The elementary computations are left to the reader. Then this converges to 0 as
R_{m}→∞. It follows that
{ }
∫ Rnf (t)e−stdt
0
_{n=1}^{∞} is a Cauchy sequence and so it
converges to I ∈ ℝ. The above computation shows that if
ˆR
_{n} also converges to ∞ as
n →∞, then
∫ Rn − st ∫ ˆRn −st
lni→m∞ 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 [38]. 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 9.3.3Let ϕ have exponential growth and have finitely manydiscontinuities on every interval
[0,R ]
and let f
(s)
≡ℒ
(ϕ)
(s)
. Then if t is a point ofcontinuity of ϕ, it follows that
k[ ( )]( )k+1
ϕ (t) = lim (−-1)- f(k) k- k- .
k→ ∞ k! 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
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
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!-e k- k-a + be dt
∫k(∞1+δ) ( )
= 1-e− ttk a + beλ(t∕k) dt for some a,b ≥ 0
k(1+δ) k!
∫ ∫
∞ -1 −t k( λ(t∕k)) k(1+δ)1- −tk ( λ(t∕k))
= 0 k!e t a+ be dt− 0 k!e t a+ be dt
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
I think the approach given above is really interesting because it gives an explicit
description of ϕ
(t)
at every point. However, there are other ways to show this. See my
single variable advanced calculus book for another approach based on the Weierstrass
approximation theorem. However, to really do it right, one should use complex variable
techniques. You can actually get the inverse Laplace transform from doing contour
integrals. It is in my book on calculus of real and complex variables. This is called the
Bromwich integral. It is actually used by computer algebra systems to invert Laplace
transforms.