Computing the Restricted Laplace Transform of a Random Variable

Question

Is there any way to calculate the restricted Laplace transform of the random variable $X$, i.e., $$ \int_{0}^{u}e^{-sx}dF(x)\ $$ $(u<\infty)$, based on its Laplace transform?

Answer 1

Let's suppose for simplicity $X$ has a continuous density $f(x)$ with $|f(x)|<Ae^{ax}$ for some real constants $A$, $a$, so that the Laplace transform $Y(s) = \int_0^\infty e^{-sx}\ f(x)\ dx$ is analytic for $\text{Re} s > a$, and $f(x) = \frac{1}{2\pi i} \int_L e^{zx} Y(z)\ dz$ where $L$ is the line $z = c + it,\ -\infty < t < \infty$ for any $c > a$. Then your restricted Laplace transform is

$$R(s) = \int_0^u e^{-xs} f(x)\ dx = \frac{1}{2\pi i} \int_L \int_0^u e^{(z-s)x} Y(z)\ dx dz = \frac{1}{2\pi i} \int_L \frac{e^{(z-s)u}-1}{z-s}\ Y(z)\ dz$$