Initial value theorem for Laplace transforms ($f(0+)={\lim }_{p\to +\infty }p\mathcal{L}f(p)$)

For a function continuous at zero and of exponential order, the displayed limit recovers its initial value from its Laplace transform. Substitute $u=pt$ to obtain $p\mathcal{L}f(p)={\int }_0^{\infty }{e}^{-u}f(u/p)du$ and apply the dominated convergence theorem; an exponential-order bound controls the tail. The limit along positive real $p$ suffices for this version.