Laplace of heaviside function with negative arguments

While using a laplace transform table I am given an original function of : - heaviside(-t) and its transform is 1/s

using the command laplace( -heaviside(-t)) it gives me the value 0, which does not match my table. I suppose the table could be wrong but it seems more likely I'm omitting something. When using negative arguments with the heaviside function do I need to directly specify the inputs to be greater than or less than specific values? I saw a thread where someone was told to type in the command assume with a value restricting the inputs to negative values, I just wasn't aware that only positive values were assumed for heaviside inputs. Can somebody clear this up for me?

Answer 1

[David Goodmanson]

Hi Darius,

since the laplace transform is

Integral{0,inf} f(t)*exp(-s*t) dt, 

the value of f(t) for t<0 makes no difference. [ It sometimes makes sense to define f(t) for t < 0 because then you can find the laplace transform of g(t) = f(t-a) for example ]. In this case since heaviside(-t) = 0 for t>0, you get 0 for a result.

It's possible that your table is using

heaviside(t) =  1,    t>0
             = -1,    t<0
             =  0,    t=0

in which case the transform you refer to is 1/s.