# Laplace Transform Determination/Miscellaneous Methods

## Solution Technique for Laplace Transform

To find the Laplace transform of a function $f$, one can evaluate it using one of the following techniques:

#### Linear Combination of Laplace Transforms

Then:

$\laptrans {\lambda \, \map f t + \mu \, \map g t} = \lambda \laptrans {\map f t} + \mu \laptrans {\map g t}$

everywhere all the above expressions are defined.

#### Laplace Transform of Exponential times Function

$\laptrans {e^{a t} \map f t} = \map F {s - a}$

everywhere that $\laptrans f$ exists, for $\map \Re s > a$

#### Laplace Transform of Function of t minus a

Let $g$ be the function defined as:

$\map g t = \begin{cases} \map f {t - a} & : t > a \\ 0 & : t \le a \end{cases}$

Then:

$\laptrans {\map g t} = e^{-a s} \map F s$

#### Laplace Transform of Constant Multiple

Let $a \in \C$ or $\R$ be constant.

Then:

$a \laptrans {\map f {a t} } = \map F {\dfrac s a}$

#### Laplace Transform of Higher Order Derivatives

 $\displaystyle \laptrans {\map {f^{\paren n} } t}$ $=$ $\displaystyle s^n \laptrans {\map f t} - \sum_{j \mathop = 1}^n s^{j - 1} \map {f^{\paren {n - j} } } 0$ $\displaystyle$ $=$ $\displaystyle s^n \map F s - s^{n - 1} \, \map f 0 - s^{n - 2} \, \map {f'} 0 - s^{n - 3} \, \map {f''} 0 - \ldots - s \, \map {f^{\paren {n - 2} } } 0 - \map {f^{\paren {n - 1} } } 0$

#### Laplace Transform of Integral

$\displaystyle \laptrans {\int_0^t \map f u \rd u} = \dfrac {\map F s} s$

wherever $\laptrans f$ exists.

#### Higher Order Derivatives of Laplace Transform

$\dfrac {\d^n} {\d s^n} \laptrans {\map f t} = \paren {-1}^n \laptrans {t^n \, \map f t}$

#### Integral of Laplace Transform

$\displaystyle \laptrans {\dfrac {\map f t} t} = \int_s^{\to \infty} \map F u \rd u$

wherever $\displaystyle \lim_{t \mathop \to 0} \dfrac {\map f t} t$ and $\laptrans f$ exist.

#### Laplace Transform of Periodic Function

Let $f$ be periodic, that is:

$\exists T \in \R_{\ne 0}: \forall x \in \R: \map f x = \map f {x + T}$

Then:

$\laptrans {\map f t} = \dfrac 1 {1 - e^{-s T} } \displaystyle \int_0^T e^{-s t} \map f t \rd t$

where $\laptrans {\map f t}$ denotes the Laplace transform.

#### Initial Value Theorem of Laplace Transform

Let $\displaystyle \lim_{t \mathop \to 0} \dfrac {\map f t} {\map g t} = 1$.

Then:

$\displaystyle \lim_{s \mathop \to \infty} \dfrac {\map F s} {\map G s} = 1$

if those limits exist.

#### Final Value Theorem of Laplace Transform

Let $\displaystyle \lim_{t \mathop \to \infty} \dfrac {\map f t} {\map g t} = 1$.

Then:

$\displaystyle \lim_{s \mathop \to 0} \dfrac {\map F s} {\map G s} = 1$

if those limits exist.