Existence of the Laplace Transform | Piecewise Continuity and Exponential Order

Not every function has a Laplace transform. Two conditions together guarantee that one exists: the function must be piecewise continuous, so that each finite piece of the integral makes sense, and of exponential order, so that the damping factor $e^{- s t}$ can overcome its growth. The conditions are sufficient but not necessary, and this section explains both halves of that statement.

Quick Reference

Item	Statement
Convergent improper integral	$lim_{R \to \infty} \int_{0}^{R} f (t) d t$ exists
Comparison test	If $0 \leq f \leq g$ on $[a, \infty)$ and $\int_{a}^{\infty} g$ converges, so does $\int_{a}^{\infty} f$
Absolutely convergent	$\int_{a}^{\infty} \| f (t) \| d t$ converges; this implies $\int_{a}^{\infty} f (t) d t$ converges
Piecewise continuous	Continuous on every $[0, R]$ except at finitely many jump discontinuities
Exponential order $a$	$\| f (t) \| < K e^{a t}$ for $t \geq M$ , for some constants $K$ , $a$ , $M$
Limit test	$f$ is of exponential order iff $lim_{t \to \infty} \frac{f (t)}{e^{a t}}$ is finite for some $a$
Existence theorem	Piecewise continuous plus exponential order $a$ $⟹$ $ℒ {f}$ exists for $s > a$
Status of the conditions	Sufficient, not necessary
Function without a transform	$f (t) = e^{t^{2}}$
Transform despite failing the test	$f (t) = t^{- 1 / 2}$

Review of Improper Integrals

Convergence, comparison, and absolute convergence

The integral $\int_{0}^{\infty} f (t) d t$ is called convergent when

lim_{R \to \infty} (\int_{0}^{R} f (t) d t)

exists, and divergent when the limit does not exist.

Comparison test for integrals with non-negative integrands. Suppose $0 \leq f (t) \leq g (t)$ for all $t \geq a$ . If $\int_{a}^{\infty} g (t) d t$ converges, then $\int_{a}^{\infty} f (t) d t$ also converges. In words, a non-negative function trapped beneath a convergent one must itself have a finite integral.

The integral $\int_{a}^{\infty} f (t) d t$ is called absolutely convergent if $\int_{a}^{\infty} | f (t) | d t$ converges.

Absolute convergence implies convergence.

Proof

We always have

- | f (t) | \leq f (t) \leq | f (t) | .

Adding $| f (t) |$ to each part,

0 \leq f (t) + | f (t) | \leq 2 | f (t) | .

Since $2 \int_{a}^{\infty} | f (t) | d t$ converges by assumption, the comparison test shows that $\int_{a}^{\infty} [f (t) + | f (t) |] d t$ converges as well, the integrand now being non-negative.

Subtracting the finite quantity $\int_{a}^{\infty} | f (t) | d t$ from that convergent integral leaves $\int_{a}^{\infty} f (t) d t$ , which is therefore convergent.

The trick is worth noting: the comparison test needs a non-negative integrand, and adding $| f |$ manufactures one.

Let us now see under what conditions we can be certain that $\int_{0}^{\infty} f (t) e^{- s t} d t$ converges, so that the Laplace transform exists.

Condition 1: Piecewise Continuity

Convergence of $\int_{0}^{\infty} g (t) d t$ requires the existence of $\int_{0}^{R} g (t) d t$ for each finite $R > 0$ . Assuming $g$ is continuous, or at least piecewise continuous, guarantees this.

A function $g$ is piecewise continuous (or sectionally continuous) on $[0, \infty)$ if it is continuous on every finite interval $[0, R]$ except possibly at a finite number of jump discontinuities, at which the function approaches different values from the left and from the right.

The two requirements do real work. "Finitely many" rules out functions oscillating infinitely often, and "jump" rules out infinite discontinuities where the function blows up.

Graph of a piecewise continuous function made of three smooth arcs separated by two vertical jumps, with the jumps drawn as dashed segments. — A typical piecewise continuous function. It is smooth on each piece, and at the two break points it jumps from one finite value to another. Both one-sided limits exist at every break, which is exactly what "jump discontinuity" means.

Which of these are piecewise continuous on the intervals indicated?

f(t) = t^{1/3}, \qquad g(t) = \sin\sqrt{t}, \qquad h(t) = \begin{cases} -1 & 0 \leq t < 1 \\ 1 & 1 \leq t \end{cases}

u (t) = \sin \frac{1}{t}, v (t) = \tan t on [0, \frac{π}{2}]

Solution

The functions $f$ , $g$ , and $h$ are piecewise continuous. The first two are in fact continuous everywhere on $[0, \infty)$ , and $h$ has a single jump discontinuity at $t = 1$ , where it leaps from $- 1$ to $1$ .

The functions $u$ and $v$ are not piecewise continuous, for two different reasons:

$u (t) = \sin \frac{1}{t}$ has infinitely many discontinuities in every interval containing $t = 0$ , since $\frac{1}{t}$ races through infinitely many periods as $t \to 0^{+}$ . The requirement of finitely many breaks fails.
$v (t) = \tan t$ has an infinite discontinuity at $t = \frac{π}{2}$ , where it grows without bound rather than jumping between two finite values.

Condition 2: Exponential Order

In the definition of the Laplace transform, the factor $e^{- s t}$ is a damping factor: for $s > 0$ it decays as $t$ increases. One may intuitively expect $\int_{0}^{\infty} e^{- s t} f (t) d t$ to converge provided $f (t)$ does not "grow too rapidly". Making that precise:

A function $f$ is of exponential order $a$ , or briefly of exponential order, as $t \to \infty$ if there exist constants $K$ , $a$ , and $M$ such that

| f (t) | < K e^{a t} when t \geq M .

The decay of $e^{- s t}$ for $s > a$ then overcomes the growth of $f (t)$ , and the integral converges.

Most functions met in practice are of exponential order. Bounded functions such as a velocity or an electric current qualify, and so do all polynomials and power functions.
Functions that are not of exponential order are easy to construct. The standard example is $f (t) = e^{t^{2}}$ , which grows much faster than any $K e^{a t}$ no matter how large $a$ and $K$ are.

Using limits to test for exponential order

Saying that $f (t)$ satisfies $| f (t) | < K e^{a t}$ for $t > M$ is equivalent to saying that

lim_{t \to \infty} \frac{f (t)}{e^{a t}}

is finite for some $a$ .

For example, $f (t) = t^{n}$ with $n = 0, 1, 2, \dots$ is of exponential order, because

lim_{t \to \infty} \frac{t^{n}}{e^{t}} = 0,

as follows from applying l'Hôpital's rule $n$ times, each application reducing the degree of the numerator by one while the denominator is unchanged.

However, $f (t) = e^{t^{2}}$ is not of exponential order, because

lim_{t \to \infty} \frac{e^{t^{2}}}{e^{a t}} = lim_{t \to \infty} e^{t^{2} - a t} = \infty

no matter how large $a$ is, since $t^{2} - a t \to \infty$ for every fixed $a$ .

The Existence Theorem

If $f (t)$ is piecewise continuous on $t \geq 0$ and of exponential order $a$ as $t \to \infty$ , then the Laplace transform of $f$ exists for all $s > a$ .

Proof

Because $f$ is of exponential order $a$ ,

| f (t) | < K e^{a t} when t \geq M

for some constants $K$ , $a$ , and $M$ . Split the defining integral at $M$ :

\int_{0}^{\infty} e^{- s t} f (t) d t = \int_{0}^{M} e^{- s t} f (t) d t + \int_{M}^{\infty} e^{- s t} f (t) d t .

Since $f$ is piecewise continuous, the first integral exists: it is over a finite interval and the integrand has only finitely many jumps. So convergence of the whole depends entirely on the second piece.

For the tail, the exponential-order bound gives

| e^{- s t} f (t) | = e^{- s t} | f (t) | \leq K e^{- s t} e^{a t} = K e^{- (s - a) t},

and

\int_{M}^{\infty} K e^{- (s - a) t} d t = {- \frac{K}{s - a} e^{- (s - a) t} |}_{M}^{\infty} = \frac{K}{s - a} e^{- (s - a) M},

which is finite provided $s > a$ .

By the comparison test, $\int_{M}^{\infty} e^{- s t} | f (t) | d t$ converges. This says the tail is absolutely convergent, and therefore $\int_{M}^{\infty} e^{- s t} f (t) d t$ converges.

Notice where each hypothesis was used: piecewise continuity handled $[0, M]$ , and exponential order handled $[M, \infty)$ . The theorem needs both.

Sufficient but Not Necessary

If the conditions of the theorem are satisfied, the existence of the Laplace transform is guaranteed. The theorem does not say that failing the conditions means the transform does not exist. When the conditions fail, the transform may or may not exist. In mathematical language, the conditions are sufficient but not necessary.

Show that $f (t) = \frac{1}{\sqrt{t}}$ fails the hypotheses of the theorem yet has a Laplace transform.

Solution

This function is not piecewise continuous on $[0, \infty)$ , because it has an infinite discontinuity at $t = 0$ : indeed $\frac{1}{\sqrt{t}} \to \infty$ as $t \to 0^{+}$ , so the required one-sided limit is not finite.

Nevertheless the transform exists. From the power rule of Section 7.1 with $α = - \frac{1}{2}$ ,

ℒ {t^{- 1 / 2}} = \frac{Γ (\frac{1}{2})}{s^{1 / 2}} = \sqrt{\frac{π}{s}} .

The reason is that the singularity, though infinite, is integrable: $\int_{0}^{1} t^{- 1 / 2} d t = 2$ is finite because the function blows up slowly enough. Piecewise continuity is a convenient hypothesis, not the true boundary of the theory.

Graph of one over the square root of t, rising steeply toward infinity as t approaches zero from the right and decaying slowly for large t. — The function $f (t) = \frac{1}{\sqrt{t}}$ . It has an infinite discontinuity at the origin, so it is not piecewise continuous, yet the area under it near zero is finite and its Laplace transform $\sqrt{π / s}$ exists.

Exercises

Determine whether each function is of exponential order.

$f (t) = t^{100}$
$f (t) = e^{3 t} \sin t$
$f (t) = t e^{t^{2}}$

Solution

Yes, of exponential order $1$ (indeed of every positive order), since $lim_{t \to \infty} \frac{t^{100}}{e^{t}} = 0$ by l'Hôpital's rule applied $100$ times.
Yes, of exponential order $3$ , since $| e^{3 t} \sin t | \leq e^{3 t}$ for all $t$ .
No. For any $a$ , $\frac{t e^{t^{2}}}{e^{a t}} = t e^{t^{2} - a t} \to \infty$ , since the exponent grows without bound. Multiplying a non-exponential-order function by $t$ does not rescue it.

Is $f (t) = \frac{1}{t}$ piecewise continuous on $[0, \infty)$ ? Does its Laplace transform exist?

Solution

It is not piecewise continuous: it has an infinite discontinuity at $t = 0$ .

Unlike $t^{- 1 / 2}$ , its transform does not exist. The power rule requires $α > - 1$ , and $α = - 1$ fails. Concretely, near the origin $\int_{0}^{1} \frac{e^{- s t}}{t} d t$ behaves like $\int_{0}^{1} \frac{d t}{t}$ , which diverges logarithmically.

Comparing this with the previous example is instructive: both functions blow up at the origin, but only one does so slowly enough to be integrable.

The function h(t) = \begin{cases} -1 & 0 \leq t < 1 \\ 1 & 1 \leq t \end{cases} satisfies both hypotheses. Compute its Laplace transform directly.

Solution

It is piecewise continuous with one jump at $t = 1$ , and bounded, hence of exponential order $0$ . Splitting the integral at the jump,

\begin{aligned} \mathscr{L}\left\{h\right\} &= \int_0^{1}(-1)e^{-st}\,dt + \int_1^{\infty}(1)e^{-st}\,dt \\ &= \left.\frac{1}{s}e^{-st}\right|_0^{1} + \left.\left(-\frac{1}{s}e^{-st}\right)\right|_1^{\infty} \\ &= \frac{e^{-s}-1}{s} + \frac{e^{-s}}{s} = \frac{2e^{-s}-1}{s} . \end{aligned}

The theorem guarantees this exists for $s > 0$ , which the answer confirms. Splitting the integral at the jump is the general method for piecewise-defined functions, and Section 7.6 packages it more efficiently using the unit step function.

Frequently Asked Questions

What is a piecewise continuous function?

One that is continuous on each finite interval except at finitely many jump discontinuities, where the left and right limits both exist but differ. Step functions, square waves, and sawtooth waves all qualify, which is precisely why the Laplace transform handles switching problems so well.

What does exponential order mean intuitively?

That the function eventually grows no faster than some exponential $K e^{a t}$ . Since the transform multiplies by $e^{- s t}$ , choosing $s > a$ makes the product decay and the integral converge. A function that outgrows every exponential, like $e^{t^{2}}$ , can never be tamed this way.

Why do I need both conditions?

They control different parts of the integral. Piecewise continuity ensures the finite piece $[0, M]$ has a finite integral; exponential order ensures the infinite tail $[M, \infty)$ converges. The proof splits the integral at $M$ for exactly this reason, and neither condition can do the other's job.

If the conditions fail, does the transform fail too?

Not necessarily. The conditions are sufficient, not necessary. The function $t^{- 1 / 2}$ is not piecewise continuous yet has the transform $\sqrt{π / s}$ . On the other hand $1 / t$ , which fails in a similar-looking way, has no transform. Failing the test means the theorem gives no answer, so you must check by hand.

How large must

s

be?

Larger than the exponential order $a$ . For $e^{3 t}$ the transform exists for $s > 3$ ; for a bounded function it exists for $s > 0$ . When several functions are combined, the valid range is the intersection, that is, $s$ greater than the largest of the individual orders.

Piecewise Continuous Functions, Exponential Order

Quick Reference

Review of Improper Integrals

Condition 1: Piecewise Continuity

Condition 2: Exponential Order

The Existence Theorem

Sufficient but Not Necessary

Exercises

Frequently Asked Questions

Quick Links

Social Media