Section 6

Cauchy's Theorem. The Cauchy-Type Integral.

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Problem 6.1

Prove that if the function f(z)f(z) is regular in the disk ∣z−a∣<R\left|z-a\right| < R and satisfies the condition ∣f(z)∣⩽M(∣z−a∣<R)\left|f(z)\right| \leqslant M \quad (\left|z-a\right| < R), then for any two points z1z_{1} and z2z_{2} of this disk the inequality holds

∣∫z1z2f(ζ)dζ∣⩽M∣z2−z1∣ \left|\int _{z_{1}}^{z_{2}} f(\zeta ) d \zeta \right| \leqslant M\left|z_{2}-z_{1}\right|
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Problem 6.2

Prove that the assertion of Problem 1 remains valid if the domain of regularity of the function f(z)f(z) is not necessarily a disk, but an arbitrary convex domain in C\mathbb {C}.

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Remark.
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A domain is called convex if, together with each pair of points belonging to this domain, it also contains the straight-line segment joining these points.

Problem 6.3

Let the function f(z)f(z) be regular in a convex domain D⊂CD \subset \mathbb {C} and satisfy the condition Re(f(z))⩾M>0(z∈D)\mathrm{Re}\left(f(z)\right) \geqslant M > 0 \quad (z \in D). Prove that for any two points z1z_{1} and z2z_{2} of this domain

∣∫z1z2f(ζ)dζ∣⩾M∣z2−z1∣ \left|\int _{z_{1}}^{z_{2}} f(\zeta ) d \zeta \right| \geqslant M\left|z_{2}-z_{1}\right|
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Problem 6.4

Prove that the assertion of Problem 3 remains valid if the condition

Re(f(z))⩾M>0(z∈D) \mathrm{Re}\left(f(z)\right) \geqslant M > 0 \quad (z \in D)

is replaced by the condition Re({eiφf(z)})⩾M\mathrm{Re}\left(\left\{ e^{i \varphi } f(z)\right\} \right) \geqslant M (the real number φ\varphi does not depend on the point zz).

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Problem 6.5

Let the function f(z)f(z) be regular in a simply connected domain D⊂CD \subset \mathbb {C}. Prove that the function

Φ(z)=∫z0zf(ζ)dζ+const(z0∈D,z∈D) \Phi (z) = \int _{z_{0}}^{z} f(\zeta ) d \zeta +\mathrm{const} \quad \left(z_{0} \in D, z \in D\right)

is an antiderivative of the function f(z)f(z).

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Problem 6.6

Let the functions f(z)f(z) and g(z)g(z) be regular in a simply connected domain D⊂CD \subset \mathbb {C}, and let F(z)F(z) and G(z)G(z) be, respectively, antiderivatives of these functions. Prove the integration-by-parts formula:

∫abF(z)g(z)dz=F(b)G(b)−F(a)G(a)−∫abf(z)G(z)dz \int _{a}^{b} F(z) g(z) d z = F(b) G(b)-F(a) G(a)-\int _{a}^{b} f(z) G(z) d z
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Problem 6.7

Find the antiderivatives of the functions:

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(1)

eaze^{a z};

(2)

ch⁡az\operatorname {ch} a z;

(3)

sh⁡az\operatorname {sh} a z;

(4)

cos⁡az\cos a z;

(5)

sin⁡az\sin a z;

(6)

eazcos⁡bze^{a z} \cos b z;

(7)

zeazz e^{a z};

(8)

z2ch⁡azz^{2} \operatorname {ch} a z;

(9)

zcos⁡azz \cos a z.

Problem 6.8

Let the function f(z)f(z) be regular in the annulus r<∣z−a∣<Rr < \left|z-a\right| < R. Prove that the integral

∮∣z−a∣=ρf(z)dz,r<ρ<R \oint _{\left|z-a\right| = \rho } f(z) d z, \quad r < \rho < R

does not depend on the number ρ\rho (the circle is traversed counterclockwise).

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Problem 6.9

Let the function f(z)f(z) be regular in the annulus r<∣z∣<Rr < \left|z\right| < R, and let a simple piecewise-smooth curve γ\gamma bound a domain that contains the disk ∣z∣⩽r\left|z\right| \leqslant r and lies in the disk ∣z∣<R\left|z\right| < R, so that while traversing the curve γ\gamma the domain remains on the left. Prove that the integral ∫γf(z)dz\int_{\gamma } f(z) d z does not depend on the choice of the curve γ\gamma (satisfying the conditions stated above).

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Problem 6.10

Let the function f(z)f(z) be regular in an arbitrary domain DD of the complex plane. Prove that a necessary and sufficient condition for the existence of an antiderivative of the function f(z)f(z) in the domain DD is that the integral of the function f(z)f(z) over any simple closed polygonal line lying in the domain DD be equal to zero.

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Problem 6.11

Prove that the following functions have no antiderivatives in the domains indicated in parentheses:

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(1)

1z(0<∣z∣<∞)\frac{1}{z} \quad (0 < \left|z\right| < \infty );

(2)

1z−1z−1(0<∣z∣<1)\frac{1}{z}-\frac{1}{z-1} \quad (0 < \left|z\right| < 1);

(3)

z1+z2(0<∣z∣<∞)\frac{z}{1+z^{2}} \quad (0 < \left|z\right| < \infty );

(4)

1z(1−z2)(0<∣z∣<1)\frac{1}{z\left(1-z^{2}\right)} \quad (0 < \left|z\right| < 1).

Problem 6.12

Let the function f(z)f(z) be regular in a bounded doubly connected domain DD, enclosed between two closed piecewise-smooth curves γ1\gamma_{1} and γ2\gamma_{2}, and continuous up to its boundary. Prove that the function f(z)f(z) has an antiderivative in the domain DD if and only if

∮γ1f(z)dz=0. \oint _{\gamma _{1}} f(z) d z = 0.
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Problem 6.13

Let the function f(z)f(z) be regular in a domain, simply connected in C‾\overline{\mathbb {C}}, containing the point z=∞z = \infty. Prove that the function f(z)f(z) has an antiderivative in this domain if and only if lim⁡z→∞zf(z)=0\lim_{z \rightarrow \infty } z f(z) = 0.

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Problem 6.14

Let the function f(z)f(z) be regular in a simply connected domain D⊂C‾D \subset \overline{\mathbb {C}} containing the point z=∞z = \infty. Denote by D′D^{\prime } the domain DD with the point z=∞z = \infty removed. Prove that the function f(z)f(z) has an antiderivative in the domain D′D^{\prime } if and only if

lim⁡z→∞z(f(z)−f(∞))=0 \lim _{z \rightarrow \infty } z(f(z)-f(\infty )) = 0
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Problem 6.15

Let the functions f(z)f(z) and g(z)g(z) be regular in a domain D⊂C‾D \subset \overline{\mathbb {C}} containing the point z=∞z = \infty, and have antiderivatives in the domain D′D^{\prime } (the domain DD with the point z=∞z = \infty removed). Prove that the functions f(z)+g(z),f(z)g(z),P(f(z)),ef(z)(P(w)f(z)+g(z), f(z) g(z), P(f(z)), e^{f(z)} (P(w) an arbitrary polynomial) are also regular in the domain DD and have an antiderivative in the domain D′D^{\prime }.

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Problem 6.16

Let the function f(z)f(z) be regular in a bounded mm-connected domain DD whose boundary consists of closed piecewise-smooth curves Γ1,Γ2,…,Γm\Gamma_{1}, \Gamma_{2}, \ldots , \Gamma_{m}. Prove that for the existence of an antiderivative of the function f(z)f(z) in the domain DD it is necessary and sufficient that

∫Γkf(z)dz=0(k=1,2,…,m−1) \int _{\Gamma _{k}} f(z) d z = 0 \quad (k = 1,2, \ldots , m-1)
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Problem 6.17

Let the function f(z)f(z) be regular in the strip −a<Im(z)<a-a < \mathrm{Im}\left(z\right) < a and satisfy the condition

f(z)→0(z→∞,−a<Im(z)<a). f(z) \rightarrow 0 \quad (z \rightarrow \infty ,-a < \mathrm{Im}\left(z\right) < a).

Prove that if the integral ∫−∞∞f(x)dx\int_{-\infty }^{\infty } f(x) d x converges, then for any α\alpha in the interval (−a,a)(-a, a) the integral ∫iα−∞iα+∞f(z)dz\int_{i \alpha -\infty }^{i \alpha +\infty } f(z) d z also converges and does not depend on α\alpha.

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Problem 6.18

Let the function f(z)f(z) be regular in the angle −a<arg⁡z<a-a < \arg z < a and satisfy the conditions

zf(z)→0(z→0,∣arg⁡z∣<a)zf(z)→0(z→∞,∣arg⁡z∣<a) \begin{array}{ll} z f(z) \rightarrow 0 & (z \rightarrow 0,\left|\arg z\right| < a) \\ z f(z) \rightarrow 0 & (z \rightarrow \infty ,\left|\arg z\right| < a) \end{array}

Prove that if the integral ∫0∞f(x)dx\int_{0}^{\infty } f(x) d x converges, then for any α\alpha in the interval (−a,a)(-a, a) the integral ∫arg⁡z=αf(z)dz\underset {\arg z = \alpha }{\int } f(z) d z also converges and does not depend on α\alpha.

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Problem 6.19

It is known from analysis that ∫−∞∞e−x2dx=π\int_{-\infty }^{\infty } e^{-x^{2}} d x = \sqrt{\pi }. Compute the integral

∫−∞∞e−x2cos⁡αxdx,−∞<α<∞ \int _{-\infty }^{\infty } e^{-x^{2}} \cos \alpha x d x, \quad -\infty < \alpha < \infty
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Problem 6.20

Compute the integral ∫0+∞e−ix2dx\int_{0}^{+\infty } e^{-i x^{2}} d x.

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Problem 6.21

Using Cauchy's integral formula, compute the integrals (all circles are traversed counterclockwise):

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(1)

∮∣z+i∣=3sin⁡zdzz+i\oint_{\left|z+i\right| = 3} \sin z \frac{d z}{z+i};

(2)

∮∣z∣=2dzz2+1\oint_{\left|z\right| = 2} \frac{d z}{z^{2}+1};

(3)

∮∣z∣=2ezz2−1dz\oint_{\left|z\right| = 2} \frac{e^{z}}{z^{2}-1} d z;

(4)

∮∣z∣=4cos⁡zz2−π2dz\oint_{\left|z\right| = 4} \frac{\cos z}{z^{2}-\pi^{2}} d z;

(5)

∮∣z+1∣=1dz(1+z)(z−1)3\oint_{\left|z+1\right| = 1} \frac{d z}{(1+z)(z-1)^{3}};

(6)

∮∣z−i∣=1cos⁡z(z−i)3dz\oint_{\left|z-i\right| = 1} \frac{\cos z}{(z-i)^{3}} d z;

(7)

∮∂Dezdzz(1−z)3\oint_{\partial D} \frac{e^{z} d z}{z(1-z)^{3}}, where DD:

  1. ∣z∣<1/2\left|z\right| < 1 / 2;

  2. ∣z∣<3/2\left|z\right| < 3 / 2;

  3. ∣z−1∣<1/2\left|z-1\right| < 1 / 2;

(8)

∮∣z∣=rdz(z−a)n(z−b)(∣a∣<r<∣b∣,n=1,2,…)\oint_{\left|z\right| = r} \frac{d z}{(z-a)^{n}(z-b)} \quad (\left|a\right| < r < \left|b\right|, n = 1,2, \ldots ).

Problem 6.22

Let the functions f(z)f(z) and g(z)g(z) be regular in the disk ∣z∣<1\left|z\right| < 1 and continuous in the closed disk ∣z∣⩽1\left|z\right| \leqslant 1. Prove that

12πi∮∣ζ∣=1[f(ζ)ζ−z+zg(ζ)zζ−1]dζ={f(z),∣z∣<1,g(1z),∣z∣>1. \frac{1}{2 \pi i} \oint _{\left|\zeta \right| = 1}\left[\frac{f(\zeta )}{\zeta -z}+\frac{z g(\zeta )}{z \zeta -1}\right] d \zeta = \begin{cases} f(z), & \left|z\right| < 1, \\ g\left(\frac{1}{z}\right), & \left|z\right| > 1. \end{cases}
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Problem 6.23

Let the function f(z)f(z) be regular in a domain D⊂C‾D \subset \overline{\mathbb {C}} containing the point z=∞z = \infty, and continuous up to its boundary. Prove that in this case Cauchy's integral formula takes the form

12πi∫∂Df(ζ)ζ−zdζ={f(z)−f(∞),z∈D,−f(∞),z∉Dˉ. \frac{1}{2 \pi i} \int _{\partial D} \frac{f(\zeta )}{\zeta -z} d \zeta = \begin{cases} f(z)-f(\infty ), & z \in D, \\ -f(\infty ), & z \notin \bar{D}. \end{cases}

and the formula for the derivatives of f(z)f(z) retains its previous form.

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Problem 6.24

Let the function f(z)f(z) be regular in the disk ∣z−a∣<R\left|z-a\right| < R and continuous in the closed disk ∣z−a∣⩽R\left|z-a\right| \leqslant R. Prove the formula

12π∫02πf(a+Reiφ)dφ=f(a) \frac{1}{2 \pi } \int _{0}^{2 \pi } f\left(a+R e^{i \varphi }\right) d \varphi = f(a)

known as the mean value theorem.

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Problem 6.25

Let the function f(z)f(z) be regular in the disk ∣z∣<R\left|z\right| < R and continuous in the closed disk ∣z∣⩽R\left|z\right| \leqslant R. Compute the integral

∬r<∣z∣<Rf(z)dxdy \iint _{r < \left|z\right| < R} f(z) d x d y
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Problem 6.26

Prove that a function, regular in some domain and not identically constant, cannot attain its greatest absolute value at an interior point of this domain. (The maximum modulus principle.)

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Problem 6.27

Let the function u(x,y)u(x, y) be harmonic in the disk ∣z−a∣⩽R\left|z-a\right| \leqslant R. Prove that

12π∫02πu(α+Rcos⁡φ,β+Rsin⁡φ)dφ=u(α,β)(a=α+iβ) \frac{1}{2 \pi } \int _{0}^{2 \pi } u(\alpha +R \cos \varphi , \beta +R \sin \varphi ) d \varphi = u(\alpha , \beta ) \quad (a = \alpha +i \beta )
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Problem 6.28

Prove that a function, harmonic in some domain and not identically constant, can attain neither its greatest nor its least value at an interior point of this domain. (The maximum principle for harmonic functions.)

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Problem 6.29

Let the function f(z)f(z) be regular in a bounded domain D⊂CD \subset \mathbb {C} and continuous up to its boundary ∂D\partial D, consisting of a finite number of closed piecewise-smooth curves. Prove the inequality

∣f(n)(z)n!∣⩽M⋅L2πρn+1(n=1,2,…) \left|\frac{f^{(n)}(z)}{n!}\right| \leqslant \frac{M \cdot L}{2 \pi \rho ^{n+1}} \quad (n = 1,2, \ldots )

where M=max⁡z∈∂D∣f(z)∣,ρM = \max_{z \in \partial D}\left|f(z)\right|, \rho is the distance from the point zz to the boundary of the domain DD, and LL is the total length of the boundary of the domain DD.

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Problem 6.30

Let the function f(z)f(z) be regular in the disk ∣z∣<R\left|z\right| < R and continuous in the closed disk ∣z∣⩽R\left|z\right| \leqslant R. Prove the inequality

∣f(n)(z)n!∣⩽MR(R−∣z∣)n+1(n=1,2,…),(∣z∣<R), \left|\frac{f^{(n)}(z)}{n!}\right| \leqslant \frac{M R}{(R-\left|z\right|)^{n+1}} \quad (n = 1,2, \ldots ), \quad (\left|z\right| < R),

where M=max⁡∣z∣=R∣f(z)∣M = \max_{\left|z\right| = R}\left|f(z)\right|.

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Problem 6.31

Let the function f(z)f(z) be regular in the disk ∣z∣<R\left|z\right| < R and continuous in the closed disk ∣z∣⩽R\left|z\right| \leqslant R. Prove the inequality

∣f(n)(z)n!∣⩽MR2π∫−ππ(R2+∣z∣2−2R∣z∣cos⁡φ)−(n+1)/2dφ(∣z∣<R) \left|\frac{f^{(n)}(z)}{n!}\right| \leqslant \frac{M R}{2 \pi } \int _{-\pi }^{\pi }\left(R^{2}+\left|z\right|^{2}-2 R\left|z\right| \cos \varphi \right)^{-(n+1) / 2} d \varphi \quad (\left|z\right| < R)

where M=max⁡∣z∣=R∣f(z)∣M = \max_{\left|z\right| = R}\left|f(z)\right|, and show that for n=1n = 1 this inequality can be written in the form

∣f′(z)∣⩽MRR2−∣z∣2(∣z∣<R) \left|f^{\prime }(z)\right| \leqslant \frac{M R}{R^{2}-\left|z\right|^{2}} \quad (\left|z\right| < R)
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Problem 6.32

Let the function f(z)f(z) be regular in the entire plane and satisfy the condition ∣f(z)∣⩽M\left|f(z)\right| \leqslant M for all zz. Prove that f(z)f(z) is identically constant. (Liouville's theorem.)

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Problem 6.33

Let the function f(z)f(z) be regular in the entire plane and satisfy the inequality

∣f(z)∣⩽M(1+∣z∣)p,p>0 \left|f(z)\right| \leqslant M(1+\left|z\right|)^{p}, \quad p > 0

Prove that f(z)f(z) is a polynomial of degree not exceeding pp.

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Problem 6.34

At all points not belonging to the curve of integration, compute the values of the following Cauchy-type integrals:

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(1)

12πi∫∣ζ∣=1dζζ(ζ−2)(ζ−z)\frac{1}{2 \pi i} \int_{\left|\zeta \right| = 1} \frac{d \zeta }{\zeta (\zeta -2)(\zeta -z)};

(2)

12πi∫01dζζ−z\frac{1}{2 \pi i} \int_{0}^{1} \frac{d \zeta }{\zeta -z};

(3)

12πi∫∂Ddζζ(2ζ2+5ζ+2)(ζ−z),D={z:∣z∣<1,∣z+23∣>13}\frac{1}{2 \pi i} \int_{\partial D} \frac{d \zeta }{\zeta \left(2 \zeta^{2}+5 \zeta +2\right)(\zeta -z)}, \quad D = \left\{ z:\left|z\right| < 1,\left|z+\frac{2}{3}\right| > \frac{1}{3}\right\}.

Problem 6.35

Let the function f(z)f(z) be regular in the annulus r<∣z−a∣<Rr < \left|z-a\right| < R. Prove that the Cauchy-type integral

12πi∮∣ζ−a∣=ρf(ζ)dζζ−z,r<ρ<R \frac{1}{2 \pi i} \oint _{\left|\zeta -a\right| = \rho } f(\zeta ) \frac{d \zeta }{\zeta -z}, \quad r < \rho < R

does not depend on the number ρ\rho for ∣z∣<r\left|z\right| < r (or for ∣z∣>R\left|z\right| > R).

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Problem 6.36

Let the function f(ζ)f(\zeta ) be regular in the annulus r<∣ζ∣<Rr < \left|\zeta \right| < R and continuous in the closed annulus r⩽∣ζ∣⩽Rr \leqslant \left|\zeta \right| \leqslant R. Denote

f1(z)=12πi∮∣ζ∣=Rf(ζ)dζζ−z,(∣z∣<R)f2(z)=−12πi∮∣ζ∣=rf(ζ)dζζ−z,(∣z∣>r) \begin{array}{ll} f_{1}(z) = \frac{1}{2 \pi i} \oint _{\left|\zeta \right| = R} f(\zeta ) \frac{d \zeta }{\zeta -z}, & (\left|z\right| < R) \\ f_{2}(z) = -\frac{1}{2 \pi i} \oint _{\left|\zeta \right| = r} f(\zeta ) \frac{d \zeta }{\zeta -z}, & (\left|z\right| > r) \end{array}

(the circles are traversed counterclockwise). Prove that for r<∣z∣<Rr < \left|z\right| < R the equality f(z)=f1(z)+f2(z)f(z) = f_{1}(z)+f_{2}(z) holds.

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Problem 6.37

Let the function f(z)f(z) be regular in the annulus r<∣z−a∣<Rr < \left|z-a\right| < R. Prove that it can be represented in the form f(z)=f1(z)+f2(z)f(z) = f_{1}(z)+f_{2}(z), where the function f1(z)f_{1}(z) is regular in the disk ∣z−a∣<R\left|z-a\right| < R, and the function f2(z)f_{2}(z) is regular for ∣z−a∣>r\left|z-a\right| > r and f2(∞)=0f_{2}(\infty ) = 0.

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Problem 6.38

Prove that the representation of the function f(z)f(z) from the previous problem is unique.

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Problem 6.39

Let a simple closed curve γ1\gamma_{1} bound a domain D1⊂CD_{1} \subset \mathbb {C}, and let a simple closed curve γ2\gamma_{2}, lying in the domain D1D_{1}, bound a domain D2⊂D1D_{2} \subset D_{1}. Prove that any function f(z)f(z), regular in the ring-shaped domain D=D1\Dˉ2D = D_{1} \backslash \bar{D}_{2}, can be represented in the form f(z)=f1(z)+f2(z)f(z) = f_{1}(z)+f_{2}(z), where the function f1(z)f_{1}(z) is regular in the domain D1D_{1}, and the function f2(z)f_{2}(z) is regular outside the domain D2D_{2}. Prove also that such a representation is unique if the condition f2(∞)=0f_{2}(\infty ) = 0 is imposed.

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Problem 6.40

Let γ\gamma be a simple closed curve bounding a domain D⊂CD \subset \mathbb {C}, and let the function φ(z)\varphi (z) be regular in some domain containing the curve γ\gamma. Prove that for the existence of a function f(z)f(z), regular in the closure of the domain DD and coinciding with the function φ(z)\varphi (z) on the curve γ\gamma, it is necessary and sufficient that

∮γφ(ζ)ζ−zdζ≡0(z∉Dˉ). \oint _{\gamma } \frac{\varphi (\zeta )}{\zeta -z} d \zeta \equiv 0 \quad (z \notin \bar{D}).
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Problem 6.41

Let γ\gamma be a simple closed piecewise-smooth curve bounding a bounded domain DD, and let the function φ(t)\varphi (t) be continuous on the curve γ\gamma. Prove that for the equality

∮γφ(t)t−zdt≡0(z∉Dˉ) \oint _{\gamma } \frac{\varphi (t)}{t-z} d t \equiv 0 \quad (z \notin \bar{D})

to hold it is necessary and sufficient that

∮γtnφ(t)dt=0(n=0,1,2,…) \oint _{\gamma } t^{n} \varphi (t) d t = 0 \quad (n = 0,1,2, \ldots )

and for the equality

∮γφ(t)t−zdt≡0(z∈D) \oint _{\gamma } \frac{\varphi (t)}{t-z} d t \equiv 0 \quad (z \in D)

to hold it is necessary and sufficient that

∮γtnφ(t)dt=0(n=−1,−2,…) \oint _{\gamma } t^{n} \varphi (t) d t = 0 \quad (n = -1,-2, \ldots )
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Problem 6.42

Let the function f(z)f(z) be regular in the strip −a<Im(z)<a-a < \mathrm{Im}\left(z\right) < a and satisfy the conditions

f(z)z→0(z→∞,∣Im(z)∣<a)∫−∞∞∣f(x)∣1+∣x∣dx<∞ \begin{aligned} \frac{f(z)}{z} \rightarrow & 0 \quad (z \rightarrow \infty ,\left|\mathrm{Im}\left(z\right)\right| < a) \\ & \int _{-\infty }^{\infty } \frac{\left|f(x)\right|}{1+\left|x\right|} d x < \infty \end{aligned}

Prove that the function f(z)f(z) can be represented in the form f1(z)+f2(z)f_{1}(z)+f_{2}(z), where the function f1(z)f_{1}(z) is regular in the half-plane Im(z)>−a\mathrm{Im}\left(z\right) > -a and satisfies the condition

f1(z)z→0(z→∞,Im(z)⩾−a+ε) \frac{f_{1}(z)}{z} \rightarrow 0 \quad (z \rightarrow \infty , \mathrm{Im}\left(z\right) \geqslant -a+\varepsilon )

(ε>0\varepsilon > 0 arbitrary), and the function f2(z)f_{2}(z) is regular in the half-plane Im(z)<a\mathrm{Im}\left(z\right) < a and satisfies the condition

f2(z)z→0(z→∞,Im(z)⩽a−ε) \frac{f_{2}(z)}{z} \rightarrow 0 \quad (z \rightarrow \infty , \mathrm{Im}\left(z\right) \leqslant a-\varepsilon )

(ε>0\varepsilon > 0 arbitrary). Prove also that such a representation is unique.

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Problem 6.43

Let the function f(ζ)f(\zeta ) be regular in the angle

−a<arg⁡ζ<a,0<a<π, -a < \arg \zeta < a, \quad 0 < a < \pi ,

continuous up to its boundary and satisfy the conditions

f(z)→0(z→∞,∣arg⁡z∣⩽a),∫0∞∣f(x)∣dx1+x<∞ f(z) \rightarrow 0 \quad (z \rightarrow \infty ,\left|\arg z\right| \leqslant a), \quad \int _{0}^{\infty }\left|f(x)\right| \frac{d x}{1+x} < \infty

Denote

f1(z)=∫arg⁡ζ=−af(ζ)dζζ−z,f2(z)=∫arg⁡ζ=af(ζ)dζζ−z f_{1}(z) = \int _{\arg \zeta = -a} f(\zeta ) \frac{d \zeta }{\zeta -z}, \quad f_{2}(z) = \int _{\arg \zeta = a} f(\zeta ) \frac{d \zeta }{\zeta -z}

Prove that:

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(1)

the function f1(z)f_{1}(z) is regular in the entire zz-plane with a cut along the ray arg⁡z=−a\arg z = -a, and the function f2(z)f_{2}(z) is regular in the entire zz-plane with a cut along the ray arg⁡z=a\arg z = a;

(2)

outside the angle ∣arg⁡z∣⩽a\left|\arg z\right| \leqslant a the equality f1=f2f_{1} = f_{2} holds;

(3)

inside the angle ∣arg⁡z∣<a\left|\arg z\right| < a the equality f1−f2=2πif_{1}-f_{2} = 2 \pi i holds.