Laplace transforms 673Im(s)Re(s)Poles of f (s)~
Fig. D.1Bromwich contour, which contains all of the poles off ̃(s). Laplace inversion em-
ploys such a contour extended to infinity in all directions.
f(t) =1
2 πi∮
Bds
αest
(s−α)(s+α). (D.16)
The integrand in eqn. (D.16) has two first-order poles ats=αands=−α. Thus, we
need to choose the leading edge of the contour to lie to the right of the points=αon
the reals-axis. Once we do this and apply the residue theorem to each of the poles,
we obtain
f(t) =
α
2 πi2 πi[
eαt
2 α−
e−αt
2 α]
=
1
2
[
eαt−e−αt]
= sinh(αt). (D.17)In Section 15.2, we showed how to solve ordinary linear differential equations using
Laplace transforms. Now let us use the result obtained in eqn. (D.17) in another
example that involves the application of the Laplace transform to the solution of an
integral equation. Consider the integral equation
x(t) =t+a^2∫t0dτ(t−τ)x(τ), (D.18)which we must solve forx(t) in terms of the constanta. We can make use of the fact
that the integral appearing in eqn. (D.18) is a convolution, and its Laplace transform
will simply be ̃x(s)/s^2 , since the Laplace transform off(t) =tisf ̃(s) = 1/s^2. Thus,
taking the Laplace transform of both sides of eqn. (D.18), we obtain the algebraic
equation