To evaluate the integral, consider $$\begin{array}{rl}\underset{s\to 0}{lim}{\int }_0^{\mathrm{\infty }}{e}^{-st}\frac{\sin t}{t}dt& =\underset{s\to 0}{lim}L\left(\frac{\sin t}{t}\right)\\ & =\underset{s\to 0}{lim}{\int }_s^{\mathrm{\infty }}L(\sin t)ds=\underset{s\to 0}{lim}{\int }_s^{\mathrm{\infty }}\frac{1}{s^2+1}ds\\ & =\underset{s\to 0}{lim}({\tan }^{-1}\mathrm{\infty }-{\tan }^{-1}s)=\underset{s\to 0}{lim}\frac{\pi }{2}-{\tan }^{-1}s\\ & =\frac{\pi }{2}\end{array}$$

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