Calculate the integral of a complex variable over a closed online. Integration of functions of a complex variable

1. Basic concepts and statements

Theorem 5.1(a sufficient condition for the existence of an integral of a function of a complex variable). Let L is a simple smooth curve on , f(z)=u(x;y)+i×v(x;y) is continuous on L. Then there exists, and the following equality holds:

Theorem 5.2. Let L is a simple smooth curve, given parametrically: L:z(t)=x(t)+i×y(t), a£ t£ b, function f(z) is continuous on L. Then the equality is true:

(where ). (5.2)

Theorem 5.3. If a f(z) analytic in the domain D function, then - analytical function and F"(z)=f(z), where the integral is taken over any piecewise smooth curve connecting the points z 0 and z.

- Newton-Leibniz formula.

2. Methods for calculating the integral

First way. Calculation of integrals of a continuous function by reducing to curvilinear integrals of functions of real variables (application of formula (5.1)).

1. Find Re f=u, Im f=v.

2. Write down the integrand f(z)dz in the form of a work ( u+iv)(dx+idy)=udx-vdy+i(udy+vdx).

3. Calculate curvilinear integrals of the form according to the rules for calculating curvilinear integrals of the second kind.

Example 5.1 . Calculate along a parabola y=x 2 from point z 1 =0 to the point z 2 =1+i.

■ Find the real and imaginary parts of the integrand. To do this, we substitute into the expression for f(z) z=x+iy:

Because y=x 2 , then dy= 2x, . That's why

The second way. Calculation of integrals from a continuous function by reducing to a definite integral in the case parametric task integration paths (application of formula (5.2)).

1. Write the parametric equation of the curve z=z(t) and determine the limits of integration: t=a corresponds to the starting point of the integration path, t=b- final.

2. Find the differential of a complex-valued function z(t): dz=z¢( t)dt.

3. Substitute z(t) into an integrand, transform the integral to the form: .

4. Calculate the resulting definite integral.

Example 5.2 . Calculate where FROM- an arc of a circle, .

■ Parametric equation of this curve: , 0£ j£ p. Then . We get

Example 5.3 . Calculate where FROM- the upper arc of the circle under the condition: a), b).

■ Setting function values ​​in the integration loop allows you to select single-valued branches of the expression , k= 0,1. Since for we have , k= 0.1, then in the first case we select a branch with k= 0, and in the second - with k= 1.

The integrand on the integration contour is continuous. Parametric equation of this curve: , 0£ j£ p. Then .

a) The branch is determined when k= 0, that is, from we get .

b) The branch is determined when k=1, that is, from we get .

The third way. Calculation of integrals of analytic functions in simply connected domains (application of formula (5.3)).

Find an antiderivative F(z) using the properties of integrals, tabular integrals, and methods known from real analysis. Apply the Newton-Leibniz formula: .

Example 5.4 . Calculate , where FROM- straight AB, z A=1-i,z B=2+i.

■ Since the integrand - analytical on the entire complex plane, then we apply the Newton-Leibniz formula

3. Basic theorems of integral calculus

functions of a complex variable

Theorem 5.4 (Cauchy). If a f(z G function, then where L- any closed loop lying in G.

Cauchy's theorem also holds for a multiply connected domain.

Theorem 5.5. Let the function f(z) is analytic in a simply connected domain D, L-an arbitrary closed piecewise-smooth contour lying in D. Then for any point z 0 lying inside the contour L, the formula is valid:

, (5.4)

where L flows in a positive direction.

Formula (5.4) is called integral Cauchy formula . It expresses the values ​​of an analytic function inside a contour in terms of its values ​​on the contour.

Theorem 5.6. Any function f(z), analytic in the domain D, has derivatives of all orders on this domain, and for " z 0 Î D the correct formula is:

, (5.5)

where L is an arbitrary piecewise-smooth closed contour lying entirely in D and containing a dot z 0 .

4. Calculation of integrals over a closed loop

from functions of a complex variable

Consider integrals of the form , where the function j(z) is analytic in , and y(z) is a polynomial that does not have zeros on a closed contour FROM.

Rule. When calculating integrals of the form, depending on the multiplicity of zeros of the polynomial y(z) and their location relative to the contour FROM 4 cases can be distinguished.

1. In the field D no polynomial zeros y(z). Then the function is analytic and by Cauchy's theorem.

2. In the field D there is one simple zero z=z 0 polynomial y(z). Then we write the fraction as , where f(z) is an analytic function in Applying the Cauchy integral formula (5.4), we obtain

. (5.6)

3. In the field D located one multiple of zero z=z 0 polynomial y(z) (multiplicity n). Then we write the fraction as , where f(z) is an analytic function in Applying formula (5.5), we obtain

4. In the field D there are two zeros of the polynomial y(z) z=z 1 and z=z 2. Then we represent the integrand as a sum of two fractions, and the integral as a sum of two integrals, each of which is calculated in accordance with item 2 or item 3.

Example 5.5 . Calculate where FROM- circle.

■ Find the zeros of the denominator - special points integrand . These are points. Next, we determine the location of the points relative to the integration contour: none of the points is included in the area bounded by a circle with a center at a point and a radius of 2 (that is, we have the first case). This can be verified by drawing or determining the distance from each of the points to the center of the circle and comparing it with the radius. For example, for , therefore does not belong to the circle.

Then the function analytic in the circle , and by Cauchy's theorem .

Note that the given integral is equal to zero for any other contour that limits the region that does not include any of the zeros of the denominator. ■

Example 5.6 . Calculate where FROM- circle.

■ Arguing as in Example 5.5, we find that only one of the zeros of the denominator is located in the circle (the second case). Therefore, we write the integrand in the form , the function analytic in a circle . Then by formula (5.6)

.■

Example 5.7 . Calculate , where FROM- circle.

1. Basic concepts

2. Calculation of integrals of functions of a complex variable

3. Examples of calculating integrals of functions of a complex variable

4. The main Cauchy theorem for a simple contour

5. Cauchy's theorem for a complex contour

6. Integral Cauchy formula

7. Calculation of integrals over a closed loop

8. Examples of calculating integrals over a closed loop

Basic concepts

1. The concept of an integral of a function of a complex variable is introduced (in the same way as in the real region) as the limit of a sequence of integral sums; the function is defined on some curve l , the curve is assumed to be smooth or piecewise smooth:

\int\limits_(l)f(z)\,dz= \lim_(\lambda\to0) \sum_(k=1)^(n)\bigl(f(\xi_k)\cdot \Delta z_k\bigr) ,\qquad\quad (2.43)

where x_k is a point selected on the arc \Delta l_k of the curve split; \Delta z_k - increment of the function argument on this part of the split, \lambda=\max_(k)|\Delta z_k|- split step, |\Delta z_k| - the length of the chord connecting the ends of the arc \Delta l_k ; the curve l is divided arbitrarily into n parts \Delta l_k,~ k=1,2,\ldots,n. A direction is chosen on the curve, i.e. start and end points are specified. In the case of a closed curve \textstyle(\left(\int\limits_(l) f(z)dz= \oint\limits_(c)f(z)dz\right)) integration occurs in the positive direction, i.e. in a direction that leaves the end region bounded by the path to the left.

Formula (2.43) defines curvilinear integral of a function of a complex variable. If we single out the real and imaginary parts of the function f(z) , i.e. write it down in the form

F(z)=u+i\,v,\qquad u=\operatorname(Re)f(z),\quad v=\operatorname(Im)f(z),\qquad u=u(x,y) ,\quad v=v(x,y),

then the integral sum can be written in the form of two terms, which will be the integral sums of curvilinear integrals of the second kind of functions of two real variables. If f(z) is assumed to be continuous on l , then u(x, y),~ v(x, y) will also be continuous on l , and hence there will be limits to the corresponding integral sums. Therefore, if the function f(z) is continuous on l , then the limit in equality (2.43) exists, i.e. there is a curvilinear integral of the function f(z) over the curve l and the formula

\int\limits_(l)f(z)\,dz= \int\limits_(l)u\,dx-v\,dy+ i \int\limits_(l)u\,dy+v\,dx\, .

Using the definition of an integral or formula (2.44) and the properties of curvilinear integrals of the second kind, it is easy to verify the validity of the following properties of a curvilinear integral of functions of a complex variable (properties known from real analysis).

\begin(aligned)&\bold(1.)~~ \int\limits_(l)\bigldz= c_1\int\limits_(l) f_1(z)\,dz+ c_2\int\limits_(l)f_2(z )\,dz\,.\\ &\bold(2.)~~ \int\limits_(AB)f(z)\,dz=- \int\limits_(BA)f(z)\,dz\, .\\ &\bold(3.)~~ \int\limits_(AB)f(z)\,dz= \int\limits_(AC)f(z)\,dz+ \int\limits_(CB)f( z)\,dz\,.\\ &\bold(4.)~~ \int\limits_(AB)|dz|= l_(AB).\\ &\bold(5.)~~ \left|\ int\limits_(l)f(z)\,dz \right|\leqslant \int\limits_(l)|f(z)|\,|dz|. \end(aligned)

in particular, \textstyle(\left|\int\limits_(AB)f(z)\,dz\right|\leqslant M\cdot l_(AB)), if the function is bounded in absolute value on the curve AB , that is |f(z)|\leqslant M,~ z\in l. This property is called the property of estimating the modulus of the integral.

\bold(6.)~~ \int\limits_(AB)dz= z_B-z_A\,.

Formula (2.44) can be considered both as a definition of a curvilinear integral of a function of a complex variable, and as a formula for its calculation through curvilinear integrals of the second kind of functions of two real variables.

To use and remember the calculation formula, we note that equality (2.44) corresponds to the formal execution on the left side under the sign of the integral of the operations of extracting the real and imaginary parts of the function f(z) , multiplying by dz=dx+i\,dy and writing the resulting product in algebraic form:

\int\limits_(l)f(z)\,dz= \int\limits_(l)(u+iv)(dx+i\,dy)= \int\limits_(l)u\,dx-v\ ,dy+i(u\,dy+v\,dx)= \int\limits_(l)u\,dx-v\,dy+ i\int\limits_(l)u\,dy+v\,dx\ ,.

Example 2.79. Calculate integrals and \int\limits_(OA)z\,dz, where the line OA

a) a line segment connecting the points z_1=0 and z_2=1+i ,
b) broken line OBA , where O(0;0),~A(1;1),~B(1;0).

▼ Solution

1. Calculate the integral \int\limits_(OA)\overline(z)\,dz. Here f(z)= \overline(z)= x-iy,~ dz=dx+i\,dy. We write the integral in terms of curvilinear integrals of the second kind:

\int\limits_(OA)\overline(z)\,dz= \int\limits_(OA) (x-iy)(dx+i\,dy)= \int\limits_(OA) x\,dx+y \,dy+ i\int\limits_(OA)x\,dy-y\,dx\,

which corresponds to formula (2.44). We calculate the integrals:

a) the integration path is a straight line segment, therefore \int\limits_(OA)\overline(z)\,dz= \int\limits_(0)^(1)2x\,dx=1.

b) the path of integration is a broken line, consisting of two segments OB= \(y=0,~ 0\leqslant x\leqslant1\) and BA= \(x=1,~ 0\leqslant y\leqslant1\). Therefore, splitting the integral into two and performing calculations, we obtain

\int\limits_(OA)\overline(z)\,dz= \int\limits_(OB)\overline(z)\,dz+ \int\limits_(BA)\overline(z)\,dz= \int\ limits_(0)^(1)x\,dx+ \int\limits_(0)^(1)y\,dy+ i\int\limits_(0)^(1) dy=1+i.

The integral of the function f(z)=\overline(z) depends on the choice of the integration path connecting the points O and A .

2. Calculate the integral \textstyle(\int\limits_(OA)z\,dz) here f(z)=z=x+iy . We write the integral in terms of curvilinear integrals of the second kind

\int\limits_(OA)z\,dz= \int\limits_(OA)x\,dx-y\,dy+ i\int\limits_(OA)x\,dy+y\,dx\,.

The integrands of the obtained integrals of the second kind are total differentials (see condition (2.30)), so it suffices to consider one case of an integration path. So, in case "a", where the equation of the segment y=x,~0 \leqslant x \leqslant1, we get the answer

\int\limits_(OA)z\,dz=i \int\limits_(0)^(1)2x\,dx=i\,.

Due to the independence of the integral from the form of the integration path, the task in this case can be formulated in a more general view: calculate integral

\int\limits_(l)z\,dz from point z_1=0 to point z_2=1+i .

In the next subsection, we consider such cases of integration in more detail.

2. Let the integral of a continuous function in some domain be independent of the form of the curve connecting two points of this domain. Let's fix the starting point, denoting z_0 . the end point is a variable, let's denote it z . Then the value of the integral will depend only on the point z, that is, it defines some function in the specified area.

Below, we will justify the assertion that in the case of a simply connected domain, the integral defines a single-valued function in this domain. We introduce the notation

\int\limits_(z_0)^(z) f(\xi)\,d\xi=F(z).

The function F(z) is an integral with a variable upper limit.

Using the definition of a derivative, i.e. considering \lim_(\Delta z\to0)\frac(\Delta F)(\Delta z), it is easy to verify that F(z) has a derivative at any point of the domain of definition, and therefore is analytic in it. In this case, for the derivative we obtain the formula

F"(z)=f(z).

The derivative of an integral with a variable upper limit is equal to the value of the integrand at the upper limit.

It follows from equality (2.46), in particular, that the integrand f(z) in (2.45) is an analytic function, since the derivative F"(z) of the analytic function F(z) by the property of such functions (see Proposition 2.28) - analytical function.

3. The function F(z) for which equality (2.46) holds is called the antiderivative for the function f(z) in a simply connected domain, and the collection of antiderivatives \Phi(z)=F(z)+c , where c=\text( const) , - indefinite integral from the function f(z) .

From points 2 and 3 we obtain the following assertion.

Statement 2.25

1. Integral with variable upper limit \textstyle(\int\limits_(z_0)^(z) f(\xi)\,d\xi) from a function analytic in a simply connected domain there is a function analytic in this domain; this function is antiderivative for the integrand.

2. Any function that is analytic in a simply connected domain has an antiderivative in it (the existence of an antiderivative).

Antiderivatives of analytic functions in simply connected domains are found, as in the case of real analysis: the properties of integrals, the table of integrals, and integration rules are used.

For example, \int e^z\,dz=e^z+c,~~ \int\cos z\,dz=\sin z+c..

Between the curvilinear integral of an analytic function and its antiderivative in a simply connected domain, there is a formula similar to the Newton-Leibniz formula from real analysis:

\int\limits_(z_1)^(z_2)f(z)\,dz= \Bigl.(F(z))\Bigr|_(z_1)^(z_2)= F(z_2)-F(z_1).

4. As in real analysis, in the complex domain, apart from integrals containing a parameter within the limits of integration (formula (2.45) gives the simplest example such integrals), integrals that depend on the parameter contained in the integrand: \textstyle(\int\limits_(l)f(\xi,z)\,d\xi). Among such integrals, an important place in theory and practice complex integration and applications takes an integral of the form \textstyle(\int\limits_(l)\dfrac(f(\xi))(\xi-z)\,d\xi).

Assuming f(z) to be continuous on the line l , we obtain that for any point z not belonging to l , the integral exists and determines, in any region not containing l , some function

\frac(1)(2\pi\,i) \int\limits_(l) \frac(f(\xi))(\xi-z)\,d\xi=F(z).

The integral (2.48) is called the Cauchy-type integral; the multiplier \frac(1)(2\pi\,i) is introduced for the convenience of using the constructed function.

For this function, as well as for the function defined by equality (2.45), it is proved that it is analytic everywhere in the domain of definition. Moreover, in contrast to the integral (2.45), it is not required here that the generating function f(z) be analytic, i.e., by formula (2.48) on the class continuous functions complex variable, a class of analytic functions is constructed. The derivative of the integral (2.48) is determined by the formula

F"(z)= \frac(1)(2\pi\,i) \int\limits_(l) \frac(f(\xi))((\xi-z)^2)\,d\xi \,.

To prove formula (2.49) and, consequently, to assert that an integral of Cauchy type is analytic, it suffices, according to the definition of a derivative, to establish the validity of the inequality

\left|\frac(\Delta F)(\Delta z)-F"(z)\right|<\varepsilon,\qquad |\Delta z|<\delta(\varepsilon)

for any \varepsilon>0 and for any z from the domain of the function F(z) .

The same method can be used to show that there exists a derivative of the function defined by equality (2.49), i.e. F""(z) , and the formula

F""(z)= \frac(1)(\pi\,i) \int\limits_(l) \frac(f(\xi))((\xi-z)^3)\,d\xi \,.

The procedure can be continued and we can prove by induction the formula for the derivative of any order of the function F(z)\colon

F^((n))(z)= \frac(n{2\pi\,i} \int\limits_{l} \frac{f(\xi)}{(\xi-z)^{n+1}}\,d\xi\,. !}

Analyzing formulas (2.48) and (2.49), it is easy to see that the derivative F(z) can be obtained formally by differentiating with respect to the parameter under the integral sign in (2.48):

F"(z)= \frac(d)(dz)\! \left(\frac(1)(2\pi\,i) \int\limits_(l) \frac(f(\xi))(\ xi-z)\,d\xi\right)= \frac(1)(2\pi\,i) \int\limits_(l) \frac(d)(dz)\!\left(\frac(f (\xi))(\xi-z)\right)\!d\xi= \frac(1)(2\pi\,i) \int\limits_(l) \frac(f(\xi))( (\xi-z)^2)\,d\xi\,.

Applying formally the rule of differentiation of an integral depending on a parameter n times, we obtain formula (2.50).

We write the results obtained in this section in the form of an assertion.

Assertion 2.26. Integral \frac(1)(2\pi\,i) \int\limits_(l) \frac(f(\xi))(\xi-z)\,d\xi from a function f(z) , continuous on the curve l , there is a function that is analytic in any domain D that does not contain l ; the derivatives of this function can be obtained by differentiating with respect to the parameter under the integral sign.

Calculation of integrals from functions of a complex variable

Above, formulas for calculating integrals of functions of a complex variable are obtained - formulas (2.44) and (2.47).

If the curve l in the formula (2.44) is set parametrically: z=z(t),~ \alpha\leqslant t\leqslant\beta or, which corresponds to the actual form: \begin(cases) x=x(t),\\ y=y(t),\end(cases)\!\!\alpha\leqslant t\leqslant\beta, then, using the rules for calculating integrals of the second kind in the case of a parametric specification of a curve, we can transform formula (2.44) to the form

\int\limits_(l)f(z)\,dz= \int\limits_(\alpha)^(\beta)f\bigl(z(t)\bigr)z"(t)\,dt\,.

The result obtained and the results obtained in the previous lecture will be written as a sequence of actions.

Methods for calculating integrals \textstyle(\int\limits_(l)f(z)\,dz).

First way. Calculation of integrals \textstyle(\int\limits_(l)f(z)\,dz) from a continuous function by reducing to curvilinear integrals of functions of real variables - the application of formula (2.44).

1. Find \operatorname(Re)f(z)=u,~ \operatorname(Im)f(z)=v.

2. Write the integrand f(z)dz as a product (u+iv)(dx+i\,dy) or, multiplying, u\,dx-v\,dy+i(u\,dy+v\,dx).

3. Calculate curvilinear integrals of the form \textstyle(\int\limits_(l)P\,dx+Q\,dy), where P=P(x,y),~ Q=Q(x,y) according to the rules for calculating curvilinear integrals of the second kind.

The second way. Calculation of integrals \textstyle(\int\limits_(l) f(z)\,dz) from a continuous function by reducing to a definite integral in the case of a parametric specification of the integration path - the application of formula (2.51).

1. Write the parametric equation of the curve z=z(t) and determine the integration limits from it: t=\alpha corresponds to the starting point of the integration path, t=\beta - to the end point.

2. Find the differential of a complex-valued function z(t)\colon\, dz=z"(t)dt.
3. Substitute z(t) into the integrand, transform the integral

\int\limits_(\alpha)^(\beta)f \bigl(z(t)\bigr)\cdot z"(t)\,dt= \int\limits_(\alpha)^(\beta)\varphi (t)\,dt\,.

4. Calculate the definite integral from the complex-valued function of a real variable obtained in Section 3.

Note that the integration of a complex-valued function of a real variable does not differ from the integration of a real-valued function; the only difference is the presence in the first case of the factor i , actions with which, of course, are considered as with a constant. For example,

\int\limits_(-1)^(1)e^(2it)dt= \left.(\frac(e^(2it))(2i))\right|_(-1)^(1)= \ frac(1)(2i)(e^(2i)-e^(-2i))= \sin2\,.

The third way. Calculation of integrals of analytic functions in simply connected domains - application of formula (2.47).

1. Find the antiderivative F(z) using the properties of integrals, tabular integrals and methods known from real analysis.

2. Apply formula (2.47): \int\limits_(z_1)^(z_2)f(z)\,dz= F(z_2)-F(z_1).

Remarks 2.10

1. In the case of a multiply connected region, cuts are made so that a single-valued function F(z) can be obtained.

2. When integrating single-valued branches of multi-valued functions, a branch is distinguished by setting the value of the function at some point of the integration curve. If the curve is closed, then the starting point of the integration path is the point at which the value of the integrand is given. The value of the integral may depend on the choice of this point.

▼ Examples 2.80-2.86 calculating integrals of functions of a complex variable

Example 2.80. Calculate \int\limits_(l)\operatorname(Re)z\,dz, where l is a line connecting the point z_1=0 with the point z_2=1+i\colon

a) l - straight line; b) l - broken line OBA , where O(0;0),~B(1;0),~A(1;1).

▼ Solution

a) We apply the first method - (formula (2.44)).

1.2. The integrand has the form \operatorname(Re)z\,dz= x(dx+i\,dy). That's why

\int\limits_(l)\operatorname(Re)z\,dz= \int\limits_(l)x\,dx+ i\int\limits_(l)x\,dy\,.

3. Calculate the integrals for y=x,~ 0\leqslant x\leqslant1(the equation of the segment OA connecting the points z_1 and z_2 ). We get

\int\limits_(l)\operatorname(Re)z\,dz= \int\limits_(l)x\,dx+ i\int\limits_(l)x\,dy= \int\limits_(0)^( 1)x\,dx+ i\int\limits_(0)^(1)x\,dx= \frac(1+i)(2)\,.

b) Since the integration path consists of two segments, we write the integral as the sum of two integrals:

\int\limits_(l)\operatorname(Re)z\,dz= \int\limits_(OB)\operatorname(Re)z\,dz+ \int\limits_(BA)\operatorname(Re)z\,dz

and each is calculated as in the previous paragraph. Moreover, for the segment OB we have

\begin(cases)y=0,\\ 0 \leqslant x \leqslant1,\end(cases) and for the segment BA\colon \begin(cases)x=1,\\ 0 \leqslant y \leqslant1.\end(cases)

We make calculations:

\int\limits_(l)\operatorname(Re)z\,dz= \int\limits_(OB)x\,dx+ i\,x\,dy+ \int\limits_(BA) x\,dx+i\, x\,dy= \int\limits_(0)^(1)x\,dx+ i \int\limits_(0)^(1)1\cdot dy= \frac(1)(2)+i.

Note that the integrand in this example is not an analytic function, so the integrals over two different curves connecting two given points can have different values, which is illustrated in this example.

Example 2.81. Calculate \int\limits_(l)|z| \overline(z)\,dz, where l is the upper semicircle |z|=1 , bypassing the curve l counterclockwise.

▼ Solution

The curve has a simple parametric equation z=e^(it),~ 0\leqslant t\leqslant\pi, so it is convenient to use the second method (formula (2.51)). The integrand here is a continuous function, it is not analytic.

1.2. For z=e^(it) we find \overline(z)=e^(-it),~ |z|=1,~ dz=i\,e^(it)dt.

3.4. Substitute in the integrand. We calculate the integral

\int\limits_(l)|z| \overline(z)\,dz= \int\limits_(0)^(\pi)1\cdot e^(-it)\cdot i\,e^(it)dt= \int\limits_(0)^ (\pi)i\,dt=i\,\pi.

Example 2.82. Compute integrals of analytic functions:

a) \int\limits_(0)^(i)\sin^2z\,dz; b) \int\limits_(-i)^(1)\frac(dz)((z-i)^2), the integration path does not pass through point i .

▼ Solution

a) Apply formula (2.47) (third rule); we find the antiderivative using methods of real analysis integration:

\int\limits_()^()\sin^2z\,dz= \frac(1)(2) \int\limits_(0)^(i)(1-\cos2z)\,dz= \left.( \frac(1)(2) \left(z-\frac(1)(2)\sin2z\right))\right|_(0)^(i)= \frac(1)(2)\,i -\frac(1)(4)\sin2i= \frac(1)(2)\,i-i\,\frac(\operatorname(sh)2)(4)= \frac(i)(4)(2- \operatorname(sh)2).

b) The integrand is analytic everywhere except for the point i . After drawing a plane cut along the ray from point i to \infty , we obtain a simply connected region in which the function is analytic and the integral can be calculated by formula (2.47). Therefore, for any curve that does not pass through the point i, the integral can be calculated using the formula (2.47), while for two given points it will have the same value.

On fig. 2.44 shows two cases of making cuts. The direction of bypassing the boundary of simply connected regions, where the integrand is analytic, is indicated by arrows. We calculate the integral:

\int\limits_(-i)^(1)\frac(dz)((z-i)^2)= \left.(\frac(-1)(z-i))\right|_(-i)^(1 )= -\frac(1)(1-i)-\frac(1)(2i)=-\frac(1+i)(2)+\frac(i)(2)= -\frac(1) (2)\,.

Example 2.83. Calculate Integral \int\limits_(0)^(1+i)z\,dz.

▼ Solution

The integrand is analytic everywhere in \mathbb(C) . We apply the third method, formula (2.47):

\int\limits_(0)^(1+i)z\,dz= \left.(\frac(z^2)(2))\right|_(0)^(1+i)= \frac( 1)(2)(1+i)^2=i.

This result is obtained in example 2.78 according to the first method.

Example 2.84. Calculate Integral \oint\limits_(C)\frac(dz)((z-a)^n), where C is a circle |z-a|=R .

▼ Solution

Let's use the second method.

1. We write the circle equation in parametric form: z-a=R\,e^(it) , or z=a+R\,e^(it),~ 0\leqslant t\leqslant2\pi.
2. Finding the differential dz=R\,i\,e^(it)\,dt.
3. Substitute z=a+R\,e^(it) and dz into the integrand:

\oint\limits_(C)\frac(dz)((z-a)^n)= \int\limits_(0)^(2\pi) \frac(R\,i\,e^(it))(R ^n e^(int))\,dt= \frac(i)(R^(n-1)) \int\limits_(0)^(2\pi) e^(it(1-n))dt\ ,.

We calculate the resulting definite integral. For n\ne1 we get

\int\limits_(0)^(2\pi) e^(it(1-n))dt= \frac(1)(i(1-n)) \Bigl.(e^(it(1-n )))\Bigr|_(0)^(2\pi)= \frac(1)((n-1)i) \bigl(1-e^(2\pi\,i(n-1)) \bigr).

Because e^(2\pi\,i(n-1))= e^(2k\pi\,i)=1, that's why \oint\limits_(C)\frac(dz)((z-a)^n) =0 for n\ne1 . For n=1 we get \oint\limits_(C)\frac(dz)(z-a)= i\int\limits_(0)^(2\pi)dt=2\pi\,i\,..

We write the result in the form of a formula:

\oint\limits_(|z-a|=R)\frac(dz)((z-a)^n)=0,\quad n\ne1;\qquad \oint\limits_(|z-a|=R)\frac(dz) (z-a)=2\pi\,i\,.

In particular, \textstyle(\oint\limits_(|z|=R)\frac(dz)(z)=2\pi i). Note that if the circle C\colon |z-a|=R bypasses the point k times, then the argument (parameter) changes from 0 to 2\pi k ( k>0 , if the circle is in the positive direction, i.e. counterclockwise, and k<0 - обход по часовой стрелке). Поэтому

\oint\limits_(C)\frac(dz)(z-a)= i \int\limits_(0)^(2\pi k)dt= 2k\pi i,\qquad \oint\limits_(C) \frac( dz)(z)=2k\pi i.

Example 2.85. Calculate the integral of a function of a complex variable \int\limits_(1)^(z)\frac(d\xi)(\xi):

a) the integration path does not pass through the point z=0 and does not bypass it, -\pi<\arg z \leqslant\pi ;

b) the integration path does not pass through the point z=0 , but goes around it n times around the circle counterclockwise.

▼ Solution

a) This integral - an integral with a variable upper limit - defines a single-valued analytic function in any simply connected domain (see 2.45)). Let's find an analytical expression for this function - antiderivative for f(z)=\frac(1)(z) . Separating the real and imaginary parts of the integral \int\limits_(l)\frac(dz)(z)(using formula (2.44)), it is easy to verify that the integrands of integrals of the second kind are total differentials and, therefore, the integral \frac(d\xi)(\xi) does not depend on the form of the curve connecting the points z_1=1 and z . Let's choose a path consisting of a segment of the Ox axis from the point z_1=1 to the point z_2=r , where r=|z| , and arcs l of the circle. connecting z_2 with z (Fig. 2.45, a).

We write the integral as a sum: \int\limits_(1)^(z) \frac(d\xi)(\xi)= \int\limits_(1)^(r) \frac(dx)(x)+ \int\limits_(l) \frac(d\xi)(\xi). To calculate the integral over a circular arc, we use formula (2.51), while the arc has the equation \xi=r\,e^(it),~ 0\leqslant t\leqslant\arg z. We get \int\limits_(l)\frac(d\xi)(\xi)= \int\limits_(0)^(\arg z) \frac(ri\,e^(it))(r\,e^ (it))\,dt=i\arg z; as a result

\int\limits_(1)^(z)\frac(d\xi)(\xi)=\ln r+i\arg z,\,-\pi<\arg z \leqslant\pi

The right side of the equality defines a single-valued function \ln z - the main value of the logarithm. We get the answer in the form

\int\limits_(1)^(z)\frac(d\xi)(\xi)=\ln z\,.

Note that the resulting equality can be taken as the definition of a single-valued function \ln z in a simply connected domain - a plane with a cut along the negative real semiaxis (-\infty;0] .

b) The integral can be written as a sum: \int\limits_(1)^(z)\frac(d\xi)(\xi)= \oint\limits_(c) \frac(dz)(z)+ \int\limits_(l)\frac(d \xi)(\xi), where c is the circle |z|=1 traversed counterclockwise n times, and l is the curve connecting the points z_1 and z and not enclosing the point z=0 (Fig. 2.45,b).

The first term is equal to 2n\pi i (see example 2.84), the second - \ln(z) - formula (2.53). We get the result \int\limits_(1)^(z)\frac(d\xi)(\xi)=\ln z+2n\pi i.

Example 2.86. Calculate Integral \int\limits_(l)\frac(dz)(\sqrt(z)) along the upper arc of the circle |z|=1 provided: a) \sqrt(1)=1 ; b) \sqrt(1)=-1 .

▼ Solution

Setting the values ​​of the function \sqrt(z) at the point of the integration contour allows you to select single-valued branches of the expression \sqrt(z)= \sqrt(|z|)\exp\!\left(\frac(i)(2)\arg z+ik\pi\right)\!,~ k=0;1(see example 2.6). The cut can be drawn, for example, along the imaginary negative semiaxis. Since for z=1 we have \sqrt(1)=e^(ik\pi),~k=0;1, then in the first case a branch with k=0 is selected, in the second - with k=1 . The integrand on the integration contour is continuous. To solve, we use formula (2.51), the curve is given by the equation z=e^(it),~0\leqslant t\leqslant\pi.

a) The branch is defined when k=0 , i.e. from z=e^(it) for the integrand we obtain \sqrt(z)=e^(\frac(i)(2)t). We calculate the integral:

\int\limits_(l)\frac(dz)(\sqrt(z))= \int\limits_(0)^(\pi) \frac(i\,e^(it))(e^(i\ ,\frac(t)(2) ))\,dt= i \int\limits_(0)^(\pi)e^(i\,\frac(t)(2))dt= \Bigl.(2 \,e^(i\,\frac(t)(2)))\Bigr|_(0)^(\pi)= 2\! \left(e^(i\,\frac(\pi)(2))-1\right)= 2(i-1).

b) The branch is determined when k=1, i.e. from z=e^(it) for the integrand we have \sqrt(z)= e^(i \left(\frac(t)(2)+\pi\right))=-e^(i\,\frac(t)(2)). We calculate the integral:

\int\limits_(l)\frac(dz)(\sqrt(z))= \int\limits_(0)^(\pi)\frac(i\,e^(it))(-e^(i \,\frac(t)(2)))\,dt= \ldots= 2(1-i).

In theory and practice, in applications of the integral calculus of functions of a complex variable, when studying the behavior of functions in bounded regions or in the vicinity of individual points, integrals are considered along closed curves - the boundaries of regions, in particular, neighborhoods of points. We will consider integrals \oint\limits_(C)f(z)dz, where f(z) is analytic in some region with the exception of individual points, C is the boundary of the region or the inner contour in this region.

Basic Cauchy theorem for a simple contour

Theorem 2.1 (Cauchy's theorem for a simple contour). If f(z) is analytic in a simply connected domain, then for any contour C belonging to this domain, the equality

\oint\limits_(C)f(z)dz=0.

The proof of the theorem is easy to obtain, based on the property of analytic functions, according to which an analytic function has derivatives of any order (see Proposition 2.28). This property ensures the continuity of partial derivatives of \operatorname(Re)f(z) and \operatorname(Im)f(z), therefore, if we use formula (2.44), then it is easy to see that for each of the integrands in curvilinear integrals of the second kind, the full differential conditions are satisfied, as are the Cauchy-Riemann conditions for analytic functions. And the integrals over closed curves of total differentials are equal to zero.

Note that all the theoretical propositions presented below are ultimately based on this important theorem, including the property of analytic functions mentioned above. So that there is no doubt about the correctness of the presentation, we note that the theorem can be proved without reference to the existence of its derivatives only on the basis of the definition of an analytic function.

Corollaries from Theorem 2.1

1. The theorem is also valid if C is the boundary of the domain D , and the function f(z) is analytic in the domain and on the boundary, i.e. in \overline(D) , since, according to the definition, analyticity in \overline(D) implies analyticity of a function in some area B containing D~(B\upset\overline(D)), while C will be an inner contour in B .

2. Integrals over various curves lying in a simply connected region of function analyticity and connecting two points of this region are equal to each other, i.e. \int\limits_(l_1)f(z)dz= \int\limits_(l_2)f(z)dz, where l_1 and l_2 are arbitrary curves connecting the points z_1 and z_2 (Fig. 2.46).

To prove it, it suffices to consider the contour C , consisting of the curve l_1 (from the point z_1 to the point z_2 ) and the curve l_2 (from the point z_2 to the point z_1 ). The property can be formulated as follows. The integral of an analytic function does not depend on the form of the integration curve connecting two points of the function's analyticity region and not leaving this region.

This justifies Statement 2.25 given above on the properties of the integral \int\limits_(z_0)^(z)f(\xi)d\xi and on the existence of an antiderivative analytic function.

Cauchy's theorem for a complex contour

Theorem 2.2 (Cauchy's theorem for a complex contour). If the function f(z) is analytical in a multiply connected region bounded by a complex contour, and on this contour, then the integral over the boundary of the region of the function is equal to zero, i.e., if C is a complex contour - the boundary of the region, then the formula (2.54 ).

Complex contour C for (n+1) - connected area consists of outer contour \Gamma and inner - C_i,~i=1,2,\ldots,n; the contours do not intersect in pairs, the bypass of the boundary is positive (in Fig. 2.47, n=3).

To prove Theorem 2.2, it suffices to draw cuts in the domain (dotted line in Fig. 2.47) so that two simply connected domains are obtained and use Theorem 2.1.

Consequences from Theorem 2.2

1. Under the conditions of Theorem 2.2, the integral over the outer contour is equal to the sum of the integrals over the inner ones; bypass on all contours in one direction (in Fig. 2.48, n=2):

\oint\limits_(\Gamma)f(z)\,dz= \sum_(k=1)^(n) \oint\limits_(C_k)f(z)\,dz\,.

2. If f(z) is analytic in a simply connected region D and on the boundary of the region, with the possible exception of the point a of this region, then the integrals over various closed curves that lie in the region D and bound the regions containing the point a are equal to among themselves (Fig. 2.49):

\oint\limits_(C_k)f(z)\,dz= \oint\limits_(C_m)f(z)\,dz\,.

The proof is obvious, since each such contour can be considered as the inner boundary of a doubly connected domain whose outer boundary is the boundary of the domain D . In accordance with formula (2.55), for n=1 any such integral is equal to the integral over the boundary D .

Comparison of the formulations of Theorem 2.2 and Corollary 1 from Theorem 2.1 allows us to make a generalization, which we write in the form of the following assertion.

Assertion 2.27. If f(z) is analytic in D , then , where C is the boundary of the domain D (simple or complex contour).

Cauchy integral formula

In the following theorem, in contrast to the two previous ones, the integral of a function is considered, which, being not analytic in the region bounded by the integration contour, has a special form.

Theorem 2.3. If the function f(z) is analytic in the domain D and on its boundary C , then for any interior point a of the domain (a\in D) the equality

F(a)= \frac(1)(2\pi i) \oint\limits_(C)\frac(f(z))(z-a)\,dz\,.

Region D can be simply connected or multiply connected, and the region boundary can be a simple or complex contour.

The proof for the case of a simply connected domain is based on the result of Theorem 2.1, and for a multiply connected one, it is reduced to the case of simply connected domains (as in the proof of Theorem 2.2) by making cuts that do not pass through the point a .

It should be noted that the point a does not belong to the boundary of the region and therefore the integrand is continuous on C and the integral exists.

The theorem is of great applied interest, namely, formula (2.57) solves the so-called boundary value problem of function theory: the values ​​of a function on the boundary of the domain are used to determine its value at any interior point.

Remark 2.11. Under the conditions of the theorem, the integral \frac(1)(2\pi i) \oint\limits_(C)\frac(f(\xi))(\xi-a)\,d\xi defines an analytic function at any point z that does not belong to the contour C , and at the points of the finite region D , bounded by the contour, it is equal to f(z) (according to the formula (2.57)), and outside \overline(D) it is equal to zero due to reasons Cauchy's theorems. This integral, called the Cauchy integral, is a special case of the Cauchy-type integral (2.48). Here the contour is closed, in contrast to the arbitrary one in (2.48), and the function f(z) is analytic, in contrast to the continuous one on l in (2.48). For the Cauchy integral, therefore, assertion 2.26, formulated for an integral of Cauchy type, on the existence of derivatives is valid. Based on this, the following assertion can be formulated.

Statement 2.28

1. An analytic function at any point of analyticity can be written as an integral

F(z)= \frac(1)(2\pi i) \oint\limits_(C)\frac(f(\xi))(\xi-z)\,d\xi,\quad z\in D \,.

2. An analytic function has derivatives of any order for which the formula

F^((n))(z)= \frac(n{2\pi i} \oint\limits_{C}\frac{f(\xi)}{(\xi-z)^{n+1}}\,d\xi\,. !}

Formula (2.59) gives an integral representation of the derivatives of an analytic function.

Calculation of integrals over a closed loop

We will consider integrals of the form \oint\limits_(C)\frac(\varphi(z))(\psi(z))\,dz, where the function \varphi(z) is analytic in D , and \psi(z) is a polynomial that has no zeros on the contour C . To calculate the integrals, the theorems from the previous lecture and their corollaries are used.

Rule 2.6. When calculating integrals of the form \oint\limits_(C)f(z)\,dz four cases can be distinguished depending on the nature (multiplicity) of the zeros of the polynomial \psi(z) and their location relative to the contour C.

1. There are no zeros of the polynomial \psi(z) in the region D. Then f(z)= \frac(\varphi(z))(\psi(z)) the function is analytic and, applying the main Cauchy theorem, we have the result \oint\limits_(C)f(z)\,dz=0.

2. In the region D there is one simple zero z=a of the polynomial \psi(z) . Then we write the fraction as \frac(f(z))(z-a) , where f(z) is a function analytic in \overline(D) . Applying the integral formula, we get the result:

\oint\limits_(C)\frac(\varphi(z))(\psi(z))\,dz= \oint\limits_(C)\frac(f(z))(z-a)\,dz= 2 \pi i\cdot f(a).

3. In the region D there is one multiple zero z=a of the polynomial \psi(z) (of multiplicity n ). Then we write the fraction in the form \frac(f(z))((z-a)^n), where f(z) is a function analytic in \overline(D) . Applying formula (2.59), we obtain the result

\oint\limits_(C)\frac(f(z))((z-a)^n)\,dz= \frac(2\pi i)((n-1)f^{(n-1)}(a). !}

4. In the region D there are two zeros of the polynomial \psi(z)\colon\,z_1=a and z_2=b . Then, using Corollary 1 from Theorem 2.2, we write the integral in the form \oint\limits_(C)\frac(dz)(z-a) , where C is an arbitrary contour bounding the region containing the point a .

▼ Solution

Consider a doubly connected region, one of whose boundary is the contour C , the other is the circle |z-a|=R . By Corollary 2 of Theorem 2.2 (see (2.56)) we have

\oint\limits_(C)\frac(dz)(z-a)= \oint\limits_(|z-a|=R)\frac(dz)(z-a)\,.

Taking into account the result of solving Example 2.84 (formula (2.52)), we obtain the answer \oint\limits_(C) \frac(dz)(z-a)=2\pi i.

Note that the solution can be obtained by applying the Cauchy integral formula with f(z)=1 . In particular, we get \oint\limits_(C)\frac(dz)(z)=2\pi i, since the contour C goes around the point z=0 once. If the contour C goes around the point z=0 k times in the positive (k>0) or negative direction (k<0) , то \oint\limits_(C)\frac(dz)(z)=2k\pi i.

Example 2.88. Calculate \oint\limits_(l)\frac(dz)(z), where l is a curve connecting points 1 and z , going around the origin once.

▼ Solution

The integrand is continuous on the curve - the integral exists. For the calculation, we use the results of the previous example and example 2.85. To do this, consider a closed loop, connecting, for example, point A with point 1 (Fig. 2.50). The integration path from point 1 to point z through point A can now be represented as consisting of two curves - a closed contour C (curve BDEFAB ) and a curve l_0 connecting points 1 and z through point A\colon

\oint\limits_(l)\frac(dz)(z)= \oint\limits_(C)\frac(dz)(z)+ \oint\limits_(l_0) \frac(dz)(z)\,.

Using the results of Examples 2.85 and 2.87, we get the answer:

\oint\limits_(l)\frac(dz)(z)= \int\limits_(1)^(z)\frac(1)(z)=\ln z+2\pi i\,.

Without changing the geometric picture, we can consider the case when the curve goes around the origin n times. Get the result

\oint\limits_(l)\frac(dz)(z)= \int\limits_(1)^(z)\frac(1)(z)=\ln z+2n\pi i\,.

The resulting expression defines a multivalued function \operatorname(Ln)z= \int\limits_(1)^(z)\frac(dz)(z), the integration path does not pass through the origin. The choice of a branch of a multi-valued expression is determined by setting the value of the function at some point.

Example 2.89. Find \ln2i= \int\limits_(1)^(2i)\frac(1)(z), if \ln1=4\pi i .

▼ Solution

We find the zeros of the denominator - the singular points of the integrand. These are the dots z_1=0,~ z_(2,3)=\pm4i. Next, you need to determine the location of the points relative to the integration contour. In both cases, none of the points is included in the area bounded by the contour. This can be verified using the drawing. Both contours are circles, the center of the first one is z_0=2+i and the radius is R=2 ; center of the second z_0=-2i and R=1 . It is possible to determine whether a point belongs to an area in a different way, namely, to determine its distance from the center of the circle and compare it with the value of the radius. For example, for the point z_2=4i this distance is equal to |4i-2-i|=|3i-2|=\sqrt(13), which is greater than the radius (\sqrt(13)>2) , so z_2=4i does not belong to the circle |z-2-i|<2 . В обоих случаях подынтегральная функция является, аналитической в соответствующих кругах. Следовательно, согласно теореме Коши (пункт 1 правил 2.6), интеграл равен нулю. Заметим, что заданный интеграл равен нулю и для любого другого контура, ограничивающего область, в которую не входят ни одна из особых точек - нулей знаменателя.

Example 2.91. Calculate in the following cases of setting the contour C\colon a) |z|=2 ; b) |z+1+i|=2 .

▼ Solution

Arguing as in the previous example, we find that in both cases only one of the singular points z_1=0 is located inside the circles. Therefore, applying clause 2 of rules 2.6 (Cauchy's integral formula), we write the integrand as a fraction \frac(1)(z)\cdot \frac(\sin z)(z^2+16), where the numerator f(z)= \frac(\sin z)(z^2+16) is a function that is analytic in the specified circles. The answer for both cases is the same:

\oint\limits_(C)\frac(\sin z)(z(z^2+16))\,dz= \left.(2\pi i \cdot \frac(\sin z)(z^2+ 16))\right|_(z=0)=0.

Example 2.92. Calculate \oint\limits_(C)\frac(\sin z)(z^3+16z)\,dz in the following cases of setting the contour C\colon a) |z+4i|=2 ; b) |z-1+3i|=2 .

▼ Solution

The integration contours are circles, as above, and in the case of "a" the center is at the point z_0=-4i,~R=2 , in the case of "b" - at the point z_0=1-3i,~R=2 .nIn both cases, one point z_0=-4i gets inside the corresponding circles. Applying clause 2 of rules 2.6, we write the integrand in the form \frac(1)(z+4i)\frac(\sin z)(z(z-4i)), where the numerator f(z)=\frac(\sin z)(z(z-4i)) is an analytic function in the domains under consideration. Applying the integral formula, we get the answer:

\oint\limits_(C)\frac(\sin z)(z^3+16z)\,dz= \left.(2\pi i\cdot \frac(\sin z)(z(z-4i)) )\right|_(z=-4i)= 2\pi i\cdot \frac(-\sin4i)(-32)= \frac(\pi i\cdot i \operatorname(sh)1)(16)= -\frac(\pi \operatorname(sh)1)(16)\,.

Example 2.93. Calculate the integral in the following cases of contour assignment: a) |z+i|=1 ; b) |z+2+i|=2 .

▼ Solution

Find singular points of the integrand - zeros of the denominator z_1=i,~z_2=-2 . We determine the belonging of the points to the corresponding areas. In case "a" into the circle |z+i|<1 не входит ни одна точка. Следовательно, интеграл в этом случае равен нулю.

In case "b" into the circle |z+2+i|<2 радиуса 2 с центром в точке z_0=-2-i входит одна точка: z=-2 . Записываем дробь в виде \frac(1)(z+2)\frac(e^z)((z-i)^2), where f(z)=\frac(e^z)((z-i)^2)- analytic function in the circle |z+2+i|<2 . Вычисляем интеграл:

\oint\limits_(C)\frac(e^z\,dz)((z-i)^2(z+2))= 2\pi i\cdot \frac(e^(-2))((2+ i)^2)= \frac(2\pi)(25)e^(-2)(4+3i).

Example 2.94. Calculate Integral \oint\limits_(C)\frac(e^z\,dz)((z-i)^2(z+2)) in the following cases of contour assignment: a) |z-i|=2 ; b) |z+2-i|=3 .

▼ Solution

a) In the circle |z-i|<2 попадает точка z=i . Записываем функцию \frac(1)((z-i)^2)\frac(e^z)(z+2) and apply clause 3 of rules 2.6 for m=2 and a=i . We calculate the integral:

\oint\limits_(C)\frac(e^z\,dz)((z-i)^2(z+2))= \left.(2\pi i \left(\frac(e^z)(z+ 2)\right)")\right|_(z=i)= \left.(2\pi i\cdot \frac(e^z(z+2)-e^z)((z+2)^ 2))\right|_(z=i)= \left.(2\pi i\cdot \frac(e^z(1+z))((z+2)^2))\right|_( z=i)= \frac(2\pi i(1+i))((2+i)^2)\,e^(i).

b) To the circle |z+2-i|<3 входят обе точки z_1=i,~z_2=-2 . Решаем в соответствии с п. 4 правил 2.6. Записываем интеграл в виде суммы двух интегралов:

\oint\limits_(C)f(z)\,dz= \oint\limits_(C_1)f(z)\,dz+ \oint\limits_(C_2) f(z)\,dz\,.

where each of the contours C_1 and C_2 covers only one of the points. In particular, as the contour C_1, you can take the circle from the previous case "a"; C_2 - circle from example 2.93 p. "b", i.e. you can use the results. Write down the answer:

\oint\limits_(C)\frac(e^z\,dz)((z-i)^2(z+2))= 2\pi i\cdot \frac(e^(-2))((2+ i)^2)+ 2\pi i\cdot \frac(1+i)((2+i)^2)\,e^(i)= \frac(2\pi i)((2+i) ^2)\bigl(e^(-2)+e^(i)(1+i)\bigl).

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Let us consider a smooth curve Γ in the complex plane given by the parametric equations

(the definition of a smooth curve is given at the beginning of §8). As noted in § 8, these equations can be written in a compact form:

When changing the parameter t from a to /3 corresponding point z(t) will move along the curve Γ. Therefore, equations (15.1) and (15.2) not only determine the points of the curve Γ, but also set the direction of going around this curve. Curve Г with a given direction of its bypass is called oriented curve.

Let in the area D C C continuous function f(r) = = u(x, y) + iv(x. y), and let the curve Γ lie in D. To introduce the concept of an integral [f(z)dz from function f(z) along the curve r, we define r

differential dz equality dz = dx + idy. The integrand is transformed to the form

Thus, the integral of the complex function f(z) along the curve Γ it is natural to define by the equality

whose right-hand side contains two real curvilinear integrals of the second kind of real functions and and and. To calculate these integrals, instead of X and at substitute functions x(t) and t/(/), but instead of dx and dy- differentials of these functions dx = x"(t) dt and dy = y"(t)dt. Then the integrals on the right-hand side of (15.3) reduce to two integrals of functions of a real variable t

We are now ready to give the following definition.

Integral along the curve G on the function of the complex variable f(z) the number is called J" f(z)dz and calculated by

where z(t) = x(t) + iy(t), a ^ t ^ ft, - equation of the curve Г, a z"(t) = = x"(t) + iy"(t).

Example 15.1. Calculate the integral of a function f(z) = (g - a) p along a circle of radius r with center a, the direction of the bypass of which is counterclockwise.

Solution: Equation of a circle z - a= g will z - a = ge a, or

When it changes t. from 0 to 2tg point z(t.) moves in a circle r counterclockwise. Then

Applying equality (15.5) and De Moivre formula (2.10), we obtain

We have obtained a result important for further presentation:

Note that the value of the integral does not depend on the radius G circles.

EXAMPLE 15.2. Calculate the integral of a function f(z) = 1 but a smooth curve Γ with origin at the point a and end at a point b.

Solution. Let the curve Γ be given by the equation z(t.) = x(t) + + iy(t), and ^ t^ /3, and a= -r(a), b = z((3). Using formula (15.5), as well as the Newton-Leibniz formula for calculating integrals of real functions, we obtain

We see that the integral f 1 dz does not depend on the type of path G, connect-

between points a and 6, and depends only on the endpoints.

Let us briefly describe another approach to the definition of the integral of the complex function f(z) along a curve, similar to the definition of an integral of a real function over a segment.

Let us partition the curve Γ in an arbitrary way into P plots points zq = a, z 1, ..., z n-th z n = b, numbered in the direction of movement from the starting point to the end (Fig. 31). Denote z - zo ==Az> ... , Zlc - Zk-l = Az/c, zn -Zn- 1 = = Azn.(Number Azk represented by a vector coming from the point zi L_i in Zk-) At each site (zk-i,Zk) we choose an arbitrary point on the curve (q- and make up the sum

This amount is called integral sum. Let us denote by L the length of the largest of the segments into which the curve G is divided. Consider a sequence of partitions for which A -? 0 (while P-* oo).

П1> units of integral sums, calculated under the condition that the length of the largest of the segments of the partition tends to zero, is called integral of the function/(G) along the curve G and is denoted by G f(z)dz:

It can be shown that this definition also leads us to formula (15.3) and is therefore equivalent to the definition (15.5) given above.

Let us establish the main properties of the integral / f(z)dz.

1°. Linearity. For any complex constants a and b

This property follows from equality (15.5) and the corresponding properties of the integral over a segment.

2°. Additivity. If the curve G divided into segments Ti m G2, then

Proof. Let the curve Γ with ends a, b is divided by a point c into two parts: a curve Гi with ends a, With and the curve Gr with ends with, b. Let Г be given by the equation z = z(t), a ^ t ^ in. and a= 2(a), b = z(ft), c = 2(7). Then the equations of the curves Г1 and Гг will be z = z(t), where a ^ t^7 for Ti and 7^ t^/? for Gg. Applying definition (15.5) and the corresponding properties of the integral over a segment, we obtain

Q.E.D.

Property 2° makes it possible to calculate integrals not only over smooth curves, but also piecewise smooth, i.e. curves that can be divided into a finite number of smooth sections.

3°. When the direction of the curve is changed, the integral changes sign.

Prove l with t in about. Let the curve Г end a and b is given by the equation r = r(?), o ^ t ^ $. A curve consisting of the same points as Γ, but differing from Γ in the direction of the detour (orientation), will be denoted by Γ. Then Г - is given by the equation z= 2i(J)> where z(t)= 2(0 -I - fi - t), Indeed, we introduce a new variable r = a + - t. When it changes t from a to (d variable r changes from (5 to a. Consequently, the point r(m) will run through the curve r.

Property 3° is proved. (Note that this property follows directly from the definition of the integral (15.8): when the orientation of the curve changes, all increments AZk change sign.)

4°. The modulus of the integral f f(z)dz does not exceed the value of the curvature G

linear integral of the modulus of the function along the length of the curve s (curvilinear integral of f(z) of the first kind):

It is easy to see that z[(t) = r" r (t)(a + - t)J = -z "t (t), dt = -dr. Using the definition (15.5) and passing to the variable r, we obtain

Proof. Let us use the fact that for the integral over a segment

(this inequality immediately follows from the definition of the integral over a segment as the limit of integral sums). From here and from (15.5) we have

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