Optimal Tauberian constant in Ingham's theorem for Laplace transforms

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.