Definition. A point at infinity in the complex plane is called isolated singular point single-valued analytic function f(z), if outside circle of some radius R,

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

To study the function at an infinitely distant point, we make the change
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 by assumption. Being analytical in this

circle (with the exception of ζ = 0), function
can be expanded in a Laurent series in powers ζ . The classification described in the previous paragraph is fully preserved.

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

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

2.
. Dot z = ∞ is an essential singular point.

§eighteen. 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 one, in the neighborhood of this point f(z) can be uniquely represented by a Laurent series:
where

Definition.deduction analytic function f(z) at an isolated singular point z 0

is called a complex number equal to the value of the integral
, taken in the positive direction along any closed contour lying in the area of ​​analyticity of the function and containing inside it the only singular point z 0 .

The residue is denoted 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 an essential singular point, the residue is equal to the coefficient with-1 Laurent row:

.

Example. Find the residue of a function
.

(Let it be easy to see that

coefficient with-1 will be obtained by multiplying the terms with n= 0:res[ f(z),i ] =
}

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

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

To calculate residues at higher order poles, multiply the function

on the
(m− order of the pole) 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
in point z= −1.

{Res[ f(z), −1] }

If some sequence converges to a finite number a , then we write
.
Earlier, we introduced infinitely large sequences into consideration. We accepted that they are convergent and denoted their limits by symbols and . These symbols represent points at infinity. They do not belong to the set of real numbers. But the concept of a limit allows one to introduce such points and provides a tool for studying their properties with the help of real numbers.

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

For any 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 the 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, the following definition takes place:

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

First, we write down the properties that the points and have. Next, we consider the question of a rigorous mathematical definition of operations for these points and the proof of these properties.

Properties of points at infinity

Sum and Difference.
; ;
; ;

Work and private.
; ; ;
;
;
; ; .

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

Undefined Operations.
; ; ; ;
; ; ;
; ;
.

Proofs for properties of points at infinity

Definition of mathematical operations

We have already given definitions for points at infinity. Now we have to define mathematical operations for them. Since we have defined these points in terms of sequences, operations on these points must also be defined in terms of 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 must not be equal to zero.
Then the difference of two points:
is the limit: .
Dot product:
is the limit: .
Private:
is the limit: .
Here and are arbitrary sequences whose limits are a and b , respectively. In the latter case, .

Property Proofs

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

Consider a 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 hold:
;
.
Then for and we have:
.
Let . Then
at ,
where .
This means that .

Other properties are proved in a similar way. As an example, we present one more proof.

Let's 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 that for any positive number M 1 the following inequalities hold:
;
.
Then for and we have:
.
Let . Then
at ,
where .
This means that .

Undefined Operations

Some of the mathematical operations with points at infinity are not defined. To show their indeterminacy, we need 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 . The limits of these sequences are equal. Amount limit

is equal to infinity.

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

equals zero.

That is, provided that and , the value of the sum limit can take on different values. Therefore, the operation is not defined.

In a similar way, the uncertainty of the remaining operations presented above can be shown.

point of infinity.

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

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

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

If has at a 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 as well.

Let the function be analytic in full space. From the analyticity of a function at a point at infinity, it follows that it is bounded in a neighborhood of this point; let at. On the other hand, analyticity in a closed circle implies its limitedness in this circle; let it in. But then the function is bounded in the whole plane: for all 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, for this point itself); underfunction deduction at infinity understand

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

It follows directly from this definition that the residue of a function at infinity is equal to the coefficient of 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 full plane, then the sum of all its residues, including the residue at infinity, is equal to zero.

Proof. Indeed, let a 1 ,…a n are the end singular points of the function and are the circle containing them all inside. By the property of integrals, the residue theorem, and the definition of a residue at an infinitely distant point, we have:

Ch.t.d.

Applications of the theory of residues to the calculation of integrals.

Let it be required to calculate the integral of a real function over some (finite or infinite) segment ( a, b) x-axis. Complement (a , b ) some curve bounding together with ( a , b ) domain, and analytically continue to.

We apply the residue theorem to the constructed analytic continuation:

(1)

If the integral over 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 equal to zero.

The following is very useful.

Lemma (Jordan). If on some sequence of arcs of circles, (, a fixed) the function tends to zero uniformly with respect to

. (2)

Proof. Denote

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

Therefore, the integral over arcs 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 the arc CE If the polar angle is counted clockwise, then the same estimate will be obtained for. In the case where the proof is simplified, since it will be redundant to estimate the integral over arcs AB and CD. The lemma is proved.

Remark 1. The sequence of arcs of circles in the lemma can be replaced arc family

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

. (3)

The proof remains valid.

Remark 2. Let's change the variable: iz=p , then the arcs of the circles of the lemma are replaced by arcs, and we get that for any function F(p ) tending to zero as uniformly with respect to and for any positive t

. (4)

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

, (5)

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

Consider examples of calculating integrals.

Example 1. .

Let's choose an auxiliary function. Because function on satisfies the inequality, then it uniformly tends to zero as, and by the Jordan 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 a helper function. The integration contour bypasses the singular point z =0. By Cauchy's theorem

It can be seen from the Jordan lemma that To estimate, consider the Laurent expansion in a neighborhood of the point z=0

where is regular at the point z =0 function. From here it is clear that

Thus, Cauchy's theorem can be rewritten as

Replacing in the first integral x on – x , we get that it is equal, so we have

In the limit at and finally:

. (7)

Example 3. Calculate the integral

We introduce an auxiliary function and choose the integration contour the same as in the previous example. Inside this contour, the logarithm admits the selection 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 a residue

According to the Reduction Theorem.

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

Similarly, for starting from some sufficiently small r , therefore

In the first integral after the 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

choose an 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, which are included in this contour, it takes the values ​​and, respectively, therefore the integrals of mutually annihilate, which makes it possible to calculate the required integral. Inside the contour there are two poles of the first order of the function with residues, respectively, equal to:

where. Applying the residue theorem, we get:

According to the above, we have:

Just as in the previous example, we will 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 a special integral

Let's choose an auxiliary function and the circuit shown in the figure. Inside the contour, the function is regular. On the lower bank of the section along the positive semiaxis. Thus, according to the Cauchy theorem:

(8).

It is obvious that with and with. Along, we have respectively and, where varies from 0 to and from to respectively. Hence,

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

whence the desired integral is equal to

Example 6. Calculate the integral

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

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

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

We will consider that branch of the function, which on the upper bank 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 the function is continuous: above the cut, below the cut.

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

For the calculation, we use the expansion of the branch 1/ in the vicinity of the point at infinity. We take the root out from under the 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 according to the binomial formula:

we find the residue of the chosen branch 1/ at an infinitely distant point: (the coefficient at 1/ z with 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. We have, i.e.

Suppose then, thus,

Inside the circle, the integrand has one pole II order minus

By the residue theorem, we have

Example 8. Similarly, we calculate the integral

After substitution we have:

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

(9)

where is the pole inside the circle. Because the right side of (9) is real, then it gives the required integral

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

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

The punctured neighborhood of the point x 0 is called the neighborhood of this point, from which the point x itself has been excluded 0 :
.

Neighborhood endpoints

At the very beginning, the definition of a neighborhood of a point was given. It is designated as . But you can explicitly specify that a 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) .
Epsilon - a 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 we can consider δ - neighborhood, σ - neighborhood, and so on.

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

The concepts of left-handed, right-handed, and punctured neighborhoods of endpoints are also widely used. We present their definitions.

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

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

Punctured Endpoint Neighborhoods

Punctured neighborhoods of the point x 0 are the same neighborhoods, from which the point itself is excluded. They are identified with a circle above the letter. We present their definitions.

Punctured neighborhood of point x 0 :
.

Punctured epsilon - neighborhood of point x 0 :
;
.

Punctured left-hand neighborhood:
;
.

Punctured right-hand neighborhood:
;
.

Neighborhoods of points at infinity

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

.
;
;
.

It was possible to determine the neighborhoods of infinitely distant points and so:
.
But instead of M, we use , so that a neighborhood with a smaller ε is a subset of a neighborhood with a larger ε , just like for neighborhoods of endpoints.

neighborhood property

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

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

The converse assertions are also true.

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

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

Theorem
The 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.
A number a is the limit of a function at a point (finite or at infinity) if for any positive numbers there exist numbers depending on and , such that for all , belongs to the corresponding neighborhood of the point a :
.

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

Proof 1 ⇒ 2

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

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

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

Notice, that .
Let be the smallest positive number and . Then, as noted above,
.
If , then .

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

Proof 2 ⇒ 1

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

Let the second definition hold. Take two positive numbers and . And let be the smallest 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,
.

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

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

The theorem has been proven.

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

We have 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 there are no other singular points of this function in some neighborhood of this point. To determine the type of this singular point, we make a change of variable , while the point z = ∞ goes to the point z 1 = 0, function w = f (z ) takes the form . Singular point type z = ∞ functions w = f (z ) we will call the type of the singular point z 1 = 0 features w = φ (z one). If the expansion of the function w = f (z ) by degrees z in the vicinity of the point z = ∞, i.e. for sufficiently large modulo values z , has the form , then, replacing z on , we get . Thus, under such a change of variable, the main and regular parts of the Laurent series are interchanged, 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 a Laurent series in powers z in the vicinity of the point z = 0. Therefore
1. Point z = ∞ is a removable singular point if there is no regular part in this expansion (with the possible exception of the term A 0);
2. Point z = ∞ - pole n -th order, if the correct part ends with a term A n · z n ;
3. Point z = ∞ is an essential singular point if the regular part contains infinitely many terms.

At the same time, the signs of 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= ∞ - 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, so z
The same result can be obtained in a different way. Let's replace z on, then . For function φ (z 1) dot z 1 = 0 is a sixth-order pole, so for f (z ) dot z = ∞ is a sixth-order pole.
2. . For this function, get the expansion in powers z difficult, so we find: ; the limit exists and is finite, so the point z
3. . The right part of the expansion in powers z contains infinitely many terms, so z = ∞ is an essential singular point. Otherwise, this fact can be established based on the fact that it does not exist.

Function residue at infinitely distant singular point.

For end singular point a , where γ - a contour containing no other than a , singular points, traversed so that the area bounded by it and containing the singular point remains on the left (counterclockwise).

Let's define it in a similar way: , where Γ − is a contour bounding 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 (end) singular points of the function must be inside the contour Γ − . Let us change the direction of bypassing the contour Γ − : . According to the main residue theorem , where the summation is over all finite singular points. Therefore, finally

,

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

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

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