Jump problem for holomorphic functions

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Check your answers in Exercise 8 against the special case

E=[−1,1] E = [-1,1]
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Let P(z)P(z) be a polynomial. Prove that any function f(z)f(z), regular in the entire complex plane with a cut along the positive part of the real axis, continuous up to the boundary of this domain, and satisfying, for all real x>0x > 0, the condition

f(x+i0)−f(x−i0)=P(x), f(x+i 0)-f(x-i 0) = P(x),

has the form

f(z)=−P(z)2πih(z)+g(z), f(z) = -\frac{P(z)}{2 \pi i} h(z)+g(z),

where g(z)g(z) is a function regular in the entire complex plane, and h(z)h(z) is a regular branch of the multivalued function Ln⁡z\operatorname {Ln} z in the plane with a cut along the positive part of the real axis.

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Find the general form of a function f(z)f(z), regular in the entire complex plane with a cut along the positive part of the real axis, continuous up to the boundary of this domain except at the point z=0z = 0, and satisfying one of the conditions (x>0)(x > 0):

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(1)

f(x+i0)−f(x−i0)=(1+x)2ln⁡xf(x+i 0)-f(x-i 0) = (1+x)^{2} \ln x;

(2)

f(x+i0)−f(x−i0)=ln⁡2xf(x+i 0)-f(x-i 0) = \ln^{2} x;

(3)

f(x+i0)−f(x−i0)=sin⁡xf(x+i 0)-f(x-i 0) = \sin \sqrt{x};

(4)

f(x+i0)−f(x−i0)=sin⁡(αln⁡x)x−1f(x+i 0)-f(x-i 0) = \frac{\sin (\alpha \ln x)}{x-1};

(5)

f(x+i0)−f(x−i0)=x+1ln⁡2x+π2f(x+i 0)-f(x-i 0) = \frac{x+1}{\ln^{2} x+\pi^{2}};

(6)

f(x+i0)−f(x−i0)=xln⁡xf(x+i 0)-f(x-i 0) = \sqrt{x} \ln x;

(7)

f(x+i0)−f(x−i0)=ln⁡xx3f(x+i 0)-f(x-i 0) = \frac{\ln x}{\sqrt[3]{x}};

(8)

f(x+i0)−f(x−i0)=xln⁡xln⁡2x+π2(x+1)f(x+i 0)-f(x-i 0) = \frac{\sqrt{x} \ln x}{\ln^{2} x+\pi^{2}}(x+1).

References

[Rud87] Rudin W. Real and Complex Analysis. 3rd ed., McGraw-Hill, 1987.
[ШПК15] Шабунин М.И., Половинкин Е.С., Карлов М.И. Сборник задач по теории функций комплексного переменного. 4-е изд., БИНОМ. Лаборатория знаний, 2015.