672 Laplace transforms
From a numerical standpoint, a forward Laplace transform can beperform straight-
forwardly, while a numerical Laplace inversion is a highly ill-posed problem mathe-
matically (Epstein and Schotland, 2008).
To see how the form of the inverse Laplace transform arises, let usassume that
f(t) exhibits an exponential divergence eγt, and let us define a functiong(t) byf(t) =
eγtg(t). Since the Laplace transform restrictstto the interval [0,∞), we will assume
g(t) = 0 fort <0.g(t) is a well-behaved function, and therefore we can define it via
its Fourier transform
g(t) =1
√
2 π∫∞
−∞dωeiωtgˆ(ω). (D.10)The Fourier transform ˆg(ω) can, itself, be expressed in terms ofg(t) as
ˆg(ω) =1
√
2 π∫∞
0dt g(t)e−iωt. (D.11)Substituting eqn. (D.11) into eqn. (D.10) yields
g(t) =1
2 π∫∞
−∞dωeiωt∫∞
0du g(u)e−iωu. (D.12)Sinceg(t) = e−γtf(t), eqn. (D.12) implies that
f(t) =
eγt
2 π∫∞
−∞dωeiωt∫∞
0du f(u)e−γue−iωu. (D.13)If we lets=γ+iω, then eqn. (D.13) becomes
f(t) =1
2 πi∫γ+i∞γ−i∞ds est∫∞
0du f(u)e−us=
1
2 πi∫γ+i∞γ−i∞dsestf ̃(s). (D.14)Eqn. (D.14) defines the Laplace inversion problem as an integral over a complex vari-
ables, which can be performed using techniques of complex integration and the cal-
culus of residues. Fort >0, eqn. (D.14) can be rewritten as a contour integral using a
contour of the type shown in Fig. D.1 whose leading edge is parallel to the imaginary
saxis and is chosen far enough to the right to enclose all of the poles off ̃(s). The
contour is then closed in the left half of the complex plane. Denoting this contour as
Bfor theBromwich contour, we can rewrite eqn. (D.14) as
f(t) =1
2 πi∮
Bdsestf ̃(s). (D.15)As an example, consider inverting the Laplace transform forf ̃(s) =α/(s^2 −α^2 ),
which we know from eqn. (D.2) to be the Laplace transform off(t) = sinh(αt). We
set up the contour integration as