September 2011

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
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
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,