13.2: Laplace transform

Definition

The Laplace transform of a function \(f(t)\) is defined by the integral

\[\mathcal{L} (f;s) = \int_{0}^{\infty} e^{-st} f(t)\ dt, \nonumber \]

for those \(s\) where the integral converges. Here \(s\) is allowed to take complex values.

Important note

The Laplace transform is only concerned with \(f(t)\) for \(t \ge 0\). Generally, speaking we can require \(f(t) = 0\) for \(t < 0\).

Standard notation

Where the notation is clear, we will use an upper case letter to indicate the Laplace transform, e.g, \(\mathcal{L} (f; s) = F(s)\).

The Laplace transform we defined is sometimes called the one-sided Laplace transform. There is a two-sided version where the integral goes from \(-\infty\) to \(\infty\).

First examples

Let’s compute a few examples. We will also put these results in the Laplace transform table at the end of these notes.

Example \(\PageIndex{1}\)

Let \(f(t) = e^{at}\). Compute \(F(s) = \mathcal{L} (f; s)\) directly. Give the region in the complex \(s\)-plane where the integral converges.

\[\begin{array} {rcl} {\mathcal{L} (e^{at} ; s)} & = & {\int_{0}^{\infty} e^{at} e^{-st}\ dt = \int_{0}^{\infty} e^{(a - s) t} \ dt = \dfrac{e^{(a - s) t}}{a - s} \vert_{0}^{\infty}} \\ {rcl} {} & = & {= \begin{cases} \dfrac{1}{s - a} & \text{ if Re} (s) > \text{Re} (a) \\ \text{divergent} & \text{ otherwise} \end{cases}} \end{array} \nonumber \]

The last formula comes from plugging \(\infty\) into the exponential. This is 0 if \(\text{Re} (a - s) < 0\) and undefined otherwise.

Example \(\PageIndex{2}\)

Let \(f(t) = b\). Compute \(F(s) = \mathcal{L} (f; s)\) directly. Give the region in the complex \(s\)-plane where the integral converges.

\[\begin{array} {rcl} {\mathcal{L} (b ; s)} & = & {\int_{0}^{\infty} be^{-st}\ dt = \dfrac{be^{- st}}{- s} \vert_{0}^{\infty}} \\ {rcl} {} & = & {= \begin{cases} \dfrac{b}{s} & \text{ if Re} (s) > 0 \\ \text{divergent} & \text{ otherwise} \end{cases}} \end{array} \nonumber \]

The last formula comes from plugging \(\infty\) into the exponential. This is 0 if \(\text{Re} (-s) < 0\) and undefined otherwise.

Example \(\PageIndex{3}\)

Let \(f(t) = t\). Compute \(F(s) = \mathcal{L} (f;s)\) directly. Give the region in the complex \(s\)-plane where the integral converges.

\[\begin{array} {rcl} {\mathcal{L} (t ; s)} & = & {\int_{0}^{\infty} te^{-st}\ dt = \dfrac{te^{- st}}{- s} - \dfrac{e^{- st}}{s^2} \vert_{0}^{\infty}} \\ {rcl} {} & = & {= \begin{cases} \dfrac{1}{s^2} & \text{ if Re} (s) > 0 \\ \text{divergent} & \text{ otherwise} \end{cases}} \end{array} \nonumber \]

Example \(\PageIndex{4}\)

Compute

\[\mathcal{L} (\cos (\omega t)). \nonumber \]

Solution

We use the formula

\[\cos (\omega t) = \dfrac{e^{i\omega t} + e^{-i \omega t}}{2}. \nonumber \]

So,

\[\mathcal{L} (\cos (\omega t); s) = \dfrac{1/(s - i\omega) + 1/(s + i\omega)}{2} = \dfrac{s}{s^2 + \omega^2}. \nonumber \]

Connection to Fourier transform

The Laplace and Fourier transforms are intimately connected. In fact, the Laplace transform is often called the Fourier-Laplace transform. To see the connection we’ll start with the Fourier transform of a function \(f(t)\).

\[\hat{f} (\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t}\ dt. \nonumber \]

If we assume \(f(t) = 0\) for \(t < 0\), this becomes

\[\hat{f} (\omega) = \int_{0}^{\infty} f(t) e^{-i \omega t}\ dt. \nonumber \]

Now if \(s = i\omega\) then the Laplace transform is

\[\mathcal{L}(f; s) = \mathcal{L} (f; i\omega) = \int_{0}^{\infty} f(t) e^{-i \omega t}\ dt. \nonumber \]

Comparing these two equations we see that \(\hat{f} (\omega) = \mathcal{L} (f; i \omega)\). We see the transforms are basically the same things using different notation –at least for functions that are 0 for \(t < 0\).