1. ## Laplace Transform

I'm looking at my old Differential Equations textbook and struggling with a sticky point regarding Laplace transforms.

He defines the Laplace transform in the standard way, saying that the Laplace transform is valid for all s such that the integral converges.

In looking at one of the exercises, he asks for the Laplace transform of the piecewise function $\displaystyle f(t)=2t+1$ for $\displaystyle 0\le t<1$ and 0 for $\displaystyle t>1$. I obtained the correct answer $\displaystyle \displaystyle\frac{1-3e^{-s}}{s}+\displaystyle\frac{2-2e^{-s}}{s^2}$, but the author says in his solutions manual that this is valid for s>0. Why not for s<0 in this case? Since the integral involved is not improper there should be no issue...am I missing something obvious?

2. Lets suppose to 'discharge' the fact that the function is 'truncated' and consider only the function $\displaystyle f(t)= 1 + 2\ t$. In that case is...

$\displaystyle \displaystyle \mathcal {L}\{f(t)\} = \int_{0}^{\infty} (1 + 2\ t)\ e^{-s\ t}\ dt = \frac{1}{s} + \frac{2}{s^{2}}$ (1)

The result (1) is valid only if the integral converges and that is true for $\displaystyle \Re (s) >0$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by Diamondlance
I'm looking at my old Differential Equations textbook and struggling with a sticky point regarding Laplace transforms.

He defines the Laplace transform in the standard way, saying that the Laplace transform is valid for all s such that the integral converges.

In looking at one of the exercises, he asks for the Laplace transform of the piecewise function $\displaystyle f(t)=2t+1$ for $\displaystyle 0\le t<1$ and 0 for $\displaystyle t>1$. I obtained the correct answer $\displaystyle \displaystyle\frac{1-3e^{-s}}{s}+\displaystyle\frac{2-2e^{-s}}{s^2}$, but the author says in his solutions manual that this is valid for s>0. Why not for s<0 in this case? Since the integral involved is not improper there should be no issue...am I missing something obvious?
Probably they are confused by their method of generating the transform. The integrals defining the transform obviously converge for all s. But if you obtained the transform by applying some of the properties of the LT to the transform of g(t)=2t+1 then you might think the restriction to s>=0 still applies, when it does not.

CB