Inverse Laplace transform where e^(st)F(s) is entire

1. Sep 13, 2014

qftqed

Heya folks,
I'm currently pondering how to decide whether a function has an inverse Laplace transform or not. In particular I am considering the function e^(-is), which I am pretty sure does not have an inverse Laplace transform. My reasoning is that when calculating the inverse by the Bromwich Integral, integration is done along the contour Re(s) = c, where c is greater than the real part of all the singularities of e^(s(t-i)). Since this has no singularities, the integral cannot be evaluated. If this is right, then any function F(s) such that e^(st)F(s) is entire will not have an inverse transform and I was wondering if this is in fact the correct conclusion or if I'm missing something important

Thanks!

EDIT: I am finding plenty of counter examples to this, e.g. F(s) = 1 with f(t) = δ(t) the Dirac delta function, but I am still not sure how this fits into the Bromwich integral picture

Last edited: Sep 13, 2014