Proof using dominated convergence theorem and assuming that function is bounded Suppose first that f is bounded, i.e. \lim_{t\to 0^+}f(t)=\alpha. A change of variable in the integral \int_0^\infty f(t)e^{-st}\,dt shows that :sF(s)=\int_0^\infty f\left(\frac ts\right)e^{-t}\,dt. Since f is bounded, the
Dominated Convergence Theorem implies that :\lim_{s\to\infty}sF(s)=\int_0^\infty\alpha e^{-t}\,dt=\alpha.
Proof using elementary calculus and assuming that function is bounded Of course we don't really need DCT here, one can give a very simple proof using only elementary calculus: Start by choosing A so that \int_A^\infty e^{-t}\,dt, and then note that \lim_{s\to\infty}f\left(\frac ts\right)=\alpha
uniformly for t\in(0,A].
Generalizing to non-bounded functions that have exponential order The theorem assuming just that f(t)=O(e^{ct}) follows from the theorem for bounded f: Define g(t)=e^{-ct}f(t). Then g is bounded, so we've shown that g(0^+)=\lim_{s\to\infty}sG(s). But f(0^+)=g(0^+) and G(s)=F(s+c), so :\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 \lim_{s\to\infty}F(s)=0. ==See also==