Alternative integration limits in a Laplace transform

Question

The unilateral Laplace transform of $f(t)$ is $\int_0^\infty e^{st} f(t) \mathrm{d}t$.

If we define the transform as $\int_{a}^\infty e^{st} f(t) \mathrm{d}t$, would it conserve all the nice properties of the true Laplace transform (e.g., the convolution theorem)?

How would its inverse be?

Answer 1

This is what you get when you have a unit step function. (Heaviside function).

If $\int_0^{\infty} f(t) e^{-st} dt = F(s)$ then $\int_a^{\infty} f(t) e^{-st} dt = e^{-as} F(s)$

Answer 2

If you mean its inverse in time domain, then it is $$u(t - a) f(t)$$ where $$u = \begin{cases} 1, & t \geq a \\ 0, & \text{ otherwise} \end{cases} $$

And yes, the convolution theorem still applies.