INTEGRAL TRANSFORMS
Find the Laplace transform ofd^2 f/dt^2.Using the definition of the Laplace transform and integrating by parts we obtain
L[
d^2 f
dt^2]
=
∫∞
0d^2 f
dt^2e−stdt=
[
df
dte−st]∞
0+s∫∞
0df
dte−stdt=−
df
dt(0) +s[sf ̄(s)−f(0)], fors> 0 ,where (13.57) has been substituted for the integral. This can be written more neatly as
L[
d^2 f
dt^2]
=s^2 f ̄(s)−sf(0)−df
dt(0), fors> 0 .In general the Laplace transform of thenth derivative is given byL[
dnf
dtn]
=snf ̄−sn−^1 f(0)−sn−^2df
dt(0)−···−dn−^1 f
dtn−^1(0), fors> 0.
(13.58)We now turn to integration, which is much more straightforward. From thedefinition (13.53),
L[∫t0f(u)du]
=∫∞0dt e−st∫t0f(u)du=[
−1
se−st∫t0f(u)du]∞0+∫∞01
se−stf(t)dt.The first term on the RHS vanishes at both limits, and so
L[∫t0f(u)du]
=1
sL[f]. (13.59)13.2.2 Other properties of Laplace transformsFrom table 13.1 it will be apparent that multiplying a functionf(t)byeathas the
effect on its transform thatsis replaced bys−a. This is easily proved generally:
L[
eatf(t)]
=∫∞0f(t)eate−stdt=∫∞0f(t)e−(s−a)tdt=f ̄(s−a). (13.60)As it were, multiplyingf(t)byeatmoves the origin ofsby an amounta.