Problem 02 | Evaluation of Integrals

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