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Properties of Laplace Transform

Integration

Not only can the Laplace transform be differentiated, but it can be integrated as
well. Again the result is another Laplace transform.

Theorem 2.8È

If f is piecewise continuous function on [0, ∞) of exponential order α ≥ 0,
∫ t
and g(t) = f(u)du, then
0
t
∫
[ ] 1
L(g(t)) =L f(τ)dτ = L[(1∗f)(t)]= L[1]L[f(t)] = L(f(t)) (Re > α).
s
0
. . . . . . .

PROOF. Since g (t) = f(t) except at points of discontinuity of f, integration by
′
parts gives

∫ [ τ ∫ ]
∞ g(t)e −st 1 τ
e −st g(t)dt = lim + e −st f(t)dt .
0 τ→∞ −s 0 s 0
Since g(0) = 0, we need only compute

g(τ)e −sτ
lim .
τ→∞ −s
To this end,

τ τ
∫ ∫
αu
|g(τ)e −sτ | ≤ e −xτ |f(u)|du ≤ Me −xτ e du =
0 0
M
= (e −(x−α)τ − e −xτ ) → 0 2
α
as τ → ∞ for x = Re(s) > α > 0. Similarly, this holds for α = 0. Hence

1
L(g(t)) = L(f(t)) (Re(s) > α).
s
. . . .
This result can be useful in finding inverse Laplace transforms since

t
∫
[ F(s) ]
−1
L = f(τ)dτ.
s
0
.

1
Example 2.16, Find the inverse Laplace transform of a)
2
s(s + 1)
. . . . .

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