Page 448 - Handbook Of Integral Equations

7.1-3. Integral Transforms

Integral transforms have the form

b
˜
f(λ)= ϕ(x, λ)f(x) dx.
a
˜
The function f(λ) is called the transform of the function f(x) and ϕ(x, λ) is called the kernel of
˜
the integral transform. The function f(x) is called the inverse transform of f(λ). The limits of
integration a and b are real numbers (usually, a =0, b = ∞ or a = –∞, b = ∞).
In Subsections 7.2–7.6, the most popular (Laplace, Mellin, Fourier, etc.) integral transforms,
applied in this book to the solution of specific integral equations, are described. These subsections
also describe the corresponding inversion formulas, which have the form

˜
f(x)= ψ(x, λ)f(λ) dλ
L
˜
and make it possible to recover f(x)if f(λ) is given. The integration path L can lie either on the
real axis or in the complex plane.
Integral transforms are used in the solution of various differential and integral equations. Figure 1
outlines the overall scheme of solving some special classes of linear integral equations by means
of integral transforms (by applying appropriate integral transforms to this sort of integral equations,
˜
one obtains first-order linear algebraic equations for f(λ)).
In many cases, to calculate definite integrals, in particular, to find the inverse Laplace, Mellin,
and Fourier transforms, methods of the theory of functions of a complex variable can be applied,
including the residue theorem and the Jordan lemma, which are presented below in Subsections 7.1-4
and 7.1-5.

7.1-4. Residues. Calculation Formulas
The residue of a function f(z) holomorphic in a deleted neighborhood of a point z = a (thus, a is an
isolated singularity of f) of the complex plane z is the number

1 2
res f(z)= f(z) dz, i = –1,
z=a 2πi
c ε
where c ε is a circle of sufficiently small radius ε described by the equation |z – a| = ε.
If the point z = a is a pole of order n* of the function f(z), then we have

1 d n–1 n
res f(z)= lim (z – a) f(z) .
z=a (n – 1)! z→a dx n–1
For a simple pole, which corresponds to n = 1, this implies

res f(z) = lim (z – a)f(z) .
z=a z→a
ϕ(z)
If f(z)= , where ϕ(a) ≠ 0 and ψ(z) has a simple zero at the point z = a, i.e., ψ(a)=0 and
ψ(z)
ψ (a) ≠ 0, then

z
ϕ(a)
res f(z)= .

z=a ψ (a)
z
* In a neighborhood of this point we have f(z) ≈ const (z – a) –n .

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 429