Does the Integral Exist? Piecewise Continuity and Exponential Order
Chapter Forty-One
Syllabus topic Module 1, "1.2 The Laplace Transform"
Pages 91 to 93 of 303
In one line
The transform exists if f is piecewise continuous and does not grow faster than an exponential, and then only for s large enough.
The question
The definition is an integral to infinity. Such an integral may fail to exist, and the two conditions below are what guarantee it does not. A question asking for "the sufficient conditions for the existence of the Laplace transform" wants them stated, so they are worth learning as sentences.
Condition one: piecewise continuity
f must be piecewise continuous on every finite interval from 0 onwards. That means: on any finite stretch it has at most finitely many breaks, and at each break the function jumps by a finite amount rather than shooting off.
A square wave is piecewise continuous. The unit step function is piecewise continuous. The function one over t is not, because at t = 0 it does not jump by a finite amount, it grows without limit.
The reason the condition is needed: a finite number of finite jumps can be integrated over by splitting the integral at the jumps, and a finite jump contributes nothing to the area. An infinite blow-up cannot be handled that way.
Condition two: exponential order
f must be of exponential order. That means there are numbers M and a such that the size of f(t) is at most M times e to the at for all large t.
In words: f may grow, but not faster than some exponential. Every function in this paper satisfies this, and it is worth seeing which functions do and which do not.
| Function | Of exponential order | Why |
|---|---|---|
| any constant | yes | take a = 0 |
| t, t squared, any power | yes | a power loses to any exponential |
| e to the 5t | yes | take a = 5 |
| sin t, cos t | yes | they never exceed 1 |
| e to the t squared | no | beats every e to the at |
| t to the power t | no | same reason |
The claim in the second row is the one from the indeterminate-forms chapter, and it is why every polynomial has a transform.
lim(t^3/e^t, t -> oo) = 0
lim(t^10/e^t, t -> oo) = 0
Why the two conditions do the job
Because together they make the integrand small enough, for large enough s.
The size of the integrand is at most e to the minus st times M e to the at, which is M times e to the minus (s minus a)t. The integral of that from zero to infinity converges whenever s minus a is positive.
integrate(e^(-(s - a) t), (t, 0, oo)) = 1/(s - a)
So the transform exists for s greater than a, and that is where the condition on s in every table entry comes from. It is not a technicality bolted on afterwards; it is the number a from the exponential-order condition.
Does the Integral Exist? Piecewise Continuity and Exponential Order
The region of convergence
Every transform therefore comes with a range of s for which it is valid, and the range is a half line.
| f(t) | F(s) | Valid for |
|---|---|---|
| 1 | 1/s | s greater than 0 |
| e to the 3t | 1/(s - 3) | s greater than 3 |
| e to the -3t | 1/(s + 3) | s greater than -3 |
| t to the n | n factorial over s to the (n+1) | s greater than 0 |
| sin at | a/(s squared + a squared) | s greater than 0 |
| e to the 5t sin 2t | 2/((s-5) squared + 4) | s greater than 5 |
Read the pattern: the boundary is always the fastest exponential growth rate in f. A bounded function gives s greater than zero; an exponential of rate a gives s greater than a; and a decaying exponential gives a negative boundary, so the transform is valid over more of the line.
In practice the condition is stated once and then not carried through the working, because every step of a transform calculation is valid on the intersection of the ranges involved. But it should be written down when a transform is derived from the definition, because that is where the marks are.
A function with no transform
The conditions are sufficient, not necessary, so a function failing them may still have a transform. But here is one that genuinely does not.
Take f(t) = e to the t squared. For any s, however large, the integrand is e to the (t squared minus st), and the exponent eventually becomes large and positive because t squared beats st. So the integrand grows without limit and the integral diverges for every s.
And one that fails condition one but has a transform anyway, which is why the conditions are only sufficient: f(t) = t to the power minus one half is unbounded at t = 0, so it is not piecewise continuous there, yet its integral converges because the blow-up is mild.
L{t^(-1/2)} = sqrt(pi)/sqrt(s)
That transform involves the square root of pi, which comes from the gamma function, and the chapter on special functions explains where.
Uniqueness, which is what makes the inverse possible
Two continuous functions with the same transform are the same function. That statement, Lerch's theorem, is what licenses the whole second half of this work: if you can find any function whose transform is the F(s) in front of you, it is the answer, and you need not worry that some other function has the same transform.
The word continuous is doing work there. Two functions that differ only at a few isolated points have the same transform, because a point contributes no area. So the inverse transform is unique up to what happens at isolated points, which never matters in this paper.
Does the Integral Exist? Piecewise Continuity and Exponential Order
Check yourself
| Function | Has a transform | The boundary on s |
|---|---|---|
| a constant 5 | yes | 0 |
| t cubed | yes | 0 |
| e to the 7t | yes | 7 |
| e to the minus 7t | yes | -7 |
| cos 4t | yes | 0 |
| e to the t cubed | no | none |
L{e^(7t)} = 1/(s - 7)
L{t^3} = 6/s^4
L{cos(4t)} = s/(s^2 + 16)
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.