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Differentiation of the Laplace Transform

In the first step we have used the procedure of completing the square. Let

{
1, t ≥ a,
u (t) =
a
0, t < a.

Theorem 2.3È

(Second Translation Theorem) (Second Shifting Theorem). If F(s) = Lf(t),
then
L(u (t)f(t − a)) = e −as F(s)(a ≥ 0).
a
. . . . . . .
This follows from the basic fact that
∫ ∫
∞ ∞
e −st (u (t)f(t − a))dt = e −st f(t − a)dt,
a
0 a
and setting τ = t − a, the right-hand integral becomes

∫ ∫
∞ ∞
e −s(τ+a) f(τ)dτ = e −as e −sτ f(τ)dτ = e −as F(s).
. 0 0

Example 2.11, Let us determine L(g(t)) for

{
0, 0 ≤ t < 1,
g(t) =
2
(t − 1) , t ≥ 1.
2
Note that g(t) is just the function f(t) = t delayed by (a =)1
unit of time. Whence

2e −s
−s
2
2
L(g(t)) = L(u (t)(t − 1) ) = e L(t ) = (Re(s) > 0).
1
s 3
. . . . .
Differentiation of the Laplace Transform

It can be proved when s is a complex variable, the Laplace transform F(s) (for
suitable functions) is an analytic function of the parameter s. When s is a real

variable, we have a formula for the derivative of F(s), which holds in the
complex case as well.
Theorem 2.4È

Let f be piecewise continuous on [0, ∞) of exponential order α and
L(f(t)) = F(s). Then

d n
n n
F(s) = L((−1) t f(t)), n = 1, 2, 3, . . . (s > α). (2.1)
ds n
. . . . . . .

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