division by t

Problem 04 | Evaluation of Integrals

Problem 04
Evaluate   $\displaystyle \int_0^\infty \dfrac{e^{-t}\sin t}{t} ~ dt$.
 

Solution 04
From Problem 01 | Division by t:
$\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right)$
 

By first shifting property:

Problem 02 | Evaluation of Integrals

Problem 02
Find the value of   $\displaystyle \int_0^\infty \dfrac{\sin t ~dt}{t}$.
 

Problem 01 | Evaluation of Integrals

Problem 01
Evaluate   $\displaystyle \int_0^\infty \dfrac{e^{-3t} - e^{-6t}}{t} ~ dt$
 

Problem 04 | Division by t

Problem 04
Find the Laplace transform of   $f(t) = \dfrac{\cos 4t - \cos 5t}{t}$.
 

Solution 04
$f(t) = \dfrac{\cos 4t - \cos 5t}{t}$

$\mathcal{L} \left\{ f(t) \right\} = \mathcal{L} \left[ \dfrac{\cos 4t - \cos 5t}{t} \right]$

$\mathcal{L} \left\{ f(t) \right\} = \mathcal{L} \left[ \dfrac{\cos 4t}{t} \right] - \mathcal{L} \left[ \dfrac{\cos 5t}{t} \right]$
 

Since
$\mathcal{L} (\cos bt) = \dfrac{s}{s^2 + b^2}$
 

Then,

Problem 03 | Division by t

Problem 03
Find the Laplace transform of   $f(t) = \dfrac{\sin^2 t}{t}$.
 

Solution 03
$f(t) = \dfrac{\sin^2 t}{t}$

$f(t) = \dfrac{\frac{1}{2}(1 - \cos 2t)}{t}$

$f(t) = \dfrac{1}{2} \left[ \dfrac{1}{t} - \dfrac{\cos 2t}{t} \right]$

$\mathcal{L} \left\{ f(t) \right\} = \dfrac{1}{2} \mathcal{L} \left[ \dfrac{1}{t} - \dfrac{\cos 2t}{t} \right]$

$\mathcal{L} \left\{ f(t) \right\} = \dfrac{1}{2} \mathcal{L} \left( \dfrac{1}{t} \right) - \dfrac{1}{2} \left( \dfrac{\cos 2t}{t} \right)$
 

Since

Problem 02 | Division by t

Problem 02
Find the Laplace transform of   $f(t) = \dfrac{e^{4t} - e^{-3t}}{t}$.
 

Solution 03
$f(t) = \dfrac{e^{4t} - e^{-3t}}{t}$

$f(t) = \dfrac{e^{4t}}{t} - \dfrac{e^{-3t}}{t}$

$\mathcal{L} \left\{ f(t) \right\} = \mathcal{L} \left\{ \dfrac{e^{4t}}{t} \right\} - \mathcal{L} \left\{ \dfrac{e^{-3t}}{t} \right\}$
 

Since
$\mathcal{L} (e^{4t}) = \dfrac{1}{s - 4}$   and

$\mathcal{L} (e^{-3t}) = \dfrac{1}{s + 3}$
 

Then,

Problem 01 | Division by t

Problem 01
Find the Laplace transform of   $f(t) = \dfrac{\sin t}{t}$.
 

Solution 01
$\mathcal{L} (\sin t) = \dfrac{1}{s^2 + 1}$
 

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \int_s^\infty \dfrac{du}{u^2 + 1}$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \big[ \arctan u \big]_s^\infty$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \lim_{a \to \infty}\big[ \arctan u \big]_s^a$

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.
 

 
 
Subscribe to RSS - division by t