A particular kind of integral transformation is known as the **Laplace transformation**, denoted by *L*. The definition of this operator is

The result—called the **Laplace transform** of *f*—will be a function of *p*, so in general,

**Example 1**: Find the Laplace transform of the function *f*( *x*) = *x*.

By definition,

Integrating by parts yields

Therefore, the function *F*( *p*) = 1/ *p* ^{2} is the Laplace transform of the function *f*( *x*) = *x*. [Technical note: The convergence of the improper integral here depends on *p* being positive, since only then will ( *x/p*) *e* ^{− px }and *e* ^{− px }approach a finite limit (namely 0) as *x* → ∞. Therefore, the Laplace transform of *f*( *x*) = *x* is defined only for *p* > 0.]

In general, it can be shown that for any nonnegative integer *n*,

Like the operators *D* and *I*—indeed, like all operators—the Laplace transform operator *L* acts on a function to produce another function. Furthermore, since

*L* is also linear.

[Technical note: Just as not all functions have derivatives or integrals, not all functions have Laplace transforms. For a function *f* to have a Laplace transform, it is sufficient that *f*( *x*) be continuous (or at least piecewise continuous) for *x* ≥ 0 and of **exponential order** (which means that for some constants *c* and λ, the inequality *x*). Any **bounded** function (that is, any function *f* that always satisfies | *f*( *x*)| ≤ *M* for some *M* ≥ 0) is automatically of exponential order (just take *c* = *M* and λ = 0 in the defining inequality). Therefore, sin *kx* and cos *kx* each have a Laplace transform, since they are continuous and bounded functions. Furthermore, any function of the form *e* ^{kx }, as well as any polynomial, is continuous and, although unbounded, is of exponential order and therefore has a Laplace transform. In short, most of the functions you are likely to encounter in practice will have Laplace transforms.]

**Example 2**: Find the Laplace transform of the function *f*( *x*) = *x* ^{3} – 4 *x* + 2.

Recall form the first statement following Example 1 that the Laplace transform of *f*( *x*) = *x* ^{n }is *F*( *p*) = *n*!/ *p* ^{n + 1 }. Therefore, since the Laplace transform operator *L* is linear,

**Example 3**: Determine the Laplace transform of *f*( *x*) = *e* ^{kx }.

Apply the definition and perform the integration:

In order for this improper integral to converge, the coefficient ( *p* – *k*) in the exponential must be positive (recall the technical note in Example 1). Thus, for *p* > *k*, the calculation yields

**Example 4**: Find the Laplace transform of *f*( *x*) = sin *kx*.

By definition,

This integral is evaluated by performing integration by parts twice, as follows:

for *p* > 0. By a similar calculation, it can be shown that

**Example 5**: Determine the Laplace transform of the function

pictured in Figure 1

Figure 1

This is an example of a **step function**. It is not continuous, but it is *piecewise* continuous, and since it is bounded, it is certainly of exponential order. Therefore, it has a Laplace transform.

Table **L**.

**Example 6**: Use Table *f*( *x*) = sin ^{2} *x*.

Invoking the trigonometric identity

**L** implies

**Example 7**: Use Table *g*( *x*) *x* ^{3} *e* ^{5x}.

The presence of the factor *e* ^{5x} suggests using the shifting formula with *k* = _{5}. Since

*f*( *x*) *e* ^{5x} = *x* ^{3} *e* ^{5x }is equal to *F*( *P* – 5). In other words, the Laplace transform of *x* ^{3} *e* ^{5x} is equal to the Laplace transform of *x* ^{3} with the argument *p* **shifted** to *p* – 5:

**Example 8**: Use Table *f*( *x*) = *e* ^{−2x} sin *x* – 3.

First, since *L* [sin *x*] = 1/( *p* ^{2} + 1), the shifting formula (with *k* = −2) says

Now, because *L*[3] = 3 · *L*[1] = 3/ *p*, linearity implies

**Example 9**: Use Table *F*( *p*) = 12/ *p* ^{5}.

This example introduces the idea of the **inverse Laplace transform operator,**, *L* ^{−1}. The operator *L* ^{−1} will “un‐do” the action of *L*. Symbolically,

If you think of the operator *L* as changing *f*( *x*) into *F*( *p*), then the operator *L* ^{−1} just changes *F*( *P*) back into *f*( *x*). Like *L*, the inverse operator *L* ^{−1} is linear.

More formally, the result of applying *L* ^{−1} a function *F*( *p*) is to recover the continuous function *f*( *x*) whose Laplace transform is the given *F*( *p*). [This situation should remind remind you of the operators *D* and *I* (which are, basically, inverses of one another). Each will un‐do the action of the other in the sense that if, say, *I* changes *f*( *x*) into *F*( *x*), then *D* will change *F*( *x*) back into *f*( *x*). In other words, *D* = *I* ^{−1}, so if you apply *I* and then *D*, you're back where you started.]

Using Table

**Example 10**: Find the continuous function whose Laplace transform is *F*( *p*) = 1/( *p* ^{2} – 1).

By partial fraction decomposition,

Therefore, by linearity of *L* ^{−1},

**Example 11**: Determine

First, note that *p* has been shifted to *p* + 2 = *p* – (‐2). Therefore, since

*k* = −2) implies

**Example 12**: Evaluate

Although *p* ^{2} – 6 *p* + 25 cannot be factored over the integers, it can be expressed as the sum of two squares:

Therefore,