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

[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

 

[jvicens]

.

Hello!

You are correct in your reasoning that the function e^(-is) does not have an inverse Laplace transform. This is because, as you mentioned, it has no singularities in the complex plane and therefore the Bromwich integral cannot be evaluated.

However, your conclusion that any function F(s) such that e^(st)F(s) is entire will not have an inverse transform is not entirely correct. While it is true that functions with no singularities do not have an inverse Laplace transform, there are also functions with singularities that do not have an inverse transform.

One important concept to consider is the region of convergence (ROC) of a Laplace transform. This is the set of complex numbers for which the integral defining the transform converges. Functions with singularities outside of the ROC will not have an inverse transform, even if they are not entire. For example, the function 1/sin(s) has singularities at s = kπ, but its ROC is the entire complex plane excluding those singularities. Therefore, it does have an inverse Laplace transform.

In the case of F(s) = 1 with f(t) = δ(t), the Dirac delta function, the inverse Laplace transform does exist, but it is not a function in the traditional sense. Instead, it is a distribution, which is a generalized function used to represent point sources. This concept is beyond the scope of the Bromwich integral, but it is an important tool in solving differential equations using Laplace transforms.

So, in summary, a function with no singularities will not have an inverse Laplace transform, but a function with singularities can have an inverse transform as long as they are within the ROC. Additionally, distributions such as the Dirac delta function can also have inverse Laplace transforms, but they are not traditional functions. I hope this helps clarify your understanding.