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Derivatives

by the fundamental theorem of calculus, and

t M
∫
ατ
αt
αt
|f(t)| ≤ M e dτ + |f(t )| ≤ e + |f(t )| ≤ Ce , t ≥ t .
0
0
0
α
t 0
Since f is continuous, the result holds for α ̸= 0, and the case α = 0 is subsumed
under this one.
To treat differential equations we will also need to know L(f ) and so forth.
′′′
Suppose that for the moment we can apply formula (2.3) to f. Then
′
′
′′
L(f (t)) = sL(f (t)) − f (0) = s(sL(f(t)) − f(0)) − f (0) = (2.5)
′
2
s L(f(t)) − sf(0) − f (0). (2.6)
′
Similarly,
2
3
′′
′′′
′
′′′
′′
L(f (t)) = sL(f (t)) − f (0) = s L(f(t)) − s f(0) − sf (0) − f (0) (2.7)
under suitable conditions.
In the general case we have the following result.
Theorem 2.7È

′
Suppose that f(t), f (t), . . . , f (n−1) (t) are continuous on (0, ∞) and of
exponential order, while f (n) (t) is piecewise continuous on [0, ∞). Then

+
n
+
L(f (n) (t)) = s L(f(t)) − s n−1 f(0 ) − s n−2 ′ (n−1) (0 ). (2.8)
f (0+) − · · · − f
. . . . . . . .
Example 2.15, Determine the Laplace transform of the Laguerre
polynomials, defined by

t
e d n
n −t
L (t) = (t e ), n = 0, 1, 2, . . .
n
n! dt n
. . . . .
.
n −t
Solution. Let y(t) = t e . Then
( )
1
L(L (t)) = L e t y (n) .
n
n!
First, we find by previous theorem, and subsequently the first translation
n
theorem coupled with L(t ) = n! ,
s n+1
n
s n!
n
L(y (n) ) = s L(y) = .
(s + 1) n+1
It follows that
( ) n
1 (s − 1)
L(L (t)) = L e t y (n) = (Re(s) > 1).
n
n! s n+1
. . . . again by the first translation theorem.

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