$$\lim_{s\rightarrow 0} \ \int_0^{+\infty} \frac{e^{-st} }{1+t^2} \ dt$$
$$\mathscr{L}\{\arctan(t)\}=\frac{F(s)}{s}$$
Final value theorem:
$$\lim_{s\rightarrow 0} \ s F(s)=\lim_{t\rightarrow+\infty}f(t)$$
$$\lim_{t\rightarrow+\infty} \ \arctan(t)=\frac{\pi}{2}$$
$$\lim_{s\rightarrow 0} \ \ s \ \frac{1}{s}F(s)=\frac{\pi}{2}$$
Is it correct?
Thanks!
Your approach is correct, but why not use Lebesgue's dominated convergence theorem, for a shorter and simpler proof? Begin by noticing that $\dfrac {\Bbb e ^{-st}} {1+t^2} \to \dfrac 1 {1+t^2}$ and that, at the same time, $\left| \dfrac {\Bbb e ^{-st}} {1+t^2} \right| \le \dfrac 1 {1+t^2}$ for all $s \ge 0$, which is integrable. Then, Lebesgue's theorem guarantees that
$$\int \limits _0 ^\infty \frac {\Bbb e ^{-st}} {1+t^2} \Bbb d t \to \int \limits _0 ^\infty \frac 1 {1+t^2} \Bbb d t = \frac \pi 2 .$$