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. Cauchy's main theorem for a simple contour
5. Cauchy's theorem for a complex contour
6. Cauchy integral 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 domain) 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 partition; \Delta z_k - increment of the function argument in this partition section,\lambda= \max_(k)|\Delta z_k| - partition step, |\Delta z_k|- length of the chord connecting the ends of the arc \Delta l_k ; curve l is divided arbitrarily into n parts \Delta l_k,~ k=1,2,\ldots,n. The direction chosen on the curve, i.e. start and end points are indicated. 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 a finite area on the left, bounded by a contour. Formula (2.43) determines
line integral of a function of a complex variable
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 on 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) along the curve l and the formula holds
\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 limited in magnitude 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 calculating it 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 integral sign of the actions of separating 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 line OA
a) a straight line connecting 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, dividing 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 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 complete differentials (see condition (2.30)), so it is enough to consider one case of the 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 more general view: calculate the integral
\int\limits_(l)z\,dz from point z_1=0 to point z_2=1+i.
In the next paragraph we will consider such cases of integration in more detail.
2. Let the integral of a continuous function in a certain region not depend on the type of curve connecting two points in this region. Let's fix it 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 determines some function in the specified area.
Below we will give a substantiation of the statement that in the case of a simply connected domain, the integral defines a single-valued function in this domain. Let us introduce the notation
\int\limits_(z_0)^(z) f(\xi)\,d\xi=F(z).
Function F(z) is an integral with a variable upper limit.
Using the definition of 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 in 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.
From equality (2.46), in particular, it follows that the integrand function 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 Statement 2.28) - analytical function.
3. The function F(z) for which equality (2.46) holds is called an antiderivative for the function f(z) in a simply connected domain, and the set 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 statement.
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 is a function analytic in this domain; this function is an antiderivative of the integrand.
2. Any analytic function 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 the rules of integration are used.
For example, \int e^z\,dz=e^z+c,~~ \int\cos z\,dz=\sin z+c..
Between curvilinear integral from 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 we consider, in addition to integrals containing a parameter within the limits of integration (formula (2.45) gives 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 important place in the theory and practice of complex integration and applications, it is occupied by 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 defines a certain function in any domain not containing l
\frac(1)(2\pi\,i) \int\limits_(l) \frac(f(\xi))(\xi-z)\,d\xi=F(z).
Integral (2.48) is called a Cauchy type integral; the factor \frac(1)(2\pi\,i) is introduced for the convenience of using the constructed function.
For this function, as for the function defined by equality (2.45), it is proved that it is analytic everywhere in the domain of definition. Moreover, unlike the integral (2.45), here it is not required that the generating function f(z) be analytic, i.e. according to 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, therefore, the statement about the analyticity of the Cauchy type integral, it is sufficient, according to the definition of the 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 definition of the function F(z) .
Using the same method, it can be shown that there is a derivative of the function defined by equality (2.49), i.e.
F""(z) , and the formula is valid
F""(z)= \frac(1)(\pi\,i) \int\limits_(l) \frac(f(\xi))((\xi-z)^3)\,d\xi \,.
The procedure can be continued and proved by induction the formula for the derivative of any order of the function F(z)\colon{2\pi\,i} \int\limits_{l} \frac{f(\xi)}{(\xi-z)^{n+1}}\,d\xi\,. !}
F^((n))(z)= \frac(n
Analyzing formulas (2.48) and (2.49), it is easy to verify 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)\! (\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 for differentiating the integral depending on the parameter n times, we obtain formula (2.50).
We write the results obtained in this section in the form of a statement. Statement 2.26. Integral\frac(1)(2\pi\,i) \int\limits_(l) \frac(f(\xi))(\xi-z)\,d\xi
of a function f(z) continuous on a curve l is a function that is analytic in any domain D not containing l; derivatives of this function can be obtained by differentiation with respect to the parameter under the integral sign.
Calculation of integrals of functions of a complex variable
Above we obtained formulas for calculating integrals of functions of a complex variable - formulas (2.44) and (2.47). If the curve l in formula (2.44) is specified parametrically: 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 definition 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\,.
We will write down the result obtained and the results obtained in the previous lecture 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 reduction to curvilinear integrals of functions of real variables - 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.
Second way. Calculation of integrals \textstyle(\int\limits_(l) f(z)\,dz) from a continuous function by reduction to a definite integral in the case of a parametric definition of the integration path - application of formula (2.51).
1. Write down the parametric equation of the curve z=z(t) and from it determine the limits of integration: t=\alpha corresponds to the starting point of the integration path, t=\beta - the final point.
2. Find the differential of a complex-valued function z(t)\colon\, dz=z"(t)dt.
3. Substitute z(t) into the integrand and 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 of the complex-valued function of a real variable obtained in step 3.
Note that integrating a complex-valued function of a real variable is no different from integrating a real-valued function; the only difference is the presence in the first case of a factor i, actions with which, naturally, 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\,.
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, table 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).
Notes 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, the branch is distinguished by specifying the value of the function at a certain point on the integration curve. If the curve is closed, then the starting point of the integration path is considered to be 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 of calculating integrals of functions of a complex variable
Example 2.80. Calculate \int\limits_(l)\operatorname(Re)z\,dz, where l is the line connecting point z_1=0 to point z_2=1+i\colon
a) l - straight; 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 at y=x,~ 0\leqslant x\leqslant1(equation of the segment OA connecting 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 we calculate each one 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 along two different curves connecting two given points can have different values, as illustrated in this example.
Example 2.81. Calculate \int\limits_(l)|z| \overline(z)\,dz, where l is the upper semicircle |z|=1, traversing the curve l counterclockwise.
▼ Solution
The curve has a simple parametric equation z=e^(it),~ 0\leqslant t\leqslant\pi, therefore it is convenient to use the second method (formula (2.51)). The integrand here is a continuous function and is not analytical.
1.2. For z=e^(it) we find \overline(z)=e^(-it),~ |z|=1,~ dz=i\,e^(it)dt.
3.4. Substitute into the integrand. 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. Calculate integrals of analytical 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 integration of real analysis:
\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 point i. By cutting the plane 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 using formula (2.47). Therefore, for any curve that does not pass through point i, the integral can be calculated using formula (2.47), and for two given points it will have the same value.
In Fig. Figure 2.44 shows two cases of making cuts. The direction of traversing 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 use 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 was 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 the circle |z-a|=R.
▼ Solution
Let's use the second method.
1. We write the equation of the circle in parametric form: z-a=R\,e^(it) , or z=a+R\,e^(it),~ 0\leqslant t\leqslant2\pi.
2. Find 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 at n\ne1 . For n=1 we get \oint\limits_(C)\frac(dz)(z-a)= i\int\limits_(0)^(2\pi)dt=2\pi\,i\,..
Let's write the result as 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 is traversed by a point k times, then the argument (parameter) changes from 0 to 2\pi k ( k>0 if the traversal 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 go around 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 - the 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 complete differentials and, therefore, the integral \frac(d\xi)(\xi) does not depend on the type of 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 a circle. connecting z_2 to 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), 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 principal 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 half-axis (-\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 a circle |z|=1 traversed counterclockwise n times, and l is a curve connecting points z_1 and z and not covering point z=0 (Fig. 2.45, b).
The first term is equal to 2n\pi i (see example 2.84), the second is \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 a point on the integration contour allows you to select unambiguous 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 made, for example, along an imaginary negative semi-axis. Since for z=1 we have \sqrt(1)=e^(ik\pi),~k=0;1, then in the first case the 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), define the curve by the equation z=e^(it),~0\leqslant t\leqslant\pi.
a) The branch is determined at 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 at 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 integral calculus of functions of a complex variable, when studying the behavior of functions in bounded areas or in the vicinity of individual points, integrals are considered over closed curves - the boundaries of areas, in particular neighborhoods of points. We will consider the integrals \oint\limits_(C)f(z)dz, where f(z) is analytical in some c region, with the exception of individual points, C is the boundary of the region or the internal contour in this region.
Cauchy's basic 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 following equality holds:
\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 Statement 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 conditions of the total differential are satisfied, like the Cauchy-Riemann conditions of analytic functions. And integrals over closed curves from total differentials are equal to zero.
Note that all theoretical positions presented below are ultimately based on this important theorem, including the above-mentioned property of analytic functions. So that there is no doubt about the correctness of the presentation, we note that the theorem can be proven 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, by definition, analyticity in \overline(D) implies analyticity of the function in some domain B containing D~(B\supset\overline(D)), and C will be the inner contour in B.
2. Integrals over various curves lying in a simply connected domain of analyticity of a function and connecting two points of this domain 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 points z_1 and z_2 (Fig. 2.46).
To prove it, it is enough to consider the contour C, consisting of curve l_1 (from point z_1 to point z_2) and curve l_2 (from point z_2 to point z_1). The property can be formulated as follows. The integral of an analytic function does not depend on the type of integration curve that connects two points in the domain of analyticity of the function and does not leave this domain.
This provides a justification for Statement 2.25 given above about the properties of the integral \int\limits_(z_0)^(z)f(\xi)d\xi and about the existence of a primitive 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 analytic in a multiply connected domain bounded by a complex contour, and on this contour, then the integral of the function over the boundary of the domain is equal to zero, i.e., if C is a complex contour - the boundary of the domain, then formula (2.54) is valid ).
A complex contour C for an (n+1) - connected region consists of an outer contour \Gamma and internal - C_i,~i=1,2,\ldots,n; the contours do not intersect in pairs, the border detour is positive (in Fig. 2.47, n=3).
To prove Theorem 2.2, it is enough to make cuts in the region (dotted line in Fig. 2.47) so that two simply connected regions are obtained and use Theorem 2.1.
Corollaries from Theorem 2.2
1. When the conditions of Theorem 2.2 are met, the integral over the outer contour is equal to the sum of the integrals over the inner contours; bypass on all circuits 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 domain D and on the boundary of the domain, with the possible exception of point a of this domain, then the integrals over various closed curves that lie in the domain D and bound the domains containing the point a are equal 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 internal boundary of a doubly connected region, the external boundary of which is the boundary of the region D. In accordance with formula (2.55) for n=1, any such integral is equal to the integral over the boundary D.
Comparing 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 statement.
Statement 2.27. If f(z) is analytic in D, then , where C is the boundary of the domain D (simple or complex contour).
Integral Cauchy formula
In the next theorem, unlike the previous two, we consider the integral of a function, which, while not being analytic in the region limited 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 internal point a of the domain (a\in D) the equality holds
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 boundary of the region 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 domain 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 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 important applied interest, namely, according to formula (2.57), the so-called boundary value problem of function theory is solved: from the values of the function on the boundary of the domain, its value at any internal point is determined.
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 not belonging to the contour C, and at points of the finite region D bounded by the contour, it is equal to f(z) (according to formula (2.57)), and outside \overline(D) it is equal to zero due to the grounds Cauchy's theorem. 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 continuous on l in (2.48). For the Cauchy integral, therefore, Statement 2.26 on the existence of derivatives, formulated for an integral of Cauchy type, is valid. Based on this, the following statement can be formulated.Statement 2.28
1. The analytical 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 analytical function has derivatives of any order, for which the formula is valid
The procedure can be continued and proved by induction the formula for the derivative of any order of the function F(z)\colon{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 the analytical function.
Calculating closed loop integrals
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 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 Depending on the nature (multiplicity) of the zeros of the polynomial \psi(z) and their location relative to the contour C, four cases can be distinguished.
1. There are no zeros of the polynomial \psi(z) in domain D. Then f(z)= \frac(\varphi(z))(\psi(z)) the function is analytic and, applying Cauchy's fundamental theorem, we have the result \oint\limits_(C)f(z)\,dz=0.
2. In region D there is one simple zero z=a of the polynomial \psi(z) . Then we write the fraction in the form \frac(f(z))(z-a) , where f(z) is a function that is 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 region D there is one multiple zero z=a of the polynomial \psi(z) (of multiple n). Then we write the fraction in the form \frac(f(z))((z-a)^n), where f(z) is a function that is 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. Region D contains 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 limiting the region containing point a .
▼ Solution
Consider a doubly connected region, one boundary of which is the contour C, the other is the circle |z-a|=R. By Corollary 2 from 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 contour C goes around point z=0 k times in a 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. Let's 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 branch of a multivalued expression is determined by specifying 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 points 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 limited by the contour. You can verify this by using the drawing. Both contours are circles, the center of the first is z_0=2+i and the radius R=2; center of the second z_0=-2i and R=1. You can determine whether a point belongs to a region differently, namely, determine its distance from the center of the circle and compare it with the value of the radius. For example, for 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 the specifications of the contour C\colon a) |z|=2 ; b) |z+1+i|=2 .
▼ Solution
Reasoning 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, using paragraph 2 of rules 2.6 (Cauchy integral formula), we write the integrand function as a fraction \frac(1)(z)\cdot \frac(\sin z)(z^2+16), where the numerator f(z)= \frac(\sin z)(z^2+16)- a function that is analytical in the specified circles. The answer is the same for both cases:
\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 specifying the contour C\colon a) |z+4i|=2 ; b) |z-1+3i|=2 .
▼ Solution
The integration contours are circles, as above, and in case “a” the center is at point z_0=-4i,~R=2, in case “b” - at point z_0=1-3i,~R=2.nIn both cases, one point z_0=-4i falls inside the corresponding circles. Applying clause 2 of rules 2.6, we write the integrand function 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 analytical function in the areas 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 specifying the contour: a) |z+i|=1 ; b) |z+2+i|=2 .
▼ Solution
We find the singular points of the integrand - the zeros of the denominator z_1=i,~z_2=-2. We determine that the points belong to the corresponding areas. In the case of "a" in the circle |z+i|<1 не входит ни одна точка. Следовательно, интеграл в этом случае равен нулю.
In case "b" in 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 specifying the contour: a) |z-i|=2 ; b) |z+2-i|=3 .
▼ Solution
a) Into 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 with 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) Into 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, the circle from the previous case “a” can be taken as the contour C_1;
C_2 - circle from example 2.93 p. "b", i.e. you can use the results obtained. We write down the answer:
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Let us consider a smooth curve Γ on the complex plane defined 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 a parameter t from A to /3 corresponding point z(t) will move along the curve Г. Therefore, equations (15.1) and (15.2) not only determine the points of the curve Г, but also specify the direction of traversal of this curve. A curve Г with a given direction of its traversal is called oriented curve.
Let in the area D C C is given a continuous function /(r) = = u(x, y) + iv(x. y), and let the curve G lie in D. To introduce the concept of integral [f(z)dz from function f(z) along the curve Г, we determine r
differential dz equality dz = dx + idy. The integrand expression is transformed to the form
Thus, the integral of the complex function f(z) along the curve Г it is natural to define by the equality
the right side of which includes 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/(/), and instead dx And dy- differentials of these functions dx = x"(t)dt And dy = y"(t)dt. Then the integrals on the right side of (15.3) will be reduced to two integrals of functions of a real variable t
We are now ready to give the following definition.
Integral along a curve G on the function of a complex variable f(z) is the number denoted by J" f(z)dz and calculated by
Where z(t) = x(t) + iy(t), a ^ t ^ ft, - equation of 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 which is counterclockwise.
Solution: Equation of a circle z - a= r will be z - a = ge a, or
When it changes t. from 0 to 2tg point z(t.) moves along the circle G counterclockwise. Then
Applying equality (15.5) and Moivre’s formula (2.10), we obtain
We have obtained a result that is important for further discussion:
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 Г with the beginning at the point A and end at a point b.
Solution. Let the curve Г be given by the equation z(t.) = x(t) + + iy(t), and ^ t^ /3, and A= -g(a), b = z((3). Using formula (15.5), as well as Newton’s 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, connecting
common points a and 6, a depends only on the end points.
Let us briefly outline another approach to determining the integral of a complex function f(z) along a curve, similar to the definition of the integral of a real function over a segment.
Let us divide the curve Г arbitrarily into P plots with dots zq = a, z 1, ..., z n-ь z n = b, numbered in the direction of movement from the starting point to the ending point (Fig. 31). Let's denote z - zo = = Az> ... , Zlc - Zk-l = Az/c, z n -Zn- 1 = = Azn.(Number Azk represented by a vector coming from the point zi L_i in Zk-) At each site (zk-i,Zk) curve, choose an arbitrary point (d- and compose the sum
This amount is called integral sum. Let us denote by A the length of the largest of the sections into which the curve G is divided. Consider the sequence of partitions for which A -? 0 (at the same time P-* oo).
The unit of integral sums, calculated under the condition that the length of the largest section of the partition tends to zero, is called integral of the function/(G) along the curve G and 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 basic 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 sections Ti m G2, That 
Proof. Let the curve G with ends a, b divided by point c into two parts: curve Gi with ends a, With and the curve GG with ends c, b. Let Γ be given by the equation z = z(t), A ^ t ^ V. and A= 2(a), b = z(ft), c = 2(7). Then the equations of 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° allows you to calculate integrals not only over smooth curves, but also over piecewise smooth, i.e. curves that can be divided into a finite number of smooth sections.
3°. When changing the direction of the curve, the integral changes sign.
Proof of l s t v o. Let the curve Г with ends 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 traversal (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, let us introduce a new variable r = a + - t. When it changes t from a to (d variable g varies from (5 to a. Consequently, the point r(t) will run along the curve Γ."
The 3° property has been proven. (Note that from the definition of the integral (15.8) this property follows directly: 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 curvilinear 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's easy to see that z[(t) = g" g (t)(a + - t)J = -z" t (t), dt = -dr. Using 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 an integral over a segment as the limit of integral sums). From here and from (15.5) we have
Theoretical minimumThere are often cases when the calculation of definite integrals by methods of complex analysis is preferable to methods
material analysis. The reasons can be very different. TFCT methods can, in some cases, make it possible to greatly reduce calculations.
Sometimes the Newton-Leibniz formula cannot be used because the indefinite integral is not expressed in elementary functions.
Methods of differentiation and integration with respect to a parameter require a very careful justification of their applicability, and sometimes the parameter
must be introduced artificially.Usually, using complex analysis methods, improper integrals are calculated - over an infinite interval or from unbounded integrals on an interval
integration of functions. The general idea is as follows. A contour integral is compiled. The integral over some parts of the contour should
coincide with the desired definite integral - at least up to a constant factor. Integrals over other parts of the contour
must be calculated. The fundamental residue theorem is then applied, which states that
,
where are the singular points of the function located inside the integration contour. Thus, a contour integral with one
side turns out to be expressed through the desired definite integral, and on the other side it is calculated using residues (which is usually
does not present any serious difficulties).The main difficulty is the choice of the integration contour. It is suggested, in principle, by the integrand function. However, without sufficient
It is difficult to master this method in practice, and therefore quite a lot of examples will be given. The most commonly used contours are composed of
elements along which it is convenient to carry out integration (straight lines, circular arcs).
integration in the complex planeExample 1. Fresnel integrals.
Let's calculate the integrals,
.
It is easy to guess that the first step is to move to exponential form, which involves considering the integral.
You just need to select the integration contour. It is clear that the semi-axis must enter the contour. Real and
the imaginary parts of the integral over this part of the contour are Fresnel integrals. Next, the calculated contour integral over the structure
the integrand resembles the Euler-Poisson integral, the value of which is known. But to obtain this integral, we need to put
, Then . And this representation of a variable is integration along a straight line passing through a point
at an angle to the real axis.
So, there are two contour elements. In order for the contour to close, we will assume that the selected two sections of the contour have a finite length and close
contour of an arc of a circle of radius. Later we will direct this radius to infinity. The result is shown in Fig. 1 circuit.
(1)
Inside the integration contour, the integrand function has no singular points, so the integral along the entire contour is equal to zero.
.
In the limit, this integral is equal to zero.
On the site you can write , then.
We substitute the obtained results into (1) and go to the limit:
Separating the real and imaginary parts, we find, taking into account the value of the Euler-Poisson integral
,
.Example 2. Selecting an integration contour containing inside a singular point of the integrand.
Let us calculate an integral similar to that considered in the first example: , where.
We will calculate the integral. Let's choose a contour similar to the one used in the first example. Only now there is no goal
reduce the calculation to the Euler-Poisson integral. Note here that when replacingthe integrand will not change.
This consideration prompts us to choose the inclined straight line of the integration contour so that it makes an angle with the real axis.
When writing the contour integral(2)
the integral along the arc of a circle tends to zero in the limit. On the site you can write:
.
Thus, from (2) when passing to the limit we find
.
Here it is taken into account that inside the integration contour the integrand has a simple pole.
From here we find the required integral:.
Example 3. Close the integration loop through the upper or lower half-plane?
Using the following fairly simple integral, we demonstrate a characteristic detail of the choice of the integration contour. Let's calculate
integral
In fact, the required integral of the function is calculated along the real axis, on which the integrand does not have
features. All that remains is to close the integration loop. Since the function under the integral has only two finite singular points, then
You can close the contour with a semicircle, the radius of which should tend to infinity. And here the question arises of how should
a semicircle should be selected: in the upper or lower half-plane (see Fig. 3 a, b). To understand this, let's write the integral over the semicircle
in both cases:
A)
b)
As can be seen, the behavior of the integral in the limit is determined by the factor .
In the case of "a", and therefore the limit will be finite under the condition .
In case “b” - on the contrary - , and therefore the limit will be finite under the condition .
This suggests that the way the loop is closed is determined by the sign of the parameter. If it is positive, then
the contour is closed through the upper half-plane, otherwise through the lower. Let's consider these cases separately.
A)
The integral over a semicircle in the limit, as we have seen, goes to zero. Inside the circuit (see Fig. 3a) there is
special point, therefore
b)
We find similarly using integration along the contour shown in Fig. 3b,
Note. It may seem strange that the integral of a complex function turns out to be real. However, this is easy to understand if in the original
in the integral, separate the real and imaginary parts. In the imaginary part, under the integral there will be an odd function, and the integral is calculated in symmetric
limits. Those. the imaginary part will go to zero, which is what happened in our calculation.Example 4. Bypassing singular points of the integrand when constructing an integration contour.
In the examples considered, the integrand either did not have singular points, or they were inside the integration contour. However
It can be convenient to choose a contour so that the singular points of the function fall on it. Such points have to be avoided. Bypass is carried out
along a circle of small radius, which then simply tends to zero. As an example, let's calculate the integral.
It may seem that the integrand does not have finite singular points, since a point is a removable singularity.
But to calculate the integral, you have to compose a contour integral from another function (to ensure that the integral goes to zero at
closing semicircle in the limit of infinite radius): . Here the integrand has a pole singularity
at point .
Thus, another integration loop is required (see Fig. 4). It is different from Fig. 3a only by the fact that the singular point goes around a semicircle,
the radius of which is expected to tend to zero in the future.
. (3)
Let us immediately note that the integral over a great semicircle in the limit of its infinitely large radius tends to zero, and inside the contour
there are no singular points, so the entire integral along the contour is zero. Next, consider the first and third terms in (3):
.
Now let's write the integral over a small semicircle, taking into account that on it. We will also immediately take into account the smallness of the radius of the semicircle:
The terms tending to zero in the limit are not written out.
We collect the terms in (3) - except for the term related to the great semicircle.
As can be seen, the terms that go to infinity at annihilate each other. Directing and , we have.
Note. For example, the Dirichlet integral is calculated in a completely similar way (recall that it differs from what was just considered by the absence
squares in the numerator and denominator).Examples of calculating definite integrals using contour
integration in the complex plane (continued)Example 5. The integrand has countless singular points.
In many cases, the choice of contour is complicated by the fact that the integrand has an infinite number of singular points. In this case it may
it turns out that the sum of the residues will actually be close, the convergence of which will still have to be proven if we sum it up
it doesn’t work (and summing series is generally a separate, rather complicated task). As an example, let's calculate the integral.
It is clear that part of the contour is the real axis. The function has no special features on it. Let's discuss how to close the loop. You should not select a semicircle.
The fact is that the hyperbolic cosine has a family of simple zeros. Therefore, inside the contour closed by a semicircle
in the limit of an infinitely large radius, there will be infinitely many singular points. How else can you close the loop? Notice, that .
It follows that you can try to include in the integration contour a segment parallel to the real axis. The circuit will close with two
vertical segments, in the limit located infinitely far from the imaginary axis (see Fig. 5).
On vertical sections of the contour. The hyperbolic cosine grows exponentially with increasing argument (in absolute value), therefore
in the limit, the integrals over the vertical sections tend to zero.
So, in the limit.
On the other hand, inside the integration contour there are two singular points of the integrand. Deductions in them
,
.
Hence,
.Example 6. The integrand of the definite and contour integrals are different.
There is a very important case of calculating definite integrals using the contour integration method. Still integrand
the contour integral function either simply coincided with the integrand of a certain integral, or passed into it by separation
real or imaginary part. But things don't always turn out to be so simple. Let's calculate the integral.
In terms of choosing a circuit, there is no particular problem. Although the function under the integral has infinitely many simple poles, we already know
Based on the experience of the previous example, a rectangular contour is needed, since . The only difference from example 5 is that
that the pole of the integrand falls on the straight line, which needs to be bypassed. Therefore, we choose the one shown
in Fig. 6 circuit.
Consider the contour integral. We will not paint it on each section of the contour, limiting ourselves to horizontal
in sections. The integral along the real axis tends to the desired value in the limit. Let us write the integrals over the remaining sections:
.
In the limit, the first two integrals will give , then they will enter the contour integral in sum
with the desired one, which differs in sign. As a result, the desired definite integral will drop out of the contour integral. It means that
the integrand was chosen incorrectly. Let's consider another integral: . We leave the outline the same.
To begin with, let us again consider integrals over horizontal sections. The integral along the real axis will transform into .
This integral is equal to zero as the integral of an odd function within symmetric limits.
In the limit, the first two brackets will vanish, again forming integrals of odd functions
within symmetrical limits. But the last bracket, up to a factor, will give the required integral. It makes sense to continue the calculation.
Similar to example 5, the integrals over the vertical sections of the contour tend to zero at . It remains to find the integral
along a semicircle, where. As in example 4, we calculate the integral, taking into account the smallness of :
.
So, we have everything to write down the contour integral in the limit:
On the other hand, inside the integration contour there was a pole of the integrand function
