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3.7 Polynomial Coefficients 113

THEOREM 3.7

Let f (t) have Laplace transform F(s) for s > b, and assume that F(s) is differentiable. Then
L[tf (t)](s) =−F (s)

for s > b.
Thus the transform of tf (t) is the negative of the derivative of the transform of f (t).
Proof Differentiate under the integral sign:
d ∞
−st
F (s) = e f (t)dt
ds 0
∞
d

= (e −st f (t))dt
ds
0
∞

= −te −st f (t)dt
0
∞

= e −st (−tf (t))dt = L[−tf (t)](s).
0
An induction argument yields the general result
d n
n
n
L[t f (t)](s) = (−1) F(s)
ds n
if F(s) can be differentiated n times.
We will also have use of the fact that, under certain conditions, the transform of f (t) has
limit 0 as s →∞.
THEOREM 3.8

Let f be piecewise continuous on [0,k] for every positive number k. Suppose there are numbers
bt
M and b such that | f (t)|≤ Me for t ≥ 0. Then,
lim F(s) = 0.
s→∞
Proof Write

∞
−st
|F(s)|= e f (t)dt

0
∞

bt
≤ e −st Me dt
0
∞
M
= e −(s−b)t
b − s
0
M
= → 0
s − b
as s →∞.
EXAMPLE 3.22
We will solve
y + 2ty − 4y = 1; y(0) = y (0) = 0.

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October 14, 2010 14:14 THM/NEIL Page-113 27410_03_ch03_p77-120