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I am having problems understanding why a Laplace transform exists or not. Here is my math and logic, hopefully someone can point out where I am wrong.

$$f(t)=e^{at} \implies ℒ[e^{at}] = F(s)=\int_0^{\infty}\exp([a-s]t)dt$$

From this point I am at a loss both mathematically and notationally, but here I go anyways:

$$\implies \frac{1}{a-s}\lim_{t \to\infty}\exp([a-s]t)$$

According to Schaum's "Laplace transform":

The Laplace transform of f(t) is said to exist if the integral converges for some value of s; otherwise it does not exist.

What does this mean for my case? The way I understand it is that $s>a$, because only then the integral will converge. Is this correct? And if it is, why can we just assume that $s>a$?

I am really frustrated, I have tried to look it up on the internet and here, but I just do not understand it.

Edited from $s>t$ to $s>a$.

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2 Answers 2

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Taking $s$ and $a$ to be real, the integral defining $F(s)$ in your case converges provided $s>a$. This specifies the domain of $F$, which is the interval $(a,\infty)$. When $s$ is real, the transform is only intended to be defined for $s$ large enough to make the integral converge.

To understand this it may help to note that the Laplace transform produces another function, which may in fact have another domain.

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  • $\begingroup$ $s$ isn't real it is complex. $\endgroup$
    – dustin
    Jan 11, 2015 at 19:55
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    $\begingroup$ @Akitirija The LT exists as a function if the integral converges for some $s$. The LT, being a function, has a domain on which the integral converges. In your case this domain is $(a,\infty)$. It is not really that we are assuming $s>a$, it is that we are never putting $s \leq a$ into $F$, because it is not in the domain of $F$. $\endgroup$
    – Ian
    Jan 11, 2015 at 20:14
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    $\begingroup$ @Akitirija That's pretty much correct. (Technically you should say something like "because this is the interval for which $\int_0^\infty \exp([a-s]t) dt$ converges", but I know what you meant.) And again, as I commented before, here we are dealing only with the case where $s$ is real. $\endgroup$
    – Ian
    Jan 11, 2015 at 20:24
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    $\begingroup$ @Akitirija you should look at this because $s$ isn't real. You will see $s\in\mathbb{C}$, $s$ is the complex frequency variable, and mentioning of convergence is about the real part of $s$. It starts on page 4. $\endgroup$
    – dustin
    Jan 11, 2015 at 20:27
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    $\begingroup$ @Akitirija yes we need $s$ is greater than $a$ but math is about being precise so what you should want to say is $\text{Re}\{s\} > a$ $\endgroup$
    – dustin
    Jan 11, 2015 at 20:40
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$s$ is complex of the form $s=\sigma +j\omega$ where $\sigma,\omega\in\mathbb{R}$ and $j=\sqrt{-1}$, so for convergence, the real part of $s$ has to be greater than zero when we have $e^{-st}$. Now, let's take your example, we have that an exponential will decay or blow up. The argument of the exponential is $a-s < 0$ for $t\in(0,\infty)$ so $a<s$ or the integral will converge when $\text{Re}\{s\} > a$.

On the Wikipedia page, you will see the formal definition and that $s\in\mathbb{C}$.

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  • $\begingroup$ It is not true that the real part of $s$ always has to be greater than zero. For example the Laplace transform of $e^{-t}$ converges for $Re(s)>-1$. Indeed this already appears later in your answer. $\endgroup$
    – Ian
    Jan 11, 2015 at 19:57
  • $\begingroup$ @Ian I just took the general form $e^{-st}$ and the example applied $e^{at}e^{-st}$. With that information, the OP should be able to set up the inequality needed for whatever case they encounter. $\endgroup$
    – dustin
    Jan 11, 2015 at 19:58
  • $\begingroup$ Allow me to be a bit more direct. "for convergence, the real part of $s$ has to be greater than zero" is in your first sentence. It is outright false. Please change it. $\endgroup$
    – Ian
    Jan 11, 2015 at 20:00
  • $\begingroup$ @Ian read the rest of sentence since we have $e^{-st}$. Then I continue saying let's look at your problem. $\endgroup$
    – dustin
    Jan 11, 2015 at 20:00
  • $\begingroup$ Where do we have $e^{-st}$, exactly? The way I'm reading your first sentence suggests that the Laplace transform of any function will never converge unless $Re(s)>0$. This is false. Please at least clarify what you mean. $\endgroup$
    – Ian
    Jan 11, 2015 at 20:01

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