The usual procedure is as follows:
$$\mathcal L \left\{ \frac {\sin t} {t}\right\}(s)=\int\limits_0^\infty e^{-st}\frac {\sin t} {t}dt $$
We have that for any $f(t)$ such that the transform exists
$$\mathcal L \left\{ \frac {f(t)} {t}\right\}(s)=\int\limits_0^\infty f(t)\frac {e^{-st}} {t}dt $$
But
$$\frac {e^{-st}} {t}=\int\limits_s^\infty e^{-mt}dm$$
Under appropriate conditions we can exchange the order of the integrands and put
$$\mathcal L \left\{ \frac {f(t)} {t}\right\}(s)=\int\limits_s^\infty \int\limits_0^\infty f(t) e^{-mt}dm dt $$
This means
$$\mathcal L \left\{ \frac {f(t)} {t}\right\}(s)=\int\limits_s^\infty F(t) dt $$
where $F$ is the transform of $f$. Using this with $\sin t$ gives
$$\mathcal L \left\{ \frac {\sin t} {t}\right\}(s)=\int\limits_s^\infty \frac{1}{1+t^2} dt = \frac{\pi}{2}-\tan^{-1}s $$
$$\int\limits_0^\infty e^{-st} \frac{\sin t}{t} dt=\int\limits_s^\infty \frac{1}{1+t^2} dt = \frac{\pi}{2}-\tan^{-1}s $$
Taking $s \to 0$
$$\lim\limits_{s \to 0} \int\limits_0^\infty e^{-st} \frac{\sin t}{t}dt=\frac{\pi}{2} $$
For the last step, you need to prove that
$$\lim\limits_{s \to 0} \int\limits_0^\infty e^{-st} \frac{\sin t}{t}dt=\int\limits_0^\infty \frac{\sin t}{t}dt $$
I know you can use the dominated convergence theorem (which is not in my personal stash), and maybe some other theorems, but I'm unable to prove it, though I know it is legitimate (the exponential usually makes things work.)