Laplace transform

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 03 | Evaluation of Integrals

Problem 03
Find the value of   $\displaystyle \int_0^\infty te^{-3t} \sin t ~ dt$
 

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

Laplace Transform of Intergrals

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

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

 

Problem 04 | Laplace Transform of Derivatives

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

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

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

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

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

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:

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

 

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,

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