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.
 

Evaluation of Integrals

If   $F(s) = \mathcal{L}\left\{ f(t) \right\}$,   then   $\displaystyle \int_0^\infty e^{-st} f(t) \, dt = F(s)$.
 

Taking the limit as   $s \to 0$,   then   $\displaystyle \int_0^\infty f(t) \, dt = F(0)$   assuming the integral to be convergent.
 

Problem 02 | Laplace Transform of Intergrals

Problem 02
Find the Laplace transform of   $\displaystyle \int_0^t (u^2 - u + e^{u}) \, du$.
 

Solution 02
$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$
 

Hence,
$\displaystyle \mathcal{L} \left[ \int_0^t (u^2 - u + e^{u}) \, du \right] = \dfrac{\mathcal{L}(t^2 - t + e^t)}{s}$

$\displaystyle \mathcal{L} \left[ \int_0^t (u^2 - u + e^{u}) \, du \right] = \dfrac{1}{s} \mathcal{L}(t^2 - t + e^t)$

Problem 01 | Laplace Transform of Intergrals

Problem 01
Find the Laplace transform of   $\displaystyle \int_0^t \dfrac{\sin u}{u} \, du$   if $\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right)$.
 

Solution 01
$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$
 

Since,
$\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right)$

$F(s) = \arctan \left( \dfrac{1}{s} \right)$
 

Then,

Pages