1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

12.5 • T HE LAPLACE TRANSFORM 545

Solution Using the integral definition for .C ( f (t)), we obtain

.C (f (t)) = f (t) e- •tdt = e-•^1 dt + e-•to dt = ==---- = -e

1

00

1c 1"" -st 1t=c l -cs

0 0 c st=O s

• EXAMPLE 12.8 Show that .C (eat) = -

1

-, where a is a real constant.

s - a

Solution We actually show that the integral defining[, ( e"t) equals the formula

F ( s) = -

1

• for values of s with Re ( s) > a and that the extension to other
  s-a
  values of s is inferred by our knowledge about the domain of a rational function.
  Using straightforward integration techniques gives

.C ( e"t) = {°" eate-•tdt = lim { R e<a-•)tdt

} 0 R-+oo} 0

e<a-s)R 1

lim +--.

R-+oo a - s s - a

Let s = <7 + ir be fixed, or where <7 > a. Then, as a-<7 is a negative real number,

we have lim e<a-s)R = 0 and use this expression in the preceding equation to

R-+oo

obtain the desired conclusion.

We can use the property of linearity to find new Laplace transforms from

known transforms.

• EXAMPLE 12 .9 Show that .C (sinh at) = 2 a 2.
  s - a

Solution Because sinhat =~eat - !e-at, we obtain

.C (sinhat) = !.c (e"t) - ~.C (e-"t) =! 1 - ~- 1 - = a.

2 2 2 s - a 2 s + a s^2 - a2

Integration by parts is also helpful in finding new Laplace transforms.