Page 588 - Handbook Of Integral Equations

ˇ
ˇ
where M – (z) and M + (z) are analytic functions in the corresponding half-planes Im z > 0 and
Im z < 0 continuous up to the real axis. Moreover,
ˇ
ˇ
M + (z) ≠ 0 for Im z ≥ 0 and M – (z) ≠ 0 for Im z ≤ 0. (6)
Relation (5) implies the formula
ˇ
k = Ind M(u).
The factorization (5) is said to be canonical provided that k =0.
In what follows we consider only functions of the form
ˇ
ˇ
M(u)=1 – T (u) (7)
ˇ
such that M(±∞) = 1. We can also assume that
ˇ
ˇ
M + (±∞)= M – (±∞) = 1. (8)
Let us state the main results concerning the factorization problem.
A function (7) admits a canonical factorization if and only if the following two conditions hold:
ˇ
ˇ
M(u) ≠ 0, Ind M(u)=0. (9)
In this case, the canonical factorization is unique. Moreover, if conditions (9) hold, then there exists
a function M(x) in the class L such that

∞
ˇ
M(u)=exp M(x)e iux dx , (10)
–∞
0
∞
ˇ
ˇ
M + (u)=exp M(x)e iux dx , M – (u)=exp M(x)e iux dx . (11)
0 –∞
ˇ
ˇ
Hence, we have M(u) ∈ Q and M ± (u) ∈ Q ± . The factors in the canonical factorization are also
described by the following formulas:
ˇ

1 ∞ ln M(τ)
ˇ
ln M + (z)= dτ, Im z > 0, (12)
2πi τ – z
–∞
ˇ

1 ∞ ln M(τ)
ˇ
ln M – (z)= – dτ, Im z < 0. (13)
2πi τ – z
–∞
In the general case of the factorization, the following assertion holds. A function (7) admits a
factorization (5) if and only if the following condition is satisfied:
ˇ
M(u) ≠ 0, –∞ < u < ∞.
In this case, relation (5) can be rewritten in the form
–k
u – i
ˇ
ˇ
ˇ
M(u)= M – (u)M + (u), –∞ < u < ∞.
u + i
The last relation implies the canonical factorization for the function
–k
u – i
ˇ
ˇ
M 1 (u)= M(u).
u + i
ˇ
ˇ
Hence, the factors M ± (u) satisfy formulas (10)–(13) if we replace M(u) in these formulas
ˇ
by M 1 (u).
Now we return to Eq. (1) for which

∞
ˇ
K(u)= K(x)e iux dx. (14)
–∞
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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