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Partial Fractions

variables the α are known as simple poles of F(s). A partial fraction
i
decomposition is

A A A
F(s) = 1 + 2 + · · · + n . (2.10)
s − α 1 s − α 2 s − α n

Multiplying both sides of (2.10) by s − α and letting s → α yield
i
i

A = lim (s − α )F(s). (2.11)
i
i
s→α i

We will see that the A are the residues of F(s) at the poles α . Therefore,
i
i

n ( ) n
∑ A i ∑
α i t
−1
f(t) = L (F(s)) = L −1 = A e .
i
s − α
i=1 i i=1

Putting in the expression (2.11) for A gives a quick method for finding the inverse:
i

n
∑
α i t
−1
f(t) = L (F(s)) = lim (s − α )F(s)e . (2.12)
i
s→α i
i=1
.

Example 2.30, Find

( )
s
−1
L .
(s − 1)(s + 2)(s − 3)

3t
t
f(t) = lim(s − 1)F(s)e + lim (s + 2)F(s)e −2t + lim F(s)e =
s→1 s→−2 s→3
1 2 3
3t
t
= − e − e −2t + e .
6 15 10
. . . . .
Most generally the process of finding the inverse Laplace transform will
require the use of the method of partial fractions. The following well-known
result on partial fractions is useful when finding inverse transforms:

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