The Inverse Laplace Transform

Definition
From   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   the value   $f(t)$   is called the inverse Laplace transform of   $F(s)$. In symbol,
 

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

where   $\mathcal{L}^{-1}$   is called the inverse Laplace transform operator.
 

To find the inverse transform, express   $F(s)$   into partial fractions which will, then, be recognizable as one of the following standard forms.
 

Table of Inverse Laplace Transform
1. $\mathcal{L}^{-1}\left\{ \dfrac{1}{s} \right\} = 1$

2. $\mathcal{L}^{-1}\left\{ \dfrac{1}{s - a} \right\} = e^{at}$

3. $\mathcal{L}^{-1}\left\{ \dfrac{1}{s^n} \right\} = \dfrac{t^{n - 1}}{(n - 1)!}; ~ ~ n = 1, ~ 2, ~ 3, ...$

4. $\mathcal{L}^{-1}\left\{ \dfrac{1}{(s - a)^n} \right\} = \dfrac{e^{at}t^{n - 1}}{(n - 1)!}$

5. $\mathcal{L}^{-1}\left\{ \dfrac{1}{s^2 + b^2} \right\} = \dfrac{1}{b}\sin bt$

6. $\mathcal{L}^{-1}\left\{ \dfrac{s}{s^2 + b^2} \right\} = \cos bt$

7. $\mathcal{L}^{-1}\left\{ \dfrac{1}{s^2 - a^2} \right\} = \dfrac{1}{a}\sinh at$

8. $\mathcal{L}^{-1}\left\{ \dfrac{s}{s^2 - a^2} \right\} = \cosh at$

9. $\mathcal{L}^{-1}\left\{ \dfrac{1}{(s - a)^2 + b^2} \right\} = \dfrac{1}{b}e^{at}\sin bt$

10. $\mathcal{L}^{-1}\left\{ \dfrac{s - a}{(s - a)^2 + b^2} \right\} = e^{at}\cos bt$

11. $\mathcal{L}^{-1}\left\{ \dfrac{s}{(s^2 + b^2)^2} \right\} = \dfrac{1}{2b}t \sin bt$

12. $\mathcal{L}^{-1}\left\{ \dfrac{1}{(s^2 + b^2)^2} \right\} = \dfrac{1}{2b^3}(\sin bt - bt\cos bt)$
 

Theorems on Inverse Laplace Transformation
Theorem 1
If   $a$   and   $b$   are constants,
$\mathcal{L}^{-1}\left\{ a~f(s) + b~g(s) \right\} = a\mathcal{L}^{-1}\left\{ f(s) \right\} + b\mathcal{L}^{-1}\left\{ g(s) \right\}$
 

Theorem 2
$\mathcal{L}^{-1}\left\{ f(s) \right\} = e^{-at} \mathcal{L}^{-1}\left\{ f(s - a) \right\}$
 

Tags
inverse transform
inverse Laplace