity converges to plus infinity.

1 the following inequalities are satisfied:
;
.
Then for and we have:
.
Let's put it. Then
at ,
Where .
This means that .

Other properties can be proved in a similar way. As an example, we give one more proof.

Let us prove that:
.
To do this we must show that
,
where and are arbitrary sequences, with limits and .

That is, we need to prove that the product of two infinitely large sequences is an infinitely large sequence.

Let's prove it. Since and , then there are some functions and , so for any positive number M 1 the following inequalities are satisfied:
;
.
Then for and we have:
.
Let's put it. Then
at ,
Where .
This means that .

Undefined operations

Part mathematical operations with points at infinity are not defined. To show their uncertainty, it is necessary to give a couple of special cases when the result of the operation depends on the choice of the sequences included in them.

Consider this operation:
.
It is easy to show that if and , then the limit of the sum of sequences depends on the choice of sequences and .

Indeed, let's take it. The limits of these sequences are . Amount limit

equals infinity.

Now let's take . The limits of these sequences are also equal. But the limit of their amount

equal to zero.

That is, provided that and , the value of the amount limit can take different meanings. Therefore the operation is not defined.

In a similar way, you can show the uncertainty of the remaining operations presented above.

Definition
Neighborhood of a real point x 0 Any open interval containing this point is called:
.
Here ε 1 and ε 2 - arbitrary positive numbers.

Epsilon - neighborhood of point x 0 is the set of points the distance from which to point x 0 less than ε:
.

A punctured neighborhood of point x 0 is the neighborhood of this point from which the point x itself is excluded 0 :
.

Neighborhoods of endpoints

At the very beginning, a definition of the neighborhood of a point was given. It is designated as . But you can explicitly indicate that the neighborhood depends on two numbers using the appropriate arguments:
(1) .
That is, a neighborhood is a set of points belonging to an open interval.

Equating ε 1 to ε 2 , we get epsilon - neighborhood:
(2) .
An epsilon neighborhood is a set of points belonging to an open interval with equidistant ends.
Of course, the letter epsilon can be replaced by any other and consider δ - neighborhood, σ - neighborhood, etc.

In limit theory, one can use a definition of neighborhood based on both set (1) and set (2). Using any of these neighborhoods gives equivalent results (see). But definition (2) is simpler, so epsilon is often used - the neighborhood of a point determined from (2).

The concepts of left-sided, right-sided and punctured neighborhoods are also widely used. endpoints. Here are their definitions.

Left neighborhood of a real point x 0 is a half-open interval located on real axis to the left of point x 0 , including the point itself:
;
.

Right-sided neighborhood of a real point x 0 is a half-open interval located to the right of point x 0 , including the point itself:
;
.

Punctured neighborhoods of endpoints

Punctured neighborhoods of point x 0 - these are the same neighborhoods from which the point itself is excluded. They are indicated with a circle above the letter. Here are their definitions.

Punctured neighborhood of point x 0 :
.

Punctured epsilon - neighborhood of point x 0 :
;
.

Pierced left side vicinity:
;
.

Punctured right side vicinity:
;
.

Neighborhoods of points at infinity

Along with end points, neighborhoods of points at infinity are also introduced. They are all punctured because there is no real number at infinity (the point at infinity is defined as the limit at infinity large sequence).

.
;
;
.

It was possible to determine the neighborhoods of points at infinity like this:
.
But instead of M, we use , so that the neighborhood with smaller ε is a subset of the neighborhood with larger ε, as for endpoint neighborhoods.

Neighborhood property

Next, we use the obvious property of the neighborhood of a point (finite or at infinity). It lies in the fact that the neighborhoods of points with smaller valuesε are subsets of neighborhoods with large values ​​of ε. Here are more strict formulations.

Let there be a final or infinitely distant point. Let it go .
Then
;
;
;
;
;
;
;
.

The converse is also true.

Equivalence of definitions of the limit of a function according to Cauchy

Now we will show that in determining the limit of a function according to Cauchy, you can use both an arbitrary neighborhood and a neighborhood with equidistant ends.

Theorem
Cauchy definitions of the limit of a function that use arbitrary neighborhoods and neighborhoods with equidistant ends are equivalent.

Proof

Let's formulate first definition of the limit of a function.
The number a is the limit of a function at a point (finite or infinitely distant), if for any positive numbers there are numbers depending on and that for all , belongs to the corresponding neighborhood of the point a:
.

Let's formulate second definition of the limit of a function.
A number a is the limit of a function at a point if for any positive number there is a number depending on that for all:
.

Proof 1 ⇒ 2

Let us prove that if a number a is the limit of a function by the 1st definition, then it is also a limit by the 2nd definition.

Let the first definition be satisfied. This means that there are functions and , so for any positive numbers the following holds:
at , where .

Since the numbers are arbitrary, we equate them:
.
Then there are such functions and , so for any the following holds:
at , where .

Notice, that .
Let be the smallest of the positive numbers and . Then, according to what was noted above,
.
If, then.

That is, we found such a function, so for any the following holds:
at , where .
This means that the number a is the limit of the function by the second definition.

Proof 2 ⇒ 1

Let us prove that if a number a is the limit of a function by the 2nd definition, then it is also a limit by the 1st definition.

Let the second definition be satisfied. Let's take two positive numbers and . And let it be the least of them. Then, according to the second definition, there is such a function , so that for any positive number and for all , it follows that
.

But according to , . Therefore, from what follows that
.

Then for any positive numbers and , we found two numbers, so for all :
.

This means that the number a is a limit by the first definition.

The theorem is proven.

References:
L.D. Kudryavtsev. Well mathematical analysis. Volume 1. Moscow, 2003.

The point at infinity.

Let the function be analytic in some neighborhood of an infinitely distant point (except for the point itself). They say it isremovable singular point, pole or essentially singular pointfunctions depending onfinite, infinite or non-existent .

Let us put and, then it will be analytic in a certain neighborhood of the point. The latter will be for a singular point of the same type as for for. The Laurent neighborhood expansion can be obtained by a simple substitution in the Laurent neighborhood expansion. But with such a replacement, the correct part is replaced by the main one, and vice versa. Thus, it is fair

Theorem 1. In the case of a removable singularity in infinitely remote point, the Laurent expansion of the function in the neighborhood of this point does not contain positive degrees, in the case of a polecontains a finite number of them, and in the caseessential feature - infinite.

If it has at the point removable feature, it is usually said that itanalytic at infinity, and accept. In this case, the function is obviously bounded in some neighborhood of the point.

Let the function be analytic in complete plane. From the analyticity of a function at a point at infinity, it follows that it is bounded in the neighborhood of this point; let at. On the other hand, from analyticity to vicious circle follows its limitation in this circle; let it be in it. But then the function is limited on the entire plane: for everyone we have. Thus, Liouville's theoremcan be given the following form.

Theorem 2. If a function is analytic in the full plane, then it is constant.

Let us now introduce the conceptresidue at infinity. Let the function be analytic in some neighborhood of a point (except, perhaps, this point itself); undersubtracting the function at infinity understand

where is a sufficiently large circle traversed clockwise (so that the circle of the point remains to the left).

From this definition it immediately follows that the residue of a function at infinity is equal to the coefficient at in its Laurent expansion in the neighborhood of a point, taken with the opposite sign:

Theorem 3. If a function has a finite number of singular points in the complete plane, then the sum of all its residues, including the residue at infinity, is equal to zero.

Proof. In fact, let a 1 ,…a n the final singular points of the function and - the circle containing them all inside. By the property of integrals, the residue theorem and the definition of residue at a point at infinity, we have:

Etc.

Applications of residue theory to the calculation of integrals.

Let it be necessary to calculate the integral of real function along some (finite or infinite) segment ( a,b) x axis. Let's add (a , b ) some curve bounding together with ( a, b ) region, and continue analytically in.

We apply the residue theorem to the constructed analytic continuation:

(1)

If the integral can be calculated or expressed in terms of the desired integral, then the calculation problem is solved.

In the case of infinite segments ( a, b ) usually consider families of infinitely expanding integration contours, which are constructed in such a way that, as a result of passing to the limit, we obtain an integral over ( a, b ). In this case, the integral over in relation (1) can not be calculated, but only its limit can be found, which often turns out to be zero.

The following is very useful:

Lemma (Jordan). If on some sequence of circular arcs,(, A fixed) the function tends to zero uniformly with respect to, then for

. (2)

Proof. Let's denote

By the conditions of the lemma, when also tends to zero, and Let a >0; on arcs AB and CD we have.

Consequently, the arc integral AB, CD tends to zero at.

Since the inequality is valid for, then on the arc BE

Therefore, and thus also tends to zero at. If on an arc SE If the polar angle is counted clockwise, then the same estimate will be obtained. In the case when the proof is simplified, because it would be unnecessary to estimate the integral over the arcs AB and CD. The lemma is proven.

Note 1. The sequence of circular arcs in the lemma can be replaced family of arcs

then, if the function at tends to zero uniformly with respect to then for

. (3)

The proof still stands.

Remark 2. Let's replace the variable: iz=p , then the arcs of circles of the lemma will be replaced by arcs, and we obtain that for any function F(p ), tending to zero as uniformly relative and for any positive t

. (4)

Replacing p in (4) with (-p ) we get that under the same conditions for

, (5)

where is the arc of a circle (see figure).

Let's look at examples of calculating integrals.

Example 1. .

Let's choose an auxiliary function. Because function satisfies the inequality, then it uniformly tends to zero as, and by Jordan’s lemma, as

For we have by the residue theorem

In the limit at we get:

Separating the real parts and using the parity of the function, we find

Example 2. To calculate the integral

Let's take an auxiliary function. The integration contour bypasses the singular point z =0. By Cauchy's theorem

From Jordan's lemma it is clear that. To estimate, consider the Laurent expansion in the neighborhood of the point z =0

where is regular at a point z =0 function. From this it is clear that

Thus, Cauchy's theorem can be rewritten as

Replacing in the first integral x by x , we find that it is equal, so we have

In the limit at and finally:

. (7)

Example 3. Calculate the integral

Let's introduce an auxiliary function and choose the integration contour the same as in the previous example. Within this contour, the logarithm allows the identification of a single-valued branch. Let denote the branch that is determined by the inequality. The function has at the point z=i second order pole with residue

By the residue theorem.

When, starting from some sufficiently large R , hence, .

Similarly for, starting from some sufficiently small r, therefore

In the first integral after replacement z=-x we get:

and thus in the limit at we have:

Comparing the real and imaginary parts gives:

, .

Example 4. For the integral

Let's select the auxiliary function and the contour shown in the figure. Inside the contour is unambiguous, if we assume that.

On the upper and lower banks of the cut, included in this contour, takes on the values ​​and, therefore, the integrals cancel each other out, which makes it possible to calculate the required integral. Inside the contour there are two poles of the first order function with residues respectively equal to:

Where. Applying the residue theorem, we get:

In accordance with the above we have:

Just as in the previous example, we prove that, and then in the limit, at we will have:

From here, comparing the imaginary parts, we get:

Example 5. Calculate the principal value of the special integral

Let's select the auxiliary function and the contour shown in the figure. Inside the contour the function is regular. On the lower bank of the cut along the positive semi-axis. Thus, according to Cauchy's theorem:

(8).

Obviously, when and when. Along, we have, respectively, and, where changes from 0 to and from to respectively. Hence,

Passing in (8) to the limit at, we thus obtain

whence the required integral is equal to

Example 6. Calculate the integral

Let's consider the function. Let's make a cut*) .

Let's put it. When going counterclockwise around a closed path (see figure, dotted line) and get an increment,

therefore, arg f (z )=( 1 +2  2 )/3 is also incremented. Thus, in the appearance of the cut, the function splits into 3 regular branches, differing from each other in the choice of the initial element of the function, i.e. value at some point.

We will consider the branch of the function that on the upper side of the cut (-1,1) takes positive values, and take the contour,

___________________

*) In fact, two cuts were made: and, however, on the axis x to the right of point x =1 function is continuous: above the cut, below the cut.

shown in the figure. On bank I we have, i.e. , on shore II (after going around the point z =1 clockwise) (i.e.), i.e. , the integrals over circles and, obviously, tend to zero**) at. Therefore, by Cauchy’s theorem for multiply connected domains

For the calculation, we use the expansion of the 1/ branch in the neighborhood of the point at infinity. Let's take it out from under the root sign, then we get where and are the branches of these functions, positive on the segment (1,) of the real axis.

on a segment of the real axis. Expanding the latter using the binomial formula:

we find the residue of the selected branch 1/ at the point at infinity: (coefficient at 1/ z with the opposite sign). But the integral is equal to this residue multiplied by, i.e. we have where finally

Example 7. Consider the integral.

__________________

**) Consider, for example, the integral over. On we have, i.e.

Let us put then, thus,

Inside a circle, the integrand has one pole II order with deduction

By the residue theorem we have

Example 8. Let us similarly calculate the integral

After substitution we have:

One of the poles of the integrand lies inside unit circle, and the other is outside it, because by the properties of the roots quadratic equation, and by virtue of the condition, these roots are real and different. Thus, by the residue theorem

(9)

where is the pole lying inside the circle. Because right part(9) is valid, then it gives the required integral

Definition. Point at infinity complex plane called isolated singular point unambiguous analytical functionf(z), If outside circle of some radius R,

those. for , there is no finite singular point of the function f(z).

To study the function at a point at infinity, we make the substitution
Function

will have a singularity at the point ζ = 0, and this point will be isolated, since

inside the circle
There are no other singular points according to the condition. Being analytical in this

circle (except for so-called ζ = 0), function
can be expanded in a Laurent series in powers ζ . The classification described in the previous paragraph remains completely unchanged.

However, if we return to the original variable z, then series in positive and negative powers z'switch' places. Those. The classification of points at infinity will look like this:

Examples. 1.
. Dot z = i − pole of the 3rd order.

2.
. Dot z = ∞ − significantly singular point.

§18. Residue of an analytic function at an isolated singular point.

Let the point z 0 is an isolated singular point of a single-valued analytic function

f(z) . According to the previous, in the vicinity of this point f(z) can be represented uniquely by the Laurent series:
Where

Definition.Deduction analytical function f(z) at an isolated singular point z 0

called complex number, equal to the value of the integral
, taken in the positive direction along any closed contour lying in the domain of analyticity of the function and containing within itself a single singular point z 0 .

The deduction is indicated by the symbol Res [f(z),z 0 ].

It is easy to see that the residue at a regular or removable singular point is equal to zero.

At a pole or essentially singular point, the residue is equal to the coefficient With-1 row Laurent:

.

Example. Find the residue of a function
.

(Let it be easy to see that

coefficient With-1 is obtained when multiplying the terms with n= 0:Res[ f(z),i ] =
}

It is often possible to calculate residues of functions over in a simple way. Let the function f(z) has incl. z 0 pole of the first order. In this case, the expansion of the function in a Laurent series has the form (§16):. Let's multiply this equality by (z−z 0) and go to the limit at
. As a result we get: Res[ f(z),z 0 ] =
So, in

In the last example we have Res[ f(z),i ] =
.

To calculate residues at higher order poles, multiply the function

on
(m− pole order) and differentiate the resulting series ( m − 1 time.

In this case we have: Res[ f(z),z 0 ]

Example. Find the residue of a function
at z= −1.

{Res[ f(z), −1] }

We defined the neighborhood of this point as the exterior of circles centered at the origin: U (∞, ε ) = {z ∈ | |z | > ε). Dot z = ∞ is an isolated singular point of the analytic function w = f (z ), if in some neighborhood of this point there are no other singular points of this function. To determine the type of this singular point, we make a change of variable, and the point z = ∞ goes to the point z 1 = 0, function w = f (z ) will take the form . Type of singular point z = ∞ functions w = f (z ) we will call the type of singular point z 1 = 0 functions w = φ (z 1). If the expansion of the function w = f (z ) by degrees z in the vicinity of a point z = ∞, i.e. at sufficiently large modulus values z , has the form , then, replacing z on , we will receive . Thus, with such a change of variable, the main and regular parts of the Laurent series change places, and the type of the singular point z = ∞ is determined by the number of terms in the correct part of the expansion of the function in the Laurent series in powers z in the vicinity of a point z = 0. Therefore
1. Point z = ∞ is a removable singular point if this expansion does not contain the correct part (except, perhaps, for the term A 0);
2. Point z = ∞ - pole n -th order if the right part ends with a term A n · z n ;
3. Point z = ∞ is an essentially singular point if the regular part contains infinitely many terms.

In this case, the criteria for the types of singular points by value remain valid: if z= ∞ is a removable singular point, then this limit exists and is finite if z= ∞ is a pole, then this limit is infinite if z= ∞ is an essentially singular point, then this limit does not exist (neither finite nor infinite).

Examples: 1. f (z ) = -5 + 3z 2 - z 6. The function is already a polynomial in powers z , the highest degree is the sixth, therefore z
The same result can be obtained in another way. We will replace z on, then . For function φ (z 1) point z 1 = 0 is a pole of sixth order, therefore for f (z ) dot z = ∞ - pole of the sixth order.
2. . For this function, obtain a power expansion z difficult, so let's find: ; the limit exists and is finite, so the point z
3. . Correct part of the power expansion z contains infinitely many terms, so z = ∞ is an essentially singular point. Otherwise, this fact can be established based on the fact that it does not exist.

Residue of a function at an infinitely distant singular point.

For the final singular point a , Where γ - a circuit containing no others except a , singular points, traversed in such a way that the area bounded by it and containing the singular point remains on the left (counterclockwise).

Let's define in a similar way: , where Γ − is the contour limiting such a neighborhood U (∞, r ) points z = ∞, which does not contain other singular points, and is traversable so that this neighborhood remains on the left (i.e., clockwise). Thus, all other (final) singular points of the function must be located inside the contour Γ − . Let's change the direction of traversing the contour Γ − : . By the main theorem on residues , where the summation is carried out over all finite singular points. Therefore, finally

,

those. residue at an infinitely distant singular point equal to the sum residues over all finite singular points, taken with the opposite sign.

As a consequence, there is total sum theorem: if function w = f (z ) is analytic everywhere in the plane WITH , with the exception of finite number singular points z 1 , z 2 , z 3 , …,z k , then the sum of residues at all finite singular points and the residue at infinity is equal to zero.

Note that if z = ∞ is a removable singular point, then the residue at it can be different from zero. So for the function, obviously, ; z = 0 is the only finite singular point of this function, so , despite the fact that, i.e. z = ∞ is a removable singular point.