  It is well known that there is an absolute constant $\mathfrak{C}>0$ such
that if the Laplace transform $G(s)=\int_{0}^{\infty}\rho(x)e^{-s
x}\:\mathrm{d}x$ of a bounded function $\rho$ has analytic continuation through
every point of the segment $(-i\lambda ,i\lambda )$ of the imaginary axis, then
$$ \limsup_{x\to\infty} \left|\int_{0}^{x}\rho(u)\:\mathrm{d}u -
G(0)\right|\leq \frac{ \mathfrak{C}}{\lambda} \: \limsup_{x\to\infty}
|\rho(x)|. $$ The best known value of the constant $\mathfrak{C}$ was so far
$\mathfrak{C}=2$. In this article we show that the inequality holds with
$\mathfrak{C}=\pi/2$ and that this value is best possible. We also sharpen
Tauberian constants in finite forms of other related complex Tauberian theorems
for Laplace transforms.
