Mathematical Methods for Physics and Engineering : A Comprehensive Guide

APPLICATIONS OF COMPLEX VARIABLES

Ideally, we would like the contribution to the integral from the circular arc Γ to

tend to zero as its radiusR→∞. Using a modified version of Jordan’s lemma,

it may be shown that this is indeed the case if there exist constantsM>0and

α>0 such that on Γ

| ̄f(s)|≤

M

Rα

.

Moreover, this condition always holds when ̄f(s)hastheform

̄f(s)=P(s)

Q(s)

,

whereP(s)andQ(s) are polynomials and the degree ofQ(s) is greater than that

ofP(s).

When the contribution from the part-circle Γ tends to zero asR→∞,we

have from the residue theorem that the inverse Laplace transform (25.25) is given

simply by

f(t)=

∑(

residues of ̄f(s)esxat all poles

)

. (25.27)

Find the functionf(x)whose Laplace transform is

̄f(s)= s

s^2 −k^2

,

wherekis a constant.

It is clear that ̄f(s) is of the form required for the integral over the circular arc Γ to tend
to zero asR→∞, and so we may use the result (25.27) directly. Now

̄f(s)esx= se

sx

(s−k)(s+k)

,

and thus has simple poles ats=kands=−k. Using (24.57) the residues at each pole can
be easily calculated as

R(k)=

kekx

2 k

and R(−k)=

ke−kx

2 k

.

Thus the inverse Laplace transform is given by

f(x)=^12

(

ekx+e−kx

)

=coshkx.

This result may be checked by computing the forward transform of coshkx.

Sometimes a little more care is required when deciding in which half-plane to

close the contourC.

Find the functionf(x)whose Laplace transform is

̄f(s)=^1

s

(e−as−e−bs),

whereaandbare fixed and positive, withb>a.

From (25.25) we have the integral

f(x)=

1

2 πi

∫λ+i∞

λ−i∞

e(x−a)s−e(x−b)s

s

ds. (25.28)