How does the answer of the Laplace transform $$\mathcal L \left\{ \frac{\sin t}{t} \right\}= \frac{\pi}{2}-\tan^{-1}(s)$$ solve the definite integral
$$\int_0^{\infty} \frac{\sin t}{t} dt = \frac{\pi}{2} $$
How are they related? why does this solve the definite integral?
Thank you.
Best Answer
Your statement is
$$\int_0^{\infty} dt \frac{\sin{t}}{t} e^{-s t} = \frac{\pi}{2} - \arctan{s} $$
Plug in $s=0$ to both sides.
There are lots of ways to prove the LT. One way to do it is to use the FT relation for the sinc term.