Initial value theorem

Suppose first that ${\displaystyle f}$ is bounded, i.e. ${\displaystyle \lim _{t\to 0^{+}}f(t)=\alpha }$. A change of variable in the integral ${\displaystyle \int _{0}^{\infty }f(t)e^{-st}\,dt}$ shows that

${\displaystyle sF(s)=\int _{0}^{\infty }f\left({\frac {t}{s}}\right)e^{-t}\,dt}$.

Since ${\displaystyle f}$ is bounded, the Dominated Convergence Theorem implies that

${\displaystyle \lim _{s\to \infty }sF(s)=\int _{0}^{\infty }\alpha e^{-t}\,dt=\alpha .}$

Of course we don't really need DCT here, one can give a very simple proof using only elementary calculus:

Start by choosing ${\displaystyle A}$ so that ${\displaystyle \int _{A}^{\infty }e^{-t}\,dt<\epsilon }$, and then note that ${\displaystyle \lim _{s\to \infty }f\left({\frac {t}{s}}\right)=\alpha }$ uniformly for ${\displaystyle t\in (0,A]}$.

The theorem assuming just that ${\displaystyle f(t)=O(e^{ct})}$ follows from the theorem for bounded ${\displaystyle f}$:

Define ${\displaystyle g(t)=e^{-ct}f(t)}$. Then ${\displaystyle g}$ is bounded, so we've shown that ${\displaystyle g(0^{+})=\lim _{s\to \infty }sG(s)}$. But ${\displaystyle f(0^{+})=g(0^{+})}$ and ${\displaystyle G(s)=F(s+c)}$, so

${\displaystyle \lim _{s\to \infty }sF(s)=\lim _{s\to \infty }(s-c)F(s)=\lim _{s\to \infty }sF(s+c)=\lim _{s\to \infty }sG(s),}$

since ${\displaystyle \lim _{s\to \infty }F(s)=0}$.