Let $f(t)$ be a function defined on the interval $[0,\,\infty)$ . The Laplace transform of $f(t)$ is the function $F(s)$ defined by $$ F(s)\,=\,\int_{0}^{\infty}e^{-st} f(t)\,dt, $$ provided that the integral converges. ¹ It suffices that $f$ be defined when $t>0$ and $s$ can be complex. We will usually denote the Laplace transform of $f$ by $\mathcal{L}\{f\}$ . Some of the most common Laplace transforms are:

1. $\displaystyle\mathcal{L}\{e^{at}\}\,=\,\frac{1}{s-a},\;\; s>a$
2. $\displaystyle\mathcal{L}\{\cos(bt)\}\,=\,\frac{s}{s^{2}+b^{2}},\;\; s>0$
3. $\displaystyle\mathcal{L}\{\sin(bt)\}\,=\,\frac{b}{s^{2}+b^{2}},\;\; s>0$
4. $\displaystyle\mathcal{L}\{t^{n}\}\,=\,\frac{\Gamma(n+1)}{s^{n+1}},\;\; s>0,\; n>-1.$
5. $\displaystyle\mathcal{L}\{f'\}\,=\, s\mathcal{L}\{f\}-\lim_{x \to 0+}f(x)$

Notice the Laplace transform is a linear transformation. It is worth noting that, if $$\int_{0}^{\infty}e^{-st}|f(t)|\,dt < \infty$$ for some $s \in \mathbb{R}$ , then $\mathcal{L}\{f\}$ is an analytic function in the complex half-plane $\{z \mid\; \Re z > s\}$ .

Much like the Fourier transform, the Laplace transform has a convolution. However, the form of the convolution used is different. $$\mathcal{L}\{f*g\} = \mathcal{L}\{f\} \mathcal{L}\{g\}$$ where $$(f*g) (t) = \int_0^t f(t-s) g(s) \, ds$$ and $$\mathcal{L}\{fg\}(s) = \int_{c - i \infty}^{c + i \infty} \mathcal{L}\{f\}(z) \mathcal{L}\{g\}(s-z) \, dz$$

The most popular usage of the Laplace transform is to solve initial value problems by taking the Laplace transform of both sides of an ordinary differential equation.

Footnotes

...http://planetmath.org/encyclopedia/LimitPoint2.html. ¹

Depending on the definition of integral one is using, one may prefer to define the Laplace transform as $\lim_{x \to 0+} \int_{x}^{\infty}e^{-st} f(t)\,dt$