Suppose f is a piecewise continuous, bounded function, meaning that it is continuous on each
[0,R]
except for finitely many points. Then from Corollary ??t → f
(t)
is integrable. So is
t → e^{−st}f
(t)
.
Definition A.2.1We say that a function defined on [0,∞) has exponential growth if for some λ ≥ 0, andC > 0,
λt
|f (t)| ≤ Ce
Note that this condition is satisfied if
|f (t)|
≤ a + be^{λt}. You simply pick C > max
(a,b)
and observe
that a + be^{λt}≤ 2Ce^{λt}.
Proposition A.2.2Let f have exponential growth and be continuous except for finitely many points in
[0,R]
for each R. Then
∫
lim R f (t) e− stdt ≡ ℒf (s)
R→∞ 0
exists for every s > λ where
|f (t)|
≤ e^{λt}. That limit is denoted as
∫ ∞
f (t)e−stdt.
0
Proof: Let R_{n}→∞. Then for R_{m}< R_{n},
||∫ Rm ∫ Rn || ∫ Rn
|| f (t)e−stdt− f (t)e−st|| ≤ |f (t)|e−stdt
| 0 0 | Rm
∫ Rn −(s−λ)t −(s−λ)R
≤ e dt ≤ e m
Rm
The elementary computations are left to the reader. Then this converges to 0 as R_{m}→∞. It follows that
{∫Rn }
0 f (t) e− stdt
_{n=1}^{∞} is a Cauchy sequence and so it converges to I ∈ ℝ. The above computation shows
that if
Rˆ
_{n} also converges to ∞ as n →∞, then
∫ ∫
Rn −st ˆRn −st
lnim→∞ 0 f (t)e = nl→im∞ 0 f (t)e
and so the limit does indeed exist and this is the definition of the improper integral ∫_{0}^{∞}f
(t)
e^{−ts}dt.
■
Certain properties are obvious. For example,
If a,b scalars and if g,f have exponential growth, then for all s large enough,
ℒ (af + bg)(s) = aℒ (f )(s)+ bℒ (g) (s)
If f^{′}
(t)
exists and has exponential growth, and so does f
(t)
then for s large enough,
′
ℒ (f )(s) = − f (0)+ sℒ(f)(s)
One can also compute Laplace transforms of many standard functions without much difficulty. That
which is most certainly not obvious is the following major theorem. This is the thing which is
omitted from virtually all ordinary differential equations books, and it is this very thing which
justifies the use of Laplace transforms. Without it or something like it, the whole method is
nonsense. I am following [27]. This theorem says that if you know the Laplace transform, this
will determine the function it came from at every point of continuity of this function. The
proof is fairly technical but only involves the theory of the integral which was presented in this
chapter.
Theorem A.2.3Let ϕ have exponential growth and have finitely many discontinuities onevery interval
[0,R]
and let f
(s)
≡ℒ
(ϕ)
(s)
. Then if t is a point of continuity of ϕ, it followsthat
[ ( ) ]( )
(−-1)k (k) k- k- k+1
ϕ(t) = kli→m∞ k! f t t .
Thus ϕ
(t)
is determined by its Laplace transform at every point of continuity.
Proof: First note that for k a positive integer, you can change the variable letting ku = t and
obtain
k+1∫ ∞ k+1 ∫ ∞ ( )k
k--- (e−uu)kdu = k--- e− t t- 1-dt
k! 0 k! 0 k k
The details involve doing this on finite intervals using the theory of the Riemann integral developed earlier
and then passing to a limit. Thus the above equals
1∫ ∞ 1 1
k! e−ttkdt = Γ (k +1) k! = k!k! = 1
0
by Proposition ??.
Now assuming that
|ϕ(u)|
≤ Ce^{λu}, then from what was just shown,
kk+1∫ ∞ ( −u )k ∫ ∞ kk+1( −u )k
-k!- e u ϕ (u)du− ϕ (1) = -k!- e u (ϕ(u)− ϕ (1))du
0 0
Assuming ϕ is continuous at 1, the improper integral is of the form
∫ 1−δkk+1 (−u )k ∫ 1+δkk+1 ( −u )k
0 k! e u (ϕ (u)− ϕ(1))du+ 1−δ k! e u (ϕ(u)− ϕ (1))du
∫ ∞ k+1( )
+ k--- e−uu k (ϕ (u)− ϕ(1))du
1+ δ k!
Consider the first integral in the above. Letting K be an upper bound for
Now this converges to 0 as k →∞. In fact, for a < 1,lim_{k→∞}
k+1
kk!-
(e−aa)
^{k} = 0 because of the ratio test
which shows that for a < 1,∑_{k}
k+1
kk!-
(e−aa)
^{k}< ∞ which implies the k^{th} term converges to 0. Here
a = 1 −δ. Next consider the last integral. This obviously converges to 0 because of the exponential growth
of ϕ. In fact,
Now changing the variable letting uk = t, and doing everything on finite intervals followed by passing to a
limit, the absolute value of the above is dominated by
( )
∫ ∞ kk+1- −t -t k 1( λ(t∕k))
k(1+ δ) k! e k k a + be dt
∫ ∞ ( )
= 1e−ttk a + beλ(t∕k) dt for some a,b ≥ 0
k(1+ δ) k!
However, the limit as k →∞ of the integral on the right equals the improper integral on the left. Thus this
converges to 0 as k →∞. Thus all that is left to consider is the middle integral in which δ was chosen such
that
|ϕ (u )− ϕ(1)|
< ε. Thus
| |
||∫ 1+δkk+1 ( −u )k || ∫ ∞ kk+1-( −u )k
|| 1−δ k! e u (ϕ(u)− ϕ (1))du|| ≤ ε 0 k! e u du = ε