From
Problem 01 | Division by t:
$\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right)$
By first shifting property:
$\mathcal{L} \left( \dfrac{e^{-t} \sin t}{t} \right) = \arctan \left( \dfrac{1}{s + 1} \right)$
Thus,
$\displaystyle \int_0^\infty \dfrac{e^{-t}\sin t}{t} ~ dt = \lim_{s \to 0} \left[ \arctan \left( \dfrac{1}{s + 1} \right) \right]$
$\displaystyle \int_0^\infty \dfrac{e^{-t}\sin t}{t} ~ dt = \arctan \left( \dfrac{1}{0 + 1} \right)$
$\displaystyle \int_0^\infty \dfrac{e^{-t}\sin t}{t} ~ dt = \arctan (1)$
$\displaystyle \int_0^\infty \dfrac{e^{-t}\sin t}{t} ~ dt = \dfrac{\pi}{4}$ answer
