Laplace Transform of Intergrals

If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then

$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$


Laplace Transform of Derivatives

For first-order derivative:
$\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$

For second-order derivative:
$\mathcal{L} \left\{ f''(t) \right\} = s^2 \mathcal{L} \left\{ f(t) \right\} - s \, f(0) - f'(0)$

For third-order derivative:
$\mathcal{L} \left\{ f'''(t) \right\} = s^3 \mathcal{L} \left\{ f(t) \right\} - s^2 f(0) - s \, f'(0) - f''(0)$

For nth order derivative:

$\mathcal{L} \left\{ f^n(t) \right\} = s^n \mathcal{L} \left\{ f(t) \right\} - s^{n - 1} f(0) - s^{n - 2} \, f'(0) - \dots - f^{n - 1}(0)$


Division by t | Laplace Transform

Division by $t$
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,

$\displaystyle \mathcal{L} \left\{ \dfrac{f(t)}{t} \right\} = \int_s^\infty F(u) \, du$


provided   $\displaystyle \lim_{t \rightarrow 0} \left[ \dfrac{f(t)}{t} \right]$   exists.

Multiplication by Power of t | Laplace Transform

Multiplication by Power of $t$
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,

$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s) = (-1)^n F^{(n)}(s)$

where   $n = 1, \, 2, \, 3, \, ...$

Change of Scale Property | Laplace Transform

Change of Scale Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,

$\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)$


Second Shifting Property | Laplace Transform

Second Shifting Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   and   $g(t)
= \begin{cases} f(t - a) & t \gt a \\ 0 & t \lt a \end{cases}$


$\mathcal{L} \left\{ g(t) \right\} = e^{-as} F(s)$


First Shifting Property | Laplace Transform

First Shifting Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   when   $s > a$   then,

$\mathcal{L} \left\{ e^{at} \, f(t) \right\} = F(s - a)$


In words, the substitution   $s - a$   for   $s$   in the transform corresponds to the multiplication of the original function by   $e^{at}$.

Linearity Property | Laplace Transform

Linearity Property
If   $a$   and   $b$   are constants while   $f(t)$   and   $g(t)$   are functions of   $t$   whose Laplace transform exists, then

$\mathcal{L} \left\{ a \, f(t) + b \, g(t) \right\} = a \, \mathcal{L} \left\{ f(t) \right\} + b \, \mathcal{L} \left\{ g(t) \right\}$


Proof of Linearity Property
$\displaystyle \mathcal{L} \left\{ a \, f(t) + b \, g(t) \right\} = \int_0^\infty e^{-st}\left[ a \, f(t) + b \, g(t) \right] \, dt$

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