# Applied Mathematics/Laplace Transforms

The Laplace transform is an integral transform which is widely used in physics and engineering. Laplace transform is denoted as ${\displaystyle \displaystyle {\mathcal {L}}\left\{f(t)\right\}}$.

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

## Definition

For a function f(t), using Napier's constant"e" and complex number "s", the Laplace transform F(s) is defined as follow:

${\displaystyle F(s)={\mathcal {L}}\left\{f(t)\right\}(s)=\int _{0}^{\infty }e^{-st}f(t)\,dt}$

The parameter s is a complex number:

${\displaystyle s=\sigma +i\omega ,\,}$ with real numbers σ and ω.

This ${\displaystyle F(s)}$ is the Laplace transform of f(t).

## Examples of Laplace transform

Examples of Laplace transform
function result of Laplace transform
${\displaystyle C}$ (constant) ${\displaystyle {\frac {C}{s}}}$
${\displaystyle t}$ ${\displaystyle {\frac {1}{s^{2}}}}$
${\displaystyle t^{n}}$ (n is natural number) ${\displaystyle {\frac {n!}{s^{n+1}}}}$
${\displaystyle {\frac {t^{n-1}}{(n-1)!}}}$ ${\displaystyle {\frac {1}{s^{n}}}}$
${\displaystyle e^{at}}$ ${\displaystyle {\frac {1}{s-a}}}$
${\displaystyle e^{-at}}$ ${\displaystyle {\frac {1}{s+a}}}$
${\displaystyle {\rm {cos}}\ \omega t}$ ${\displaystyle {\frac {s}{s^{2}+{\omega }^{2}}}}$
${\displaystyle {\rm {sin}}\ \omega t}$ ${\displaystyle {\frac {\omega }{s^{2}+{\omega }^{2}}}}$
${\displaystyle {\frac {t^{n-1}}{\Gamma (n)}}}$ ${\displaystyle {\frac {1}{s^{n}}}}$ (n>0)
${\displaystyle \delta (t-a)}$ (Delta function) ${\displaystyle e^{-as}}$
${\displaystyle H(t-a)}$ (Heaviside function) ${\displaystyle {\frac {e^{-as}}{s}}}$

## Examples of calculation

(1)Suppose ${\displaystyle f(t)=C}$ (C = constant)
${\displaystyle \int _{0}^{\infty }e^{-st}C\,dt}$
${\displaystyle ={\frac {C}{s}}}$
${\displaystyle =F(s)}$

(2)Suppose ${\displaystyle f(t)=e^{-at}}$
${\displaystyle \int _{0}^{\infty }e^{-st}\cdot e^{-at}\,dt}$
${\displaystyle =\int _{0}^{\infty }e^{-(s+a)t}\,dt}$
${\displaystyle =\left[{\frac {-e^{-(s+a)t}}{s+a}}\right]_{0}^{\infty }}$
${\displaystyle ={\frac {1}{s+a}}}$