Mathematical Methods for Physics and Engineering : A Comprehensive Guide

25.5 INVERSE LAPLACE TRANSFORM

a b

1

f(x)

x

Figure 25.8 The result of the Laplace inversion of ̄f(s)=s−^1 (e−as−e−bs)with

b>a.

Now, despite appearances to the contrary, the integrand has no poles, as may be confirmed
by expanding the exponentials as Taylor series abouts= 0. Depending on the value ofx,
several cases arise.
(i) Forx<aboth exponentials in the integrand will tend to zero as Res→∞. Thus
we may closeLwith a circular arc Γ in therighthalf-plane (λcan be as small as desired),
and we observe thats×integrand tends to zero everywhere on Γ asR→∞.Withno
poles enclosed and no contribution from Γ, the integral alongLmust also be zero. Thus

f(x)=0 forx<a. (25.29)

(ii) Forx>bthe exponentials in the integrand will tend to zero as Res→−∞,and
so we may closeLin the left half-plane, as in figure 25.7(a). Again the integral around Γ
vanishes for infiniteR, and so, by the residue theorem,

f(x)=0 forx>b. (25.30)

(iii) Fora<x<bthe two parts of the integrand behave in different ways and have to
be treated separately:

I 1 −I 2 ≡

1

2 πi

∫

L

e(x−a)s

s

ds−

1

2 πi

∫

L

e(x−b)s

s

ds.

The integrand ofI 1 then vanishes in the far left-hand half-plane, but does now have a
(simple) pole ats=0.ClosingLin the left half-plane, and using the residue theorem, we
obtain

I 1 = residue ats=0ofs−^1 e(x−a)s=1. (25.31)

The integrand ofI 2 , however, vanishes in the far right-hand half-plane (and also has
a simple pole ats= 0) and is evaluated by a circular-arc completion in that half-plane.
Such a contour encloses no poles and leads toI 2 =0.
Thus, collecting together results (25.29)–(25.31) we obtain

f(x)=







0forx<a,

1fora<x<b,

0forx>b,

as shown in figure 25.8.