Page 517 - Handbook Of Integral Equations

Thus, in this setting, the first Sokhotski–Plemelj formula (a representation of an arbitrary function
in the form of the difference of boundary values of analytic functions) is an obvious consequence of
the decomposition of a Fourier integral into the right and the left integral. The second formula can
also be rewritten as follows:
∞ ∞
1 Y(τ) –1 1 Y(τ)
F{y(x) sign x} = dτ, F dτ = y(x) sign x. (13)
πi τ – u πi τ – u
–∞ –∞
10.4-3. The Analytic Continuation Theorem and the Generalized Liouville Theorem

Below is the analytic continuation theorem and the generalized Liouville theorem combined into a
single statement, which will be used in Chapters 10 and 11.
Let functions Y 1 (z) and Y 2 (z) be analytic in the upper and lower half-planes, respectively,
possibly except for a point z ∗ ≠ ∞, at which these functions have a pole. If Y 1 (z) and Y 2 (z) are
bounded at infinity, the principal parts of their expansions in a neighborhood of z ∗ have the form

c 1 c 2 c m P m–1 (z)
+ + ··· + ≡ ,
(z – z ∗ ) 2 (z – z ∗ ) m (z – z ∗ ) m
z – z ∗
and if the functions themselves coincide on the real axis, then these functions represent a single
rational function on the entire plane:

P m–1 (z)
Y(z)= c 0 + ,
(z – z ∗ ) m
where c 0 is a constant. The pole z ∗ can belong either to the open half-planes or to the real axis.
Let us also give a more general version of the above statement.
If functions Y 1 (z) and Y 2 (z) are analytic in the upper and lower half-planes, respectively, possibly except for finitely
many points z 0 = ∞, z k (k =1, ... , n), at which these functions can have poles, if the principal parts of the expansions of
these functions in a neighborhood of a pole have the form
0 0 2 0 m 0
c 1 z + c 2 z + ··· + c m 0 z ≡ P 0 (z) at the point z 0 ,
k
c k c k c m k P m k –1 (z)
1
+ 2 + ··· + ≡ at the points z k ,
z – z k (z – z k ) 2 (z – z k ) m k (z – z k ) m k
and if the functions themselves coincide on the real axis, then these functions represent a single rational function on the entire
plane:
n
P m k –1 (z)
Y(z)= C + P 0 (z)+
(z – z k ) m k
k=1
where C is a constant. The poles z k can belong either to the open half-planes or to the real axis.
10.4-4. The Riemann Boundary Value Problem
The solution of the Riemann problem in this section differs from the traditional one, because it is
expressed not by means of integrals of the Cauchy type (see Subsection 12.3-7) but by means of
Fourier integrals. To solve equations of convolution type under consideration, the Fourier integral
technique is more convenient.
By the index of a continuous complex-valued nonvanishing function M(u)(M(u)= M 1 (u)+
iM 2 (u), –∞ < u < ∞, M(–∞)= M(∞)) we mean the variation of the argument of this function
on the real axis expressed in the number of full rotations:

∞
1
1 ∞ 1 ∞
Ind M(u)= arg M(u) = ln M(u) = d ln M(u).
2π –∞ 2πi –∞ 2πi
–∞
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 499