Otherwise no, sometimes $F(s)$ extends analytically to an entire function with the integral converging absolutely iff $\Re(s)>1$ and converging conditionally iff $\Re(s)>0$.įor example $\phi(t)=1_dt$$ converges iff $\Re(s)>0$. The Laplace transform is usually understood as conditionally convergent, meaning that it converges in the former instead of the latter sense. Communications in Statistics - Simulation and Computation: Vol. It can be any kind of singularity (including essential singularity, branch point and natural boundary). The abscissa of convergence of the laplace-stieltjes transform of a ph-distribution. By default, the independent variable is t, and the transformation variable is s. for h(t) et, (1) is convergent for Re(s) > 1, but h(s) 1 1+sextends analytically to any s6 1). Compute the Laplace transform of exp(-at). In case ais nite, the function h(s) may extend analytically to a domain larger than the region of convergence (e.g. case of the Laplace - Stieltjes transform, the domain of absolute convergence is. The real constant a is known as the abscissa of absolute convergence. If $f\ge 0$ then yes there is a singularity on the real axis at the abscissa of convergence. If a probability distribution of phase type has an irreducible representation (,T), the abscissa of convergence of its Laplace-Stieltjes transform is shown. If the integral ( 1 ) has a nonnegative abscissa of convergence oc. (This follows from the dominated convergence theorem.) The constant a is known as the abscissa of absolute convergence, and depends on the growth behavior of f(.
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