Applied Mathematics/Laplace Transforms

The Laplace transform is an integral transform which is widely used in physics and engineering.

Laplace Transforms involve a technique to change an expression into another form that is easier to work with using an improper integral. We usually introduce Laplace Transforms in the context of differential equations, since we use them a lot to solve some differential equations that can't be solved using other standard techniques. However, Laplace Transforms require only improper integration techniques to use. So you may run across them in first year calculus.

Notation: The Laplace Transform is denoted as $$\displaystyle\mathcal{L} \left\{f(t)\right\}$$.

The Laplace transform is named after mathematician and astronomer Pierre-Simon Laplace.

Definition
For a function $$f(t)$$, using Napier's constant $$e$$ and a complex number $$s$$, the Laplace transform $$F(s)$$ is defined as follows:
 * $$F(s) = \mathcal{L} \left\{f(t)\right\}(s)=\int_0^{\infty} e^{-st} f(t) \,dt $$

The parameter $$s$$ is a complex number.


 * $$s = \sigma + i \omega, \, $$ with real numbers $$\sigma$$ and $$\omega$$.

This $$F(s)$$ is the Laplace transform of $$f(t)$$.

Explanation
Here is what is going on.

Examples of Laplace transform
In the above table,
 * 1) $$C$$ and $$a$$ are constants
 * 2) $$n$$ is a natural number
 * 3) $$\delta(t-a)$$ is the Delta function
 * 4) $$H(t-a)$$ is the Heaviside function



Examples
1. Calculate $$\mathcal{L}\{C\} $$ (where $$C$$ is a constant) using the integral definition. $$ \begin{array}{rcl} \displaystyle{\int_0^{\infty} e^{-st} C \,dt} & = & C \displaystyle{\int_0^{\infty} e^{-st} \,dt} \\ & = & \displaystyle{ C \lim_{b \to \infty}{ \int_0^{b} e^{-st} \,dt} } \\ & = & \displaystyle{ \left. C \lim_{b \to \infty}{ \frac{e^{-st}}{-s} } \right|_{t=0}^{t=b} } \\ & = & \displaystyle{ -\frac{C}{s}\left[ \lim_{b \to \infty} {e^{-bs}} - e^0 \right] } \\ & = & \displaystyle{ -\frac{C}{s} [ 0 - 1 ] } \\ & = & \displaystyle{ \frac{C}{s} } \end{array} $$

$$\therefore \displaystyle\mathcal{L} \left\{C\right\} = \frac{C}{s}$$

2. Calculate $$\mathcal{L}\{e^{-at}\} $$ using the integral definition.

$$ \begin{array}{rcl} \displaystyle\int_0^{\infty} e^{-st} \cdot e^{-at} \,dt & = & \displaystyle\int_0^{\infty} e^{-(s+a)t} \,dt \\ & = & \displaystyle\lim_{b \to \infty}{ \displaystyle\int_0^{b} e^{-(s+a)t} \,dt } \\ & = & \displaystyle\lim_{b \to \infty} \left[\displaystyle\frac{-e^{-(s+a)t}}{s+a}\right]_{t=0}^{t=b} \\ & = & \displaystyle\lim_{b\to\infty} \left[\displaystyle\frac{-e^{-(s+a)b}}{s+a} - \displaystyle\frac{-e^{-(s+a)0}}{s+a}\right] \\ & = & \displaystyle\frac{1}{s+a}

\end{array} $$

$$ \therefore \mathcal{L}\{e^{-at}\} = \displaystyle\frac{1}{s+a} $$