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,

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

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

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

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

Problem 03 | Laplace Transform of Derivatives

Problem 03
Find the Laplace transform of   $f(t) = e^{5t}$   using the transform of derivatives.
 

Solution 03
$f(t) = e^{5t}$       ..........       $f(0) = 1$

$f'(t) = 5e^{5t}$
 

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

$\mathcal{L} (5e^{5t}) = s \, \mathcal{L} (e^{5t}) - 1$

$1 = s \, \mathcal{L} (e^{5t}) - \mathcal{L} (5e^{5t})$

$s \, \mathcal{L} (e^{5t}) - 5 \, \mathcal{L} (e^{5t}) = 1$

Problem 02 | Laplace Transform of Derivatives

Problem 02
Find the Laplace transform of   $f(t) = \sin^2 t$   using the transform of derivatives.
 

Solution 02
$f(t) = \sin^2 t$       ..........       $f(0) = 0$

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

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

$\mathcal{L} (\sin 2t) = s \, \mathcal{L} (\sin^2 t) - 0$

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

$s \, \mathcal{L} (\sin^2 t) = \dfrac{2}{s^2 + 4}$

Problem 01 | Laplace Transform of Derivatives

Problem 01
Find the Laplace transform of   $f(t) = t^3$   using the transform of derivatives.
 

Solution 01
$f(t) = t^3$       ..........       $f(0) = 0$

$f'(t) = 3t^2$       ..........       $f'(0) = 0$

$f''(t) = 6t$       ..........       $f''(0) = 0$

$f'''(t) = 6$
 

$\mathcal{L} \left\{ f'''(t) \right\} = s^3 \mathcal{L} \left\{ f(t) \right\} - s^2 f(0) - s \, f'(0) - f''(0)$

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:

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

 

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