# 6.6: The Laplace Transform

- Page ID
- 396

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

We have seen many techniques of solving differential equations that involve using a substitution. There is a special type of substitution, called an integral transform that simplifies the task of solving differential equations.

Definition: Integral Transform

Let \(K(s,t)\) be a function of two variables. The *Integral Transform with Kernel \(K\)*, is defined as the mapping that takes functions to functions by the rule

\[ f(x) \rightarrow \int_a^b K(s,t) \, f(t)\, dt .\]

Note: \(a\) and \(b\) can be any real numbers or even infinity or negative infinity.

The most important integral transform in the field of differential equations is when \(a\) is 0, \(b\) is infinity, and \(K(s,t)\) is \(e^{-st}\).

Definition: Laplace Transform

Let \(f(x)\) be a function. Then the Laplace Transform of \(f(x)\) is

\[ L(s)= \int_0^\infty e^{-st} f(t) \, dt. \label{laplace}\]

Example \(\PageIndex{1}\)

Find the Laplace Transform of \(f(x) = x^2\).

**Solution**

We just work out the integral in equation \ref{laplace}

\[ L(s) = \int_0^{\infty} e^{-st}\, t^2\,dt. \nonumber\]

This can be worked out by integrating by parts twice.

\[\begin{align*} \int_o^{\infty} e^{-st}\,t^2 \, dt &= \left. \dfrac{-1}{s}e^{-st}\,t^2 \right|_0^{\infty} + \dfrac{2}{s}\int_0^{\infty}e^{-st}\,t\,dt \\[5pt] & = \dfrac{2}{s} \left[ \left. \dfrac{-1}{s}e^{-st}\,t \right|_0^{\infty} + \dfrac{1}{s}\int_0^{\infty}e^{-st}\,dt \right] \\[5pt] &=\left.\dfrac{-2}{s^3}e^{-st}\right|_0^{\infty} \\[5pt] &= \dfrac{2}{s^3}. \end{align*}\]

We used the fact that

\[ \lim_{t \rightarrow \infty} t^n\,e^{-st} = 0 \nonumber \]

for all \(s > 0\), which is easily proven using induction and L'Hopital's rule.

We can see that finding the Laplace transform of a function is just a matter of integration. We will usually see integration by parts enter into the picture.

Exercise \(\PageIndex{1}\)

Find the Laplace transform of the following functions.

- \( f(x) = \sin(at) \)
- \(f(x) = \cos(at) \)
- \(f(x) = e^{at}\)
- \(f(x) = t^n\) with n a positive integer
- \(f(x) = e^{at} \sin(bt)\)

Our next question to ask is when the Laplace transform of a function is defined. Since the Laplace transform of a function is defined as an improper integral, the integral may not converge. Fortunately most of the functions that we know and love have convergent Laplace transforms. More generally we have the following theorem:

Theorem: Existence

Let \(f(t)\) be a piecewise continuous function on \(0 \le t \le A\) for any \(A\). Also let

\[ |f(t)| \le Ke^{at} \]

for all \(t \ge M\) with \(K\), and \(M\) real constants and \(K\) and \(M\) positive.

Then the Laplace transform \(\mathcal{L}\{f(t)\} = F(s)\) exists for \(s \ge a\).

Proof

First, since \(f(t)\) is piecewise continuous, for large values of \(t\), \(f(t)\) is continuous. Since convergence only depends on the behavior for large values of \(t\), we can assume that \(f\) is continuous. To finish the proof we substitute the larger \(Ke^{at}\) for the smaller \(f(t)\) to get

\[\begin{align*} \int_0^{\infty} e^{-st}\, f(t) \,dt &\le \int _0^{\infty} e^{-st}\, K\,e^{at}\,dt \\[5pt] & = K \int_0^{\infty} e^{(-s+a)t}\,dt \\[5pt] &= \left.\dfrac{K}{a-s} e^{(a-s)t} \right|_0^{\infty} \end{align*}\]

This converges whenever \(a \le s\).

Another important fact about Laplace Transforms is that the Laplace transform is a linear transformation from "nice" functions to functions, where "nice" means functions that have a Laplace transform. That is,

\[ \mathcal{L}\{f(t) + g(t)\} = \mathcal{L}\{f(t)\} + \mathcal{L}\{g(t)\} \nonumber\]

and

\[ \mathcal{L}\{cf(t)\} = c\mathcal{L}\{f(t)\}. \nonumber \]

\(\square\)

### Contributors

- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.