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I don't understand why the laplace transform of some function, say f(t), has to be "piecewise continuous" and not "continuous". Is "piecewise continuous" sort of like the minimum requirement? This troubles me because I don't think f(t)=t is piecewise continuous, it's simply continuous...

Also, I don't understand what's so special about a function being of "exponential order". Can someone explain to me why this property is so special and apparently makes the Laplace Transform exist?

Also, my teacher says that f(t) has to be of exponential order from [0,infinity] and other sources say that it at least has to be of exponential order from [T,infinity] where T>0.

Thanks.

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  • $\begingroup$ Yeah, I know that's what the theorem states but I'd like to know the actual reason for that. thanks $\endgroup$ – DLV Sep 27 '14 at 17:06
  • $\begingroup$ Maybe i'm mathematically blind but I dont see why it is the case that if a function is of exponential order then the integral will converge. I'd like to know what makes one entail the other. $\endgroup$ – DLV Sep 27 '14 at 17:16
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The Laplace transform is defined as:

$F(s)=\int _0 ^{+\infty} e ^{-st} f(t) dt$

Your first question: As one can see the limit of the integral is from $0$ to $\infty$. So, it is inherently assumed that $f(t)$ is zero for $t<0$. As a result, when we talk about $f(t)=t$, it is actually $f(t)=t, t\geq0$, which is a piecewise continuous function.

Second question: A function $f(t)$ is said of exponential order if there exists a constant $a$ and positive constants $t_0$ and $M$ such that $|f(t)|<Me^{at}$, for all $t>t_0$ at which $f(t)$ is defined. This condition should hold because otherwise the Laplace transform will not exist. To prove it, lets first split the integral into two parts:

$F(s)=\int _0 ^{+\infty} e ^{-st} f(t) dt = \int _0 ^{t_0} e ^{-st} f(t) dt+\int _{t_0} ^{\infty} e ^{-st} f(t) dt$

It is easy to show that the first part exists. Now, we need to show that the existence of the laplace transform depends on the convergence of the second part:

$|\int _{t_0} ^{\infty} e ^{-st} f(t) dt|\leq \int _{t_0} ^{\infty} |e ^{-st} f(t)| dt \leq \int _{t_0} ^{\infty} Me ^{-st} e^{at} dt = M\int _{t_0} ^{\infty} e^{-(s-a)t} dt$

This integral converges if $s>a$. So, we conclude that if $f(x)$ is exponentially ordered, there exists a constant $a$ in which $F(s)$ exists for $s>a$.

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  • $\begingroup$ Nice thanks so much. However, do you have anything to say about the thing I mentioned where some people say it has to be exponentially ordered after some T>0 or if it has to be exponentially ordered for ALL t>0? $\endgroup$ – DLV Sep 27 '14 at 21:01
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    $\begingroup$ I mostly agree with the first people: it has to be exponentially ordered after some T. As a counter example for the second group, take the delta dirac function delte(t-t0), which has a Laplace transform but doesn't satisfy the condition for ALL t>0. $\endgroup$ – Sina Sep 27 '14 at 21:30
  • $\begingroup$ I think will should be replaced by may in otherwise the Laplace transform will not exist. $\endgroup$ – Birendra Singh Mar 27 at 6:31
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Any continuous function can be said to be piecewise continuous. Saying only "piecewise continuous" is just a way to use fewer words.

In fact, we can relax the requirement of piecewise continuity, and just require that $f(t)$ be integrable on any finite subset of $[0,\infty)$. An example of such a function which is not continuous on $[0,\infty)$ is $f(t)=t^{-1/2}$.

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