Definition:Laplace Transform

Definition
Let $f: \left [{0 \,.\,.\, \to} \right) \to \mathbb F$ be a function of a real variable $t$, where $\mathbb F \in \left\{ {\R, \C}\right\}$.

The Laplace transform of $f$, denoted $\mathcal L \left\{{f}\right\}$ or $F$, is defined as:


 * $\displaystyle \mathcal L \left\{{f \left({t}\right)}\right\} = F \left({s}\right) = \int_0^{\to +\infty} e^{-s t} f \left({t}\right) \rd t$

whenever this improper integral exists. Here $\mathcal L \left\{{f}\right\}$ is a complex function of the variable $s$.

Discontinuity at zero
Let $f: \left ({0 \,.\,.\, \to} \right) \to \mathbb F$ be a function of a real variable $t$, where $\mathbb F \in \left\{ {\R, \C}\right\}$.

Let $f$ be discontinuous or not defined at $t = 0$.

Then the Laplace transform of $f$ is defined as:


 * $\displaystyle \mathcal L \left\{{f \left({t}\right)}\right\} = F \left({s}\right) = \int_{0^+}^{\to +\infty} e^{-s t} f \left({t}\right) \rd t = \lim_{\epsilon \to 0^+}\int_{\epsilon}^{\to +\infty} e^{-s t} f \left({t}\right) \rd t$

whenever this improper integral exists.

Here the integral is improper both because of its lower bound and because of its upper bound.

Application in Physics
In the field of Signal Processing, the domain of $f$ is called the time domain, and the domain of $\mathcal L \left\{{f}\right\}$ is called the frequency domain.

Notation
Also denoted as:


 * $\mathcal L \left[{f \left({t}\right)}\right]$


 * $\mathscr L \left\{ {f \left({t}\right)}\right\}$


 * $\tilde f \left({s}\right)$

Graphical Interpretation
Define $\gamma$ as the integrand of $\displaystyle \int_0^{\to +\infty} e^{-s t} f \left({t}\right) \rd t$ as a function of $f\left({t}\right)$, $s$, and $t$:


 * $\gamma\left({f\left({t}\right), t; s}\right) = f\left({t}\right)e^{-s t}$

For any particular function $f$, holding $s$ fixed, the integrand of the Laplace Transform $\kappa$ can be interpreted as a contour.

That is, for a given function $f$ and a particular complex number $s_0$ held constant:


 * $\kappa\left({t}\right): \left[{0 \, .\, .\, \to }\right) \to \C$


 * $\kappa\left({t}\right) = \gamma\left({f\left({t}\right), t; s_0}\right)$

is a parameterization of a contour.

The following are three examples of the contours defined by the integrand defining the Laplace transform of $\cos\left({t}\right)$.

For $\mathcal L \left\{{\cos \left({t}\right)}\right\} \left({3-9i}\right)$, notice how the contour spirals towards the origin.

Intuitively, the integral converges because as $t$ increases without bound, the contour "shrinks" as it continues its path into the origin.


 * LCost 3minus9i.png

For $\mathcal L \left\{{\cos \left({t}\right)}\right\} \left({\dfrac 1 2 + 4i}\right)$, though the contour is not simple, it still "shrinks" as the parameter $t$ increases without bound:


 * LCost halfPlus4i.png

From Laplace Transform of Cosine, $\mathcal L \left\{{\cos \left({t}\right)}\right\} \left({-3+12i}\right)$ doesn't exist. Notice how the contour spirals outward as $t$ increases without bound, never settling at a "stable point":


 * LCost minus3plus12i.png

Restriction to Reals
Although the definition of the Laplace transform has $s$ be a complex variable, sometimes the restriction of $\mathcal L \left\{{f}\right\}\left({s}\right)$ to wholly real $s$ is sufficient to solve a particular differential equation.

Therefore, elementary textbooks introducing the Laplace transform will often write something like the following:


 * A profound understanding of the workings of the Laplace transform requires considering it to be a so-called analytic function of a complex variable, but in most of this book we shall assume that the variable $s$ is real.


 * -- : $\S 3.1$

Also see

 * Definition:Inverse Laplace Transform