# Alternative integration limits in a Laplace transform

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?

• forget the uni-lateral Laplace transform $F(s) = \int_0^\infty f(t) e^{-st} dt$ and learn only the bilateral Laplace tranform $$\mathcal{L}[h(t)](s) = \int_{-\infty}^\infty h(t) e^{-st} dt$$ hence $$\int_0^\infty f(t) e^{-st} dt = \mathcal{L}[f(t)1_{t >0}](s)$$ while $$\int_a^\infty f(t) e^{-st} dt = \mathcal{L}[f(t)1_{t >a}](s)$$ – reuns May 23 '16 at 18:10
• (and your teacher is bad) – reuns May 23 '16 at 18:10
• I'm self-taught :,-( ! – altroware May 29 '16 at 7:53

If $\int_0^{\infty} f(t) e^{-st} dt = F(s)$ then $\int_a^{\infty} f(t) e^{-st} dt = e^{-as} F(s)$
If you mean its inverse in time domain, then it is $$u(t - a) f(t)$$ where $$u = \begin{cases} 1, & t \geq a \\ 0, & \text{ otherwise} \end{cases}$$