Laplace transform of a finite duration signal Consider the following signal:
$$ x(t) = e^{-2t}[u(t) - u(t-5)] $$

This signal exists only from 0 to 5 time units. Elsewhere, it is zero.

Now, let's find the laplace transform of this signal using Linearity and Time shift properties.

$$ e^{-2t}u(t) \leftrightarrow \frac{1}{s+2} \ , \ \ Re \{s \} &gt; -2 $$
Also,
$$ e^{-2(t-5)}u(t-5) \leftrightarrow \frac{e^{-5s}}{s+2} \ , \ \ Re \{s \} &gt; -2 $$
$$ \Rightarrow e^{-2t}u(t-5) \leftrightarrow \frac{e^{-5s}e^{-10}}{s+2} \ , \ \ Re \{s \} &gt; -2 $$

Thus, by linearity property, 
$$ e^{-2t}[u(t) - u(t-5)] = \frac{1 - e^{-5(s+2)}}{s+2} \ , \ \ Re \{s \} &gt; -2$$

Note: The time shifting property doesn't alter the ROC;

However, the textbooks that i am refering (Oppenheim and Schaum series) both tell that the ROC of a finite duration signal is the entire S-plane, possibly zero and infinity (in some cases).

But the above signal being of finite-duration, possess ROC that is not the entire s-plane. Please help me figure this conceptual error. 

Note: The above problem is from Schaum series. Here are the images of the textbook's section relevant to the above question.

Source of the Question and its solution:


Property of finite duration signals: 

In Schaum's outline series:

In oppenheim:
The property claimed by Schaum and Oppenheim is also true for the given example. Note that the Laplace transform

$$X(s)=\frac{1-e^{-5(s+2)}}{s+2}\tag{1}$$

has no pole at $s=-2$:

$$\lim_{s\to -2}X(s)=\frac{1-(1-5(s+2))}{s+2}\Big{|}_{s=-2}=5$$

So the ROC is indeed the entire $s$-plane. Even though the ROCs of the two individual signals in your solution have the same right half-plane as ROC, their sum has the entire $s$-plane as ROC because the two signals cancel everywhere except in a finite interval.