Newman Tauberian theorem (${\int }_0^{\infty }f(t)dt=F(0)$)

For bounded locally integrable $f$ on the nonnegative real line, suppose its Laplace transform $F$ extends holomorphically to an open neighborhood of the closed right half-plane. Then the ordinary improper integral converges to $F(0)$. Contour damping for bounded Laplace transforms bounds the difference from the finite-interval transform by $2\Vert f{\Vert }_{\infty }/R$ in the limit for every radius $R$. Extension merely to the open right half-plane is insufficient, as the constant function and transform $1/z$ show.

The displayed kernel has residue one at zero and vanishes at the endpoints of a radius-$R$ imaginary-axis segment. On either radius-$R$ semicircle, $|1+z^2/R^2|=2|\mathrm{Re}z|/R$, cancelling the inverse real-part bound on a bounded function's Laplace transform tail. Use the full transform only on a right semicircle and a thin analytic leftward path; deform the finite-interval entire transform to the large left semicircle. This proves the Newman Tauberian theorem without assuming analytic continuation across an entire left half-disk.