Answer

To evaluate the integral, consider

\[ \begin{aligned} \lim_{s \to 0} \int_{0}^{\infty} e^{-st}\frac{\sin t}{t} dt &= \lim_{s \to 0} L\roundbr{\frac{\sin t}{t}}\newline &= \lim_{s \to 0} \int_{s}^{\infty}L\roundbr{\sin t} ds = \lim_{s \to 0} \int_{s}^{\infty}\frac{1}{s^{2} + 1} ds\newline &= \lim_{s \to 0} \roundbr{\tan^{-1} \infty - \tan^{-1} s} = \lim_{s \to 0} \frac{\pi}{2} - \tan^{-1} s\newline &= \frac{\pi}{2} \end{aligned} \]

where we have used the formula for integral of Laplace Transform.