$\mathcal{L} (\sin t) = \dfrac{1}{s^2 + 1}$
Therefore,
$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \int_0^\infty \dfrac{du}{u^2 + 1}$
$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \Big[ \arctan u \Big]_0^\infty$
$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \infty - \arctan 0$
$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \frac{1}{2}\pi - 0$
$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \frac{1}{2}\pi$
Therefore,
$\displaystyle \int_0^\infty \dfrac{\sin t ~dt}{t} = \frac{\pi}{2}$ answer
