The Mathematics of Financial Modelingand Investment Management

4-PrincipCalculus Page 135 Friday, March 12, 2004 12:39 PM

Principles of Calculus 135

if the improper integral exists.

Laplace transforms “project” a function into a different function

space, that of their transforms. Laplace transforms exist only for func-

tions that are sufficiently smooth and decay to zero sufficiently rapidly

when x → ∞. The following conditions ensure the existence of the

Laplace transform:

■ f(x) is piecewise continuous.

■ f(x) is of exponential order as x → ∞, that is, there exist positive real

constants K, a, and T, such that fx()≤ Keax, for x > T.

Note that the above conditions are sufficient but not necessary for

Laplace transforms to exist. It can be demonstrated that, if they exist,

Laplace transforms are unique in the sense that if two functions have

the same Laplace transform they coincide pointwise. As a consequence,

the Laplace transforms are invertible in the sense that the original func-

tion can be fully recovered from its transform. In fact, it is possible to

define the inverse Laplace transform as the operator L–1(F(s)) such that

L–1[L(s)] = f(x)

The inverse Laplace transform can be represented as a Bromwich

integral, that is, an integral defined on a contour in the complex plane

that leaves all singularities of the transform to the left:

γ+ i∞

1

fX()= --------- ∫ esxLs()sd

2 πiγ– i∞

The following conditions ensure the existence of an inverse Laplace

transform:

lim Fs()= 0

s → ∞

limsF s()is finite

s → ∞

We will now list (without proof) some key properties of Laplace

transforms; both the one-sided and two-sided Laplace transforms have

similar properties. The Laplace transform is a linear operator in the

sense that, if f,g are real-valued functions that have Laplace transforms

and a,b are real-valued constants, then the following property holds: