Laplace transform

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 04
Find the Laplace transform of   $f(t) = t \, \sin t$   using the transform of derivatives.
 

Solution 04
$f(t) = t \, \sin t$       ..........       $f(0) = 0$

$f'(t) = t \, \cos t + \sin t$       ..........       $f'(0) = 0$

$f''(t) = (-t \, \sin t + \cos t) + \cos t = -t \, \sin t + 2\cos t$
 

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

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
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
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,