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Consider the Laplace transform: $$\mathscr{L}(1) = \int_0^\infty e^{-st}dt = -\left.\frac{1}{s}e^{-st}\right|_0^\infty = \frac1s$$ Math textbooks usually state that this is only valid for the condition $s > 0$ (because of the $\infty$ limit in integration).

Now, the question is: how do we know that this condition is satisfied?

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    $\begingroup$ Actually, if you allow complex numbers, the integral converges for $\text{Re}(s) > 0$. But what do you mean by "how do we know that this condition is satisfied"? The Laplace transform of a function is a function of a new variable $s$. Like any variable, it has whatever value you give it. If you want $s > 0$, make sure you only ever give it values $> 0$. $\endgroup$ Commented Apr 20, 2015 at 17:42
  • $\begingroup$ The confusion I have is that when we solve differential equations we always use such transformation and never worry about the validity of the $s > 0$ assumption. $\endgroup$ Commented Apr 20, 2015 at 17:49
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    $\begingroup$ $$\mathscr{L}(1) (s)=\frac1s$$ $\endgroup$ Commented Apr 20, 2015 at 17:58
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    $\begingroup$ Whenever you use Laplace transforms to solve differential equations, an implicit "For $\text{Re}(s)$ sufficiently large" applies to all the calculations. The point is that to recover a continuous function from its Laplace transform, it suffices to know the Laplace transform for $\text{Re}(s)$ greater than some arbitrary real number. $\endgroup$ Commented Apr 20, 2015 at 18:33

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You don't really ''know'' it is satisfied, you more like demand that it has to be satisfied.
If E.g. you solve a differential equation by using the laplace transform, then the solution you get should only be defined for positive $s$.
This is not always stated explicitly because either the professor assumes you know that is the case or the function you found is actually defined for $s<0$ and seems to be a solution on the the set of points on which it is defined.

Often DE's are solved using techniques that might not (yet) have been proven to give correct answers but then by testing the solution and using uniqueness theorems it might actually turn out that you did find the correct answer. So trespassing the allowed range might give you problems in certain kinds of problems but if you manage to find a general solution that way then in the end it doesn't really matter how you found it, as long as you've got the solution.

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