Problem 03 | Evaluation of Integrals

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

By "first shifting property" of Laplace transformation:
$\mathcal{L} (e^{-3t} \sin t) = \dfrac{1}{(s + 3)^2 + 1}$

$\mathcal{L} (e^{-3t} \sin t) = \dfrac{1}{(s^2 + 6s + 9) + 1}$

$\mathcal{L} (e^{-3t} \sin t) = \dfrac{1}{s^2 + 6s + 10}$
 

From "multiplication by power of t":
$\mathcal{L} (te^{-3t} \sin t) = (-1)^1\dfrac{d}{ds} \left(\dfrac{1}{s^2 + 6s + 10} \right)$

$\mathcal{L} (te^{-3t} \sin t) = - ~ \dfrac{-1(2s + 6)}{(s^2 + 6s + 10)^2}$

$\mathcal{L} (te^{-3t} \sin t) = \dfrac{2s + 6}{(s^2 + 6s + 10)^2}$
 

Therefore,
$\displaystyle \int_0^\infty te^{-3t} \sin t ~ dt = \lim_{s \to 0} \left[ \dfrac{2s + 6}{(s^2 + 6s + 10)^2} \right]$

$\displaystyle \int_0^\infty te^{-3t} \sin t ~ dt = \dfrac{2(0) + 6}{[ \, 0^2 + 6(0) + 10 \, ]^2}$

$\displaystyle \int_0^\infty te^{-3t} \sin t ~ dt = \dfrac{6}{10^2}$

$\displaystyle \int_0^\infty te^{-3t} \sin t ~ dt = \dfrac{6}{100}$

$\displaystyle \int_0^\infty te^{-3t} \sin t ~ dt = \dfrac{3}{50}$           answer