The Laplace Transf arm Method 233improper integrals. Since Ix: lf(x)le-sx dx converges for s >a, the integral
Ix: f(x )csx dx converges for s > a. Thus, the second integral on the right-
hand side of equation (8) converges for s > a. Since both improper integrals
on the right-hand side of (8) converge for s > a, the Laplace transform of
f(x), .C[f(x)], exists for s >a.
The function f(x) = 1/ ft is of exponential order 0 as x --> +oo, but it is
not piecewise continuous on [O, b] for any b > 0 since limx_,o+ f(x) = +oo.
Thus, f ( x) = 1 /ft does not satisfy the first hypothesis of Theorem 5 .1;
however, the Laplace transform exists, as the following calculations show. By
definition
.C [ )x] = 1= e~x dx.
Making the change of variable t = sx, we find
The value of the definite integral on the right-hand side of the last equation
was obtained from a table of integrals.
Early in calculus, you discovered that many functions have the same deriva-
tive. Letting D denote the differential operator as we did earlier, we find that
D[x3] = D[x^3 + 1] = D[x^3 + C] = 3x^2 , where C is an arbitrary constant.
Just as many functions have the same derivative, many functions have the
same Laplace transform. At the beginning of this section, we calculated the
Laplace transform of the function f ( x) = 1 and found .C [f ( x)] = 1 / s = F ( s)
provided s > 0. The Laplace transform of the piecewise continuous function
{1,g(x ) =
3,x =I- 2x=2
is
1
.C[g(x)] = = le-sx dx = 12 le-sx dx + 1= le-sx dx = - 1 = F(s),
0 0 2 sprovided s >a. The function g(x) could have been chosen to differ from f(x)
at any finite set of values of x, or even at an infinite set of values such as the set
{1, 2, 3, ... }. In calculus, you learned if D[f(x)] = D[g(x)] on some interval
[a, b], then f(x) = g(x) + C on [a, b] for some constant C. The following
theorem, which we state without proof, is an analogous theorem for Laplace
transforms.